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This chapter is all about cardinality of sets. At first this looks like a
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1 CHAPTER Cardinality of Sets This chapter is all about cardinality of sets At first this looks like a very simple concept To find the cardinality of a set, just count its elements If A = { a, b, c, d }, then A = ; if B = { n Z : n }, then B = In this case A < B What could be simpler than that? Actually, the idea of cardinality becomes quite subtle when the sets are infinite The main point of this chapter is to explain how there are numerous different kinds of infinity, and some infinities are bigger than others Two sets A and B can both have infinite cardinality, yet A < B Sets with Equal Cardinalities We begin with a discussion of what it means for two sets to have the same cardinality Up until this point we ve said A = B if A and B have the same number of elements: Count the elements of A, then count the elements of B If you get the same number, then A = B Although this is a fine strategy if the sets are finite (and not too big!), it doesn t apply to infinite sets because we d never be done counting their elements We need a new approach that applies to both finite and infinite sets Here it is: Definition Two sets A and B have the same cardinality, written A = B, if there exists a bijective function f : A B If no such bijective function exists, then the sets have unequal cardinalities, that is, A B A a b c d e f B 0 The above picture illustrates our definition There is a bijective function f : A B, so A = B The function f matches up A with B Think of f as describing how to overlay A onto B so that they fit together perfectly
2 8 Cardinality of Sets On the other hand, if A and B are as indicated in either of the following figures, then there can be no bijection f : A B (The best we can do is a function that is either injective or surjective, but not both) Therefore the definition says A B in these cases A a b c d f B 0 A a b c d e f B 0 Example The sets A = { n Z : 0 n } and B = { n Z : n 0 } have the same cardinality because there is a bijective function f : A B given by the rule f (n) = n Several comments are in order First, if A = B, there can be lots of bijective functions from A to B We only need to find one of them in order to conclude A = B Second, as bijective functions play such a big role here, we use the word bijection to mean bijective function Thus the function f (n) = n from Example is a bijection Also, an injective function is called an injection and a surjective function is called a surjection We emphasize and reiterate that Definition applies to finite as well as infinite sets If A and B are infinite, then A = B provided there exists a bijection f : A B If no such bijection exists, then A B Example This example shows that N = Z To see why this is true, notice that the following table describes a bijection f : N Z n f (n) Notice that f is described in such a way that it is both injective and surjective Every integer appears exactly once on the infinitely long second row Thus, according to the table, given any b Z there is some natural number n with f (n) = b, so f is surjective It is injective because the way the table is constructed forces f (m) f (n) whenever m n Because of this bijection f : N Z, we must conclude from Definition that N = Z Example may seem slightly unsettling On one hand it makes sense that N = Z because N and Z are both infinite, so their cardinalities are both infinity On the other hand, Z may seem twice as large as
3 Sets with Equal Cardinalities N because Z has all the negative integers as well as the positive ones Definition settles the issue Because the bijection f : N Z matches up N with Z, it follows that N = Z We summarize this with a theorem Theorem There exists a bijection f : N Z Therefore N = Z The fact that N and Z have the same cardinality might prompt us compare the cardinalities of other infinite sets How, for example, do N and R compare? Let s turn our attention to this In fact, N R This was first recognized by Georg Cantor (8 8), who devised an ingenious argument to show that there are no surjective functions f : N R (This in turn implies that there can be no bijections f : N R, so N R by Definition ) We now describe Cantor s argument for why there are no surjections f : N R We will reason informally, rather than writing out an exact proof Take any arbitrary function f : N R Here s why f can t be surjective: Imagine making a table for f, where values of n in N are in the lefthand column and the corresponding values f (n) are on the right The first few entries might look something as follows In this table, the real numbers f (n) are written with all their decimal places trailing off to the right Thus, even though f () happens to be the real number 0, we write it as , etc n f (n)
4 0 Cardinality of Sets There is a diagonal shaded band in the table For each n N, this band covers the n th decimal place of f (n): The st decimal place of f () is the st entry on the diagonal The nd decimal place of f () is the nd entry on the diagonal The rd decimal place of f () is the rd entry on the diagonal The th decimal place of f () is the th entry on the diagonal, etc The diagonal helps us construct a number b R that is unequal to any f (n) Just let the nth decimal place of b differ from the nth entry of the diagonal Then the nth decimal place of b differs from the nth decimal place of f (n) In order to be definite, define b to be the positive number less than whose nth decimal place is 0 if the nth decimal place of f (n) is not 0, and whose nth decimal place is if the nth decimal place of f (n) equals 0 Thus, for the function f illustrated in the above table, we have b = and b has been defined so that, for any n N, its nth decimal place is unequal to the nth decimal place of f (n) Therefore f (n) b for every natural number n, meaning f is not surjective Since this argument applies to any function f : N R (not just the one in the above example) we conclude that there exist no bijections f : N R, so N R by Definition We summarize this as a theorem Theorem There exists no bijection f : N R Therefore N R This is our first indication of how there are different kinds of infinities Both N and R are infinite sets, yet N R We will continue to develop this theme throughout this chapter The next example shows that the intervals (0, ) and (0,) on R have the same cardinality P f (x) 0 x Figure A bijection f : (0, ) (0,)
5 Sets with Equal Cardinalities Example Show that (0, ) = (0,) To accomplish this, we need to show that there is a bijection f : (0, ) (0,) We describe this function geometrically Consider the interval (0, ) as the positive x-axis of R Let the interval (0,) be on the y-axis as illustrated in Figure, so that (0, ) and (0,) are perpendicular to each other The figure also shows a point P = (,) Define f (x) to be the point on (0,) where the line from P to x (0, ) intersects the y-axis By similar triangles, we have x + = f (x) x, and therefore f (x) = x x + If it is not clear from the figure that f : (0, ) (0,) is bijective, then you can verify it using the techniques from Section (Exercise 6, below) It is important to note that equality of cardinalities is an equivalence relation on sets: it is reflexive, symmetric and transitive Let us confirm this Given a set A, the identity function A A is a bijection, so A = A (This is the reflexive property) For the symmetric property, if A = B, then there is a bijection f : A B, and its inverse is a bijection f : B A, so B = A For transitivity, suppose A = B and B = C Then there are bijections f : A B and g : B C The composition g f : A C is a bijection (Theorem ), so A = C The transitive property can be useful If, in trying to show two sets A and C have the same cardinality, we can produce a third set B for which A = B and B = C, then transitivity assures us that indeed A = C The next example uses this idea Example Show that R = (0, ) Because of the bijection g : R (0, ) where g(x) = x, we have R = (0, ) Also, Example shows that (0, ) = (0,) Therefore R = (0,) So far in this chapter we have declared that two sets have the same cardinality if there is a bijection between them They have different cardinalities if there exists no bijection between them Using this idea, we showed that Z = N R = (0, ) = (0, ) So, we have a means of determining when two sets have the same or different cardinalities But we have neatly avoided saying exactly what cardinality is For example, we can say that Z = N, but what exactly is Z, or N? What exactly are these things that are equal? Certainly not numbers, for they are too big
6 Cardinality of Sets And saying they are infinity is not accurate, because we now know that there are different types of infinity So just what kind of mathematical entity is Z? In general, given a set X, exactly what is its cardinality X? This is a lot like asking what a number is A number, say, is an abstraction, not a physical thing Early in life we instinctively grouped together certain sets of things (five apples, five oranges, etc) and conceived of as the thing common to all such sets In a very real sense, the number is an abstraction of the fact that any two of these sets can be matched up via a bijection That is, it can be identified with a certain equivalence class of sets under the "has the same cardinality as" relation (Recall that this is an equivalence relation) This is easy to grasp because our sense of numeric quantity is so innate But in exactly the same way we can say that the cardinality of a set X is what is common to all sets that can be matched to X via a bijection This may be harder to grasp, but it is really no different from the idea of the magnitude of a (finite) number In fact, we could be concrete and define X to be the equivalence class of all sets whose cardinality is the same as that of X This has the advantage of giving an explicit meaning to X But there is no harm in taking the intuitive approach and just interpreting the cardinality X of a set X to be a measure the size of X The point of this section is that we have a means of deciding whether two sets have the same size or different sizes Exercises for Section A Show that the two given sets have equal cardinality by describing a bijection from one to the other Describe your bijection with a formula (not as a table) R and (0, ) R and (, ) R and (0,) The set of even integers and the set of odd integers A = { k : k Z } and B = { k : k Z } 6 N and S = { n : n N} Z and S = {, 8,,,,,,8,6,} 8 Z and S = { x R : sin x = } { 0, } N and N 0 { 0, } N and Z [0,] and (0,) N and Z (Suggestion: use Exercise 8 of Section ) P(N) and P(Z) (Suggestion: use Exercise, above) N N and { (n, m) N N : n m } B Answer the following questions concerning bijections from this section Find a formula for the bijection f in Example (page 8) 6 Verify that the function f in Example is a bijection
7 Countable and Uncountable Sets Countable and Uncountable Sets Let s summarize the main points from the previous section A = B if and only if there exists a bijection A B N = Z because there exists a bijection N Z N R because there exists no bijection N R Thus, even though N, Z and R are all infinite sets, their cardinalities are not all the same The sets N and Z have the same cardinality, but R s cardinality is different from that of both the other sets This means infinite sets can have different sizes We now make some definitions to put words and symbols to this phenomenon In a certain sense you can count the elements of N; you can count its elements off as,,,,, but you d have to continue this process forever to count the whole set Thus we will call N a countably infinite set, and the same term is used for any set whose cardinality equals that of N Definition Suppose A is a set Then A is countably infinite if N = A, that is, if there exists a bijection N A The set A is uncountable if A is infinite and N A, that is, if A is infinite and there exists no bijection N A Thus Z is countably infinite but R is uncountable This section deals mainly with countably infinite sets Uncountable sets are treated later If A is countably infinite, then N = A, so there is a bijection f : N A You can think of f as counting the elements of A The first element of A is f (), followed by f (), then f () and so on It makes sense to think of a countably infinite set as the smallest type of infinite set, because if the counting process stopped, the set would be finite, not infinite; a countably infinite set has the fewest elements that a set can have and still be infinite It is common to reserve the special symbol ℵ 0 to stand for the cardinality of countably infinite sets Definition The cardinality of the natural numbers is denoted as ℵ 0 That is, N = ℵ 0 Thus any countably infinite set has cardinality ℵ 0 (The symbol ℵ is the first letter in the Hebrew alphabet, and is pronounced aleph The symbol ℵ 0 is pronounced aleph naught ) The summary of facts at the beginning of this section shows Z = ℵ 0 and R ℵ 0 Example Let E = { k : k Z } be the set of even integers The function f : Z E defined as f (n) = n is easily seen to be a bijection, so we have Z = E Thus, as N = Z = E, the set E is countably infinite and E = ℵ 0
8 Cardinality of Sets Here is a significant fact: The elements of any countably infinite set A can be written in an infinitely long list a, a, a, a, that begins with some element a A and includes every element of A For example, the set E in the above example can be written in list form as 0,,,,,6, 6,8, 8, The reason that this can be done is as follows Since A is countably infinite, Definition says there is a bijection f : N A This allows us to list out the set A as an infinite list f (), f (), f (), f (), Conversely, if the elements of A can be written in list form as a, a, a,, then the function f : N A defined as f (n) = a n is a bijection, so A is countably infinite We summarize this as follows Theorem A set A is countably infinite if and only if its elements can be arranged in an infinite list a, a, a, a, As an example of how this theorem might be used, let P denote the set of all prime numbers Since we can list its elements as,,,,,,, it follows that the set P is countably infinite As another consequence of Theorem, note that we can interpret the fact that the set R is not countably infinite as meaning that it is impossible to write out all the elements of R in an infinite list (After all, we tried to do that in the table on page, and failed!) This raises a question Is it also impossible to write out all the elements of Q in an infinite list? In other words, is the set Q of rational numbers countably infinite or uncountable? If you start plotting the rational numbers on the number line, they seem to mostly fill up R Sure, some numbers such as, π and e will not be plotted, but the dots representing rational numbers seem to predominate We might thus expect Q to be uncountable However, it is a surprising fact that Q is countable The proof presented below arranges all the rational numbers in an infinitely long list Theorem The set Q of rational numbers is countably infinite Proof To prove this, we just need to show how to write the set Q in list form Begin by arranging all rational numbers in an infinite array This is done by making the following chart The top row has a list of all integers, beginning with 0, then alternating signs as they increase Each column headed by an integer k contains all the fractions (in reduced form) with numerator k For example, the column headed by contains the fractions,,,,, and so on It does not contain,, 6, etc, because those are not reduced, and in fact their reduced forms appear in the column headed by You should examine this table and convince yourself that it contains all rational numbers in Q
9 Countable and Uncountable Sets Next, draw an infinite path in this array, beginning at 0 and snaking back and forth as indicated below Every rational number is on this path
10 6 Cardinality of Sets Beginning at 0 rational numbers: and following the path, we get an infinite list of all 0,,,,,,,,,,,,,,,,,,,, By Theorem, it follows that Q is countably infinite, that is, Q = N It is also true that the Cartesian product of two countably infinite sets is itself countably infinite, as our next theorem states Theorem If A and B are both countably infinite, then so is A B Proof Suppose A and B are both countably infinite By Theorem, we know we can write A and B in list form as A = { a, a, a, a, }, B = { b, b, b, b, } Figure shows how to form an infinite path winding through all of A B Therefore A B can be written in list form, so it is countably infinite B b (a, b ) (a, b ) (a, b ) (a, b ) (a, b ) (a 6, b ) (a, b ) b 6 (a, b 6 ) (a, b 6 ) (a, b 6 ) (a, b 6 ) (a, b 6 ) (a 6, b 6 ) (a, b 6 ) b (a, b ) (a, b ) (a, b ) (a, b ) (a, b ) (a 6, b ) (a, b ) b (a, b ) (a, b ) (a, b ) (a, b ) (a, b ) (a 6, b ) (a, b ) b (a, b ) (a, b ) (a, b ) (a, b ) (a, b ) (a 6, b ) (a, b ) b (a, b ) (a, b ) (a, b ) (a, b ) (a, b ) (a 6, b ) (a, b ) b (a, b ) (a, b ) (a, b ) (a, b ) (a, b ) (a 6, b ) (a, b ) a a a a a a 6 a A Figure A product of two countably infinite sets is countably infinite
11 Countable and Uncountable Sets As an example of a consequence of this theorem, notice that since Q is countably infinite, the set Q Q is also countably infinite Recall that the word corollary means a result that follows easily from some other result We have the following corollary of Theorem Corollary Given n countably infinite sets A, A, A,, A n, with n, the Cartesian product A A A A n is also countably infinite Proof The proof is by induction on n For the basis step, notice that when n = the statement asserts that for countably infinite sets A and A, the product A A is countably infinite, and this is true by Theorem Assume that for k, any product A A A A k of countably infinite sets is countably infinite Consider a product A A A A k+ of k + countably infinite sets It is easily confirmed that the function f : A A A A k A k+ (A A A A k ) A k+ f (x, x,, x k, x k+ ) = ( ) (x, x,, x k ), x k+ is bijective, so A A A A k A k+ = (A A A A k ) A k+ By the induction hypothesis, (A A A A k ) A k+ is a product of two countably infinite sets, so it is countably infinite by Theorem As noted above, A A A A k A k+ has the same cardinality, so it too is countably infinite Theorem 6 If A and B are both countably infinite, then A B is countably infinite Proof Suppose A and B are both countably infinite By Theorem, we know we can write A and B in list form as A = { a, a, a, a, }, B = { b, b, b, b, } We can shuffle A and B into one infinite list for A B as follows A B = { a, b, a, b, a, b, a, b, } (We agree not to list an element twice if it belongs to both A and B) Therefore, by Theorem, it follows that A B is countably infinite
12 8 Cardinality of Sets Exercises for Section Prove that the set A = { ln(n) : n N } R is countably infinite Prove that the set A = { (m, n) N N : m n } is countably infinite Prove that the set A = { (n, n) : n Z } is countably infinite Prove that the set of all irrational numbers is uncountable (Suggestion: Consider proof by contradiction using Theorems and 6) Prove or disprove: There exists a countably infinite subset of the set of irrational numbers 6 Prove or disprove: There exists a bijective function f : Q R Prove or disprove: The set Q 00 is countably infinite 8 Prove or disprove: The set Z Q is countably infinite Prove or disprove: The set { 0, } N is countably infinite 0 Prove or disprove: The set A = { n : n N} countably infinite Describe a partition of N that divides N into eight countably infinite subsets Describe a partition of N that divides N into ℵ 0 countably infinite subsets Prove or disprove: If A = {X N : X is finite}, then A = ℵ 0 Suppose A = { (m, n) N R : n = πm } Is it true that N = A? Theorem implies that N N is countably infinite Construct an alternate proof of this fact by showing that the function ϕ : N N N defined as ϕ(m, n) = n (m ) is bijective Comparing Cardinalities At this point we know that there are at least two different kinds of infinity On one hand, there are countably infinite sets such as N, of cardinality ℵ 0 Then there is the uncountable set R Are there other kinds of infinity beyond these two kinds? The answer is yes, but to see why we first need to introduce some new definitions and theorems Our first task will be to formulate a definition for what we mean by A < B Of course if A and B are finite we know exactly what this means: A < B means that when the elements of A and B are counted, A is found to have fewer elements than B But this process breaks down if A or B is infinite, for then the elements can t be counted The language of functions helps us overcome this difficulty Notice that for finite sets A and B it is intuitively clear that A < B if and only if there exists an injective function f : A B but there are no surjective functions f : A B The following diagram illustrates this:
13 Comparing Cardinalities A a b c d f B 0 We will use this idea to define what is meant by A < B and A B For emphasis, the following definition also restates what is meant by A = B Definition Suppose A and B are sets () A = B means there is a bijection A B () A < B means there is an injection A B, but no surjection A B () A B means A < B or A = B For example, consider N and R The function f : N R defined as f (n) = n is clearly injective, but it is not surjective because given the element R, we have f (n) for every n N In fact, Theorem of Section asserts that there is no surjection N R Definition yields N < R () Said differently, ℵ 0 < R Is there a set X for which R < X? The answer is yes, and the next theorem explains why It implies R < P(R) (Recall that P(A) denotes the power set of A) Theorem If A is any set, then A < P(A) Proof Before beginning the proof, we remark that this statement is obvious if A is finite, for then A < A = P(A) But our proof must apply to all sets A, both finite and infinite, so it must use Definition We prove the theorem with direct proof Let A be an arbitrary set According to Definition, to prove A < P(A) we must show that there is an injection f : A P(A), but no surjection f : A P(A) To see that there is an injection f : A P(A), define f by the rule f (x) = { x } In words, f sends any element x of A to the one-element set { x } P(A) Then f : A P(A) is injective, as follows Suppose f (x) = f (y) Then { x } = { y } Now, the only way that { x } and { y } can be equal is if x = y, so it follows that x = y Thus f is injective Next we need to show that there exists no surjection f : A P(A) Suppose for the sake of contradiction that there does exist a surjection
14 0 Cardinality of Sets f : A P(A) Notice that for any element x A, we have f (x) P(A), so f (x) is a subset of A Thus f is a function that sends elements of A to subsets of A It follows that for any x A, either x is an element of the subset f (x) or it is not Using this idea, define the following subset B of A: B = { x A : x f (x) } A Now since B A we have B P(A), and since f is surjective there is an a A for which f (a) = B Now, either a B or a B We will consider these two cases separately, and show that each leads to a contradiction Case If a B, then the definition of B implies a f (a), and since f (a) = B we have a B, which is a contradiction Case If a B, then the definition of B implies a f (a), and since f (a) = B we have a B, again a contradiction Since the assumption that there is a surjection f : A P(A) leads to a contradiction, we conclude that there are no such surjective functions In conclusion, we have seen that there exists an injection A P(A) but no surjection A P(A), so Definition implies that A < P(A) Beginning with the set A = N and applying Theorem over and over again, we get the following chain of infinite cardinalities ℵ 0 = N < P(N) < P(P(N)) < P(P(P(N))) < () Thus there is an infinite sequence of different types of infinity, starting with ℵ 0 and becoming ever larger The set N is countable, and all the sets P(N), P(P(N)), etc, are uncountable In the next section we will prove that P(N) = R Thus N and R are the first two entries in the chain () above They are are just two relatively tame infinities in a long list of other wild and exotic infinities Unless you plan on studying advanced set theory or the foundations of mathematics, you are unlikely to ever encounter any types of infinity beyond ℵ 0 and R Still you will in future mathematics courses need to distinguish between countably infinite and uncountable sets, so we close with two final theorems that can help you do this Theorem 8 infinite An infinite subset of a countably infinite set is countably Proof Suppose A is an infinite subset of the countably infinite set B Because B is countably infinite, its elements can be written in a list
15 Comparing Cardinalities b, b, b, b, Then we can also write A s elements in list form by proceeding through the elements of B, in order, and selecting those that belong to A Thus A can be written in list form, and since A is infinite, its list will be infinite Consequently A is countably infinite Theorem If U A, and U is uncountable, then A is uncountable Proof Suppose for the sake of contradiction that U A, and U is uncountable but A is not uncountable Then since U A and U is infinite, then A must be infinite too Since A is infinite, and not uncountable, it must be countably infinite Then U is an infinite subset of a countably infinite set A, so U is countably infinite by Theorem 8 Thus U is both uncountable and countably infinite, a contradiction Theorems 8 and can be useful when we need to decide whether a set is countably infinite or uncountable They sometimes allow us to decide its cardinality by comparing it to a set whose cardinality is known For example, suppose we want to decide whether or not the set A = R is uncountable Since the x-axis U = { (x,0) : x R } R has the same cardinality as R, it is uncountable Theorem implies that R is uncountable Other examples can be found in the exercises Exercises for Section Suppose B is an uncountable set and A is a set Given that there is a surjective function f : A B, what can be said about the cardinality of A? Prove that the set C of complex numbers is uncountable Prove or disprove: If A is uncountable, then A = R Prove or disprove: If A B C and A and C are countably infinite, then B is countably infinite Prove or disprove: The set { 0, } R is uncountable 6 Prove or disprove: Every infinite set is a subset of a countably infinite set Prove or disprove: If A B and A is countably infinite and B is uncountable, then B A is uncountable 8 Prove or disprove: The set { (a, a, a,) : a i Z} of infinite sequences of integers is countably infinite Prove that if A and B are finite sets with A = B, then any injection f : A B is also a surjection Show this is not necessarily true if A and B are not finite 0 Prove that if A and B are finite sets with A = B, then any surjection f : A B is also an injection Show this is not necessarily true if A and B are not finite
16 Cardinality of Sets The Cantor-Bernstein-Schröeder Theorem An often used property of numbers is that if a b and b a, then a = b It is reasonable to ask if the same property applies to cardinality If A B and B A, is it true that A = B? This is in fact true, and this section s goal is to prove it This will yield an alternate (and highly effective) method of proving that two sets have the same cardianlity Recall (Definition ) that A B means that A < B or A = B If A < B then (by Definition ) there is an injection A B On the other hand, if A = B, then there is a bijection (hence also an injection) A B Thus A B implies that there is an injection f : A B Likewise, B A implies that there is an injection g : B A Our aim is to show that if A B and B A, then A = B In other words, we aim to show that if there are injections f : A B and g : B A, then there is a bijection h : A B The proof of this fact, though not particularly difficult, is not entirely trivial, either The fact that f and g guarantee that such an h exists is called the the Cantor-Bernstein- Schröeder theorem This theorem is very useful for proving two sets A and B have the same cardinality: it says that instead of finding a bijection A B, it suffices to find injections A B and B A This is useful because injections are often easier to find than bijections We will prove the Cantor-Bernstein-Schröeder theorem, but before doing so let s work through an informal visual argument that will guide us through (and illustrate) the proof Suppose there are injections f : A B and g : B A We want to use them to produce a bijection h : A B Sets A and B are sketched below For clarity, each has the shape of the letter that denotes it, and to help distinguish them the set A is shaded A B Figure The sets A and B The injections f : A B and g : B A are illustrated in Figure Think of f as putting a copy f (A) = { f (x) : x A } of A into B, as illustrated This copy, the range of f, does not fill up all of B (unless f happens to be surjective) Likewise, g puts a copy g(b) of B into A Because they are
17 The Cantor-Bernstein-Schröeder Theorem not necessarily bijective, neither f nor g is guaranteed to have an inverse But the map g : B g(b) from B to g(b) = {g(x) : x B} is bijective, so there is an inverse g : g(b) B (We will need this inverse soon) f g g Figure The injections f : A B and g : B A Consider the chain of injections illustrated in Figure On the left, g puts a copy of B into A Then f puts a copy of A (containing the copy of B) into B Next, g puts a copy of this B-containing-A-containing-B into A, and so on, always alternating g and f f f f g g g Figure An infinite chain of injections The first time A occurs in this sequence, it has a shaded region A g(b) In the second occurrence of A, the shaded region is (A g(b)) (g f )(A g(b)) In the third occurrence of A, the shaded region is (A g(b)) (g f )(A g(b)) (g f g f )(A g(b)) To tame the notation, let s say (g f ) = (g f ) (g f ), and (g f ) = (g f ) (g f ) (g f ), and so on Let s also agree that (g f ) 0 = ι A, that is, it is the identity function on A Then the shaded region of the nth occurrence of A in the sequence is n (g f ) k (A g(b)) k=0 This process divides A into gray and white regions: the gray region is G = (g f ) k (A g(b)), k=0
18 Cardinality of Sets and the white region is A G (See Figure 6) Figure 6 suggests our desired bijection h : A B The injection f sends the gray areas on the left bijectively to the gray areas on the right The injection g : g(b) B sends the white areas on the left bijectively to the white areas on the right We can thus define h : A B so that h(x) = f (x) if x is a gray point, and h(x) = g (x) if x is a white point A B f g f g Figure 6 The bijection h : A B This informal argument suggests that given injections f : A B and g : B A, there is a bijection h : A B But it is not a proof We now present this as a theorem and tighten up our reasoning in a careful proof, with the above diagrams and ideas as a guide Theorem 0 (The Cantor-Bernstein-Schröeder Theorem) If A B and B A, then A = B In other words, if there are injections f : A B and g : B A, then there is a bijection h : A B Proof (Direct) Suppose there are injections f : A B and g : B A Then, in particular, g : B g(b) is a bijection from B onto the range of g, so it has an inverse g : g(b) B (Note that g : B A itself has no inverse g : A B unless g is surjective) Consider the subset G = (g f ) k (A g(b)) A k=0
19 The Cantor-Bernstein-Schröeder Theorem Let W = A G, so A = G W is partitioned into two sets G (think gray) and W (think white) Define a function h : A B as { f (x) if x G h(x) = g (x) if x W Notice that this makes sense: if x W, then x G, so x A g(b) G, hence x g(b), so g (x) is defined To finish the proof, we must show that h is both injective and surjective For injective, we assume h(x) = h(y), and deduce x = y There are three cases to consider First, if x and y are both in G, then h(x) = h(y) means f (x) = f (y), so x = y because f is injective Second, if x and y are both in W, then h(x) = h(y) means g (x) = g (y), and applying g to both sides gives x = y In the third case, one of x and y is in G and the other is in W Say x G and y W The definition of G gives x = (g f ) k (z) for some k 0 and z A g(b) Note h(x) = h(y) now implies f (x) = g (y), that is, f ((g f ) k (z)) = g (y) Applying g to both sides gives (g f ) k+ (z) = y, which means y G But this is impossible, as y W Thus this third case cannot happen But in the first two cases h(x) = h(y) implies x = y, so h is injective To see that h is surjective, take any b B We will find an x A with h(x) = b Note that g(b) A, so either g(b) W or g(b) G In the first case, h(g(b)) = g (g(b)) = b, so we have an x = g(b) A for which h(x) = b In the second case, g(b) G The definition of G shows g(b) = (g f ) k (z) for some k > 0, and z A g(b) Thus g(b) = (g f ) (g f ) k (z) Rewriting this, ( g(b) = g f ( (g f ) k (z) )) Because g is injective, this implies b = f ( (g f ) k (z) ) Let x = (g f ) k (z), so x G by definition of G Observe that h(x) = f (x) = f ( (g f ) k (z) ) = b We have now seen that for any b B, there is an x A for which h(x) = b Thus h is surjective Since h : A B is both injective and surjective, it is also bijective
20 6 Cardinality of Sets Here are some examples illustrating how the Cantor-Bernstein-Schröeder theorem can be used This includes a proof that R = P(N) Example 6 The intervals [0, ) and (0, ) in R have equal cardinalities Surely this fact is plausible, for the two intervals are identical except for the endpoint 0 Yet concocting a bijection [0,) (0,) is tricky (Though not particularly difficult: see the solution of Exercise of Section ) For a simpler approach, note that f (x) = + x is an injection [0,) (0,) Also, g(x) = x is an injection (0,) [0,) The Cantor-Bernstein-Schröeder theorem guarantees a bijection h : [0,) (0,), so [0,) = (0,) Theorem The sets R and P(N) have the same cardinality Proof Example shows that R = (0, ), and Example 6 shows (0,) = [0,) Thus R = [0,), so to prove the theorem we just need to show that [0, ) = P(N) By the Cantor-Bernstein-Schröeder theorem, it suffices to find injections f : [0,) P(N) and g : P(N) [0,) To define f : [0,) P(N), we use the fact that any number in [0,) has a unique decimal representation 0b b b b, where each b i one of the digits 0,,,,, and there is not a repeating sequence of s at the end (Recall that, eg, 0 = 060, etc) Define f : [0,) P(N) as f ( 0b b b b ) = { 0b, 0 b, 0 b, } For example, f (0) = { 0,00,000,0000,00000, }, and f (00) = { 0,00 } Also f (0) = f (00) = { 0,0 } To see that f is injective, take two unequal numbers 0b b b b and 0d d d d in [0,) Then b i d i for some index i Hence b i 0 i f (0b b b b ) but b i 0 i f (0d d d d ), so f (0b b b b ) f (0d d d d ) Consequently f is injective Next, define g : P(N) [0,), where g(x) = 0b b b b is the number for which b i = if i X and b i = 0 if i X For example, g ({, }) = 00000, and g ({,,6,8, }) = Also g( ) = 0 and g(n) = 0 To see that g is injective, suppose X Y Then there is at least one integer i that belongs to one of X or Y, but not the other Consequently g(x) g(y ) because they differ in the ith decimal place This shows g is injective From the injections f : [0,) P(N) and g : P(N) [0,), the Cantor- Bernstein-Schröeder theorem guarantees a bijection h : [0, ) P(N) Hence [0,) = P(N) As R = [0,), we conclude R = P(N)
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INCIDENCE-BETWEENNESS GEOMETRY MATH 410, CSUSM. SPRING 2008. PROFESSOR AITKEN This document covers the geometry that can be developed with just the axioms related to incidence and betweenness. The fullMore information


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"Love Scott and my dogs do, too. He's really good with them; knowledgeable and kind. "
My name is Andrea Nuss. I worked as veterinary technician for 10 years when I younger. I now work full time in retail. I have three dogs of my own and enjoy spending time playing with them.
Hi my name is Scott. I love working with dogs and all kinds of other animals. I worked in law-enforcement for over 20 years and was working with the k-9 unit. I have had dogs all my life and all sizes. Dogs are our special babies. Thank you for letting me take care of your baby!
"Love Scott and my dogs do, too. He's really good with them; knowledgeable and kind. "
"Scott was great with Indy. He was nice tired out but not overheated. Also he handled Indy really well since he's a. Oh puppy."
"Thanks Scott for taking them out. I enjoyed the videos and they enjoyed the extra belly rubs! "
"Kelsey did such a great job with my boys!! They thoroughly enjoyed their walk with her!!!"
"Kelsey was super Friendly, took them for a good long walk, they came back tired and happy "
"ELIZABETH was extremely friendly she hit it off with Bear immediately and her love for animals truly shows I would highly recommend her"
"Excellent walker! Very nice person. Took excellent care with my dogs. I really like his credentials as my dad was once a policeman who also worked with the K-9 unit for a time. "
Addison is a vibrant city just north of Dallas. Although small in size, this Texan town is home to a vast assortment of restaurants, hotels and retail shops. The combination of small town community and big city energy make this a pawsome place for dogs and people alike.
Dogs living in Addison have a pupperific array of parks to choose from. Addison Town Park is a delightful place to take a stroll with your furry friend. Quorum Park has a beautiful fountain that lights up at night, as well as a paved walking path. For a heavenly experience, visit Celestial Park. Inspiring quotations, poetry and a human sundial make this a one of a kind park.
If your pup likes a night out on the town, he's in luck. Addison has 29 pet friendly restaurants, so go ahead and schedule a dinner date with your furry BFF. The Londoner's name says it all. This pub serves British cuisine classics like Scotch eggs, shepherd's pie and chicken curry. Choose a seat on the dog friendly patio so your pooch can join you. Lazy Dog is not to be missed. Not only do they have delicious food for humans, they have a dog menu with items like grilled chicken breast and brown rice.
Les Lacs Dog Park is just calling your pup's name. Located in the Marsh Lane and Beltway Drive neighborhood, this park provides a place for your furry pal to let loose and run wild. He'll love meeting new friends while also getting some much needed exercise. A separate small dog area allows smaller pups to play at their own speed. Drinking water is available, so make sure your pooch stays hydrated, especially during the hot Texan summer days. Les Lacs Linear Park is connected to the Dog Park, and features plenty of amenities, including a lake and a walking trail to keep your pup busy.
There are nine pet friendly hotels in Addison, so if you and Fido are just passing through, you'll have no problem finding a place to stay. Red Roof Inn and Suites Addison is located in the Town Center. This dog loving hotel allows one pet per room for no additional fee. Hyatt House Dallas Addison is also in the area. For a nominal fee, two pets up to 50 pounds can bunk here for the night. Residence Inn by Marriott welcomes two dogs of any size for an additional fee. La Quinta Inn and Suites also allows two dogs, but no fee is charged. Sounds like a tail waggin' deal!
The Addison Road and Airport Parkway neighborhood is home to Redding Trail Dog Park. Bring your pup over so she can romp and play with friends new and old. She'll love the chance to ditch the leash and run around carefree. A small dog section is provided for petite pooches, and each area has drinking fountains for two legged and four legged park patrons. You can also access the Redding Trail here which branches off from the park. For an extra treat, clip on your pal's leash and take a stroll down the beautifully landscaped path. You can also head over to North Addison Park. It's a little off the beaten path, so you may have the walking trail all to yourselves.
It's no secret dogs and parks go hand in hand. Clip on your pup's leash and head over to Vitruvian Park. Located in the Marsh Lane and Spring Valley Road area, this 12 acre park is sure to make your pooch's tail wag. Public art, live performances and water features all make this park one happening place. Let your pal strut his stuff on the waterfront walkway. This path also connects to other regional trails, so feel free to explore the area. There are also two loops around the park; one clocks in at .72 and the other is .68. Your canine buddy will love feeling the sunshine on her back while getting her exercise.
Just two miles west of Addison is Wagging Tail Dog Park. Situated in the Chalfont Place neighborhood of north Dallas, this dog park is deluxe. It spreads over 6.9 acres and has separate areas for large and small dogs. Your four legged pal can dart and dash all over, or play a game of chase with some new friends. A drinking fountain and a bowl are provided so dogs can stay hydrated while they play. There is also an observation deck that offers scenic views of a waterfall and White Rock Creek. A walking trail loops the park, so when he's done playing with his furry friends, your buddy can talk a walk with you.

subset: commoncrawl


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Books ordered may be returned for a full refund if they are not as described. Delivery is guaranteed - or your money back.
Order online or by phoneAUS 1800 067 877 NZL 0800 771 773 INT +61 2 4861 7666
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London, United Kingdom: Methuen, 1930. "The strange story of her Imperial Highness Princess Olga Borisovna - fiction based on fact of a cross section of European society from 1866 - 1924. CONDITION Green waxed cloth bds with bright Gilt title (fresh not worn) pp 239 plus ads to rear (tight unread state, not inscribed, has foxed ends and edges) Original Orange wrapper is bright with tanned spine and sl. chipped spine head. An excellent copy.. First English. Cloth. Very Good++/Near Fine. 8vo - over 7¾" - 9¾" tall.


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In November Halifax lost one of it's finest contributors to the film & television community, Matt Chisholm.
I'm going to do a terrible job of eulogizing him because he was a very good friend and it still sucks way too hard to think about the fact that he's gone now, but (here we go) he was a writer, actor, crew-member and honest to god, one of the funniest people I've ever met. Seriously, so goddamn funny. Basically everyone in Halifax has a story (or five) about Matt's kindness and generousity. I'm lucky to have known him. We all were.
I've worked with Matt a million times in my career, but I'm most proud of our work together on-screen in FOAM DRIVE RENEGADES, which Matt co-wrote alongside Adam Deviller.
This past summer, over the course of a few drinks, I told Matt that FOAM DRIVE is honestly one of the things I'm most proud of in my career, and I really mean that, I love it's blend of absurd comedy and tension, and mostly, I loved acting with Matt. It really sucks to know that I'll never share the screen with him again, but I am so thankful that I'll always have FOAM DRIVE to remember those cold nights in Dartmouth we all spent making this damn thing.
If you didn't know him, I'm sorry, give the flick a watch and hopefully that'll help, but you missed out on knowing one helluva human being…
Love ya, brother.


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Mix a few parts country twang, a spattering of hipster appeal and mix it in a pot of neo-soul-jam-rock and you've got the alternative rock group known to Pittsburgh as The Harlan Twins.
There's a humble demeanor in the band's performances. No one seems to be the main lead, as the two lead vocalists have this cute shyness to them as they drink their beer too fast and talk about their bangs between songs. But the sound is far from something you'd expect from a wallflower group - it is loud, gritty, in-your-face rock-dance-party music.
I'm always impressed with the caliber of music at WYEP's events, and these third-thursdays are no exception. There's always a nice balance of empty-next couples looking to get back into the live music scenes now that their kids have flown the coop, young and hip college-aged trendsetters who are interested in this month's radio station favorite, and people of all walks of life who are interested in a good - free - thursday night event.
As is typical with my experience with the scenesters around, I'm disappointed with the lack of dancing to this very danceable band. What does it take to get people dancing these days? I guess cocaine made for a better party drug than vicodin, though neither really seem appropriate for such a pristine and welcoming venue as the WYEP live studio room. I would eat off the floors, if prompted (and perhaps on an entirely different kind of party drug), but it isn't uncomfortable at all. All the volunteers are great, the hosts are welcoming, and it's nice to see that many of the WYEP staff are present at these events. It shows a degree of pride in something that is truly a wonderful series.
Pittsburgh's Ear for Music
A project of Tarsier Music Network.
Serve the communities you love.
WYEP Third Thursdays with the Harlan Twins
In this entry: live reviews, The Harlan Twins, wyep
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Washington: Facebook founder Mark Zuckerberg has told his investors that his company was established to accomplish a social mission- to make the world more open and connected, ahead of the biggest five billion dollar initial public offering.
"Facebook was not originally created to be a company. It was built to accomplish a social mission -- to make the world more open and connected. We think it`s important that everyone who invests in Facebook understands what this mission means to us, how we make decisions and why we do the things we do," The Los Angles Times quoted Zuckerberg, as saying in a letter included in the company`s S-1 filing with the SEC.
He said Facebook was not established simply to want to make money, but rather as a company that wants to make money to keep improving itself.
"Most great people care primarily about building and being a part of great things, but they also want to make money. Through the process of building a team -- and also building a developer community, advertising market and investor base -- I`ve developed a deep appreciation for how building a strong company with a strong economic engine and strong growth can be the best way to align many people to solve important problems.
"Simply put: We don`t build services to make money; we make money to build better services," he added.
Stating the purpose of an initial public offering, he said it aimed to keep the promise made to employees over the past eight years instead of cashing in on its profits.
Zuckerberg described Facebook culture as the 'Hacker Way' and added that it played a crucial role in maintaining the success of the social network.
Facebook uses the Hacker Way as "an approach to building that involves continuous improvement and iteration," he said.


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Deer Head Glass Art
Some deer hunters dream of a rack like this. This is 20" wide by 29" tall and is framed in an old 1948 frame that was found in an old farm barn. It again exhibits that your fantasy can become reality in glass art. Attitudes are worth sharing!
Call 281-850-2995 for more information.


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- JotForm SupportChriistianAnswered on October 28, 2015 at 11:48 AM
Are you referring to the form embedded to this page in your site - http://www.flwga.com/#!membership-application-/c1ln3? I checked the form but I was not able to find the link you are referring to. Can you provide which part did you add the link? If you can provide us a screenshot of the issue then that would be great.
We will wait for your response.


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On April 29, 2008, JAMES F.; beloved husband of the late Doris M. Wilson (nee Brown); beloved fahter of John P. Wilson, Karen A. Glacken and Tim E. Wilson; brother of Kate Klinefelter, Betty Barr and Pat Cates. Also survived by four grandchildren and three great-grandchildren. A memorial service will be held Sunday, 4 P.M. at Myers-Durboraw Funeral Home, 91 Willis Street, Westminster. Interment will be private. The family will receive friends Sunday, 3 P.M. until time of service at the funeral home. Contributions may be made in his name to Redeemer Lutheran Church, 27015 Ridge Road, Damascas, MD 20872, Carroll's, Gills United Methodist Church, 11525 Greenspring Avenue, Lutherville, MD 21093 or the American Heart Association, P. O. Box, 50216 Glen Allen, VA, 23058-5216
The Baltimore Sun
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Tuesday, February 1, 2011
Simple pillow cover tutorial
Bored and feeling unproductive, I took an hour tonight to make a cover for pillow form that was well, naked. I am going to try to do a super quick tutorial so you can make one yourself if you wish! My friend Rebecca at Painting the Roses White asked me long ago if I was going to do a tutorial after I posted this pretty little cover last summer. So, it's a little late, but here is a mini tutorial!
sewing machine
1 yard of fabric (I did not use it all for this size)
straight pins
measure tape or quilters ruler
Rotary cutter (or scissors)
Self healing cutting mat (not necessary is not using rotary cutter)
The pillow form I had was a 16x16" so here is how I measured and cut the fabric! The same principles apply regardless of the form size.
1) My form was 16"x16" So for the FRONT of the pillow cover, add 1" for seam allowance around all sides. Cut your fabric to 17"x17". Again, this will be the FRONT of the pillow.
2) For the back side measurement, you are going to add 4" to the WIDTH of the front size. This new measurement for the BACK now becomes 17"(h)x 21"(w). This additional 4" is for the extra fabric you need to make an envelope fold to slip the pillow form into.
3) Next, cut your BACK piece {measuring 17"(h)x 21'(w)} exactly in half to make the two pieces that will 'flap', making the opening slit for the pillow. So, I cut my back fabric at 10 1/2" at the center point of the fabric.
4) Now, take the front piece and place it PATTERN SIDE facing up at you. Then, place the two half pieces pattern facing down on top of that. They will overlap in the center, making the opening for your pillow form. Ultimately, the pretty pattern sides are touching now. Make sure your edges line up quite well.
5) Once you check that the edges on all four sides match up, take one of the BACK pieces and iron the inside edge (in the middle of the pillow) at 1/4" and then over again on itself again. Press press press! Then pin the fold before sewing with a simple straight stitch to secure. Repeat this on the other BACK piece.
6) Now you place the back pieces pretty side down on top of the front piece (that is pretty side up) and pin all around the edges (see my picture below-this is the overlapping flap the pillow goes in)
7) Now sew with a straight stitch at 1/4' all the way around the pillow! I cut the corners off (without cutting the threads!) to make for a nice crisp corner. Turned her right side out and stuffed my pillow form inside!
Oh man, I just reread that and I DO NOT claim to be a good tutorial-ist! SO terrible. If you have any questions, lemme know. In retrospect, next time I am going to add an inch to the measurements to make the cover a bit more loose. This one was too tight for my liking. I like it to look more relaxed. But it looks great on my guest bed in the Dolly Room!
Love and Light,


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- vcphotoAsked on January 29, 2017 at 03:50 PM
I created a form where individuals could each upload a total of 5 hi-res jpg's and I have a total of 141 (or 142) submissions.
When in the "View Submissions" page, I've tried clicking the "Download All Uploads" function, hoping it will download the nearly 700 submitted images at once. It tells you to check your email and follow the link to download. I get that email, and I click it to download, it goes through its downloading process and tries to generate a .zip folder (in this case it looks like it creates a part 1 and part 2 since there are so many) but when I try to extract the files once its finished, it says that it's not open-able.
The zipped file shows up as: "vcphoto-2017-01-29-12-36-pm.part1.rar" in my downloads folder - which leads me to believe that it isn't finished downloading (since you usually see that "part" note on files when it's still in the process of downloading) but my browser shows that the download is complete.
Thoughts? Suggestions? I've love to figure this out..
Thank you!
- JotForm SupportChriistianAnswered on January 29, 2017 at 09:36 PM
There does seem to be a lot of files, which is why it takes some time for the download to complete. Can you please try partially downloading the submissions instead? Here's a guide on How to Download the Submission Files
1. Click the gear icon
2. Select custom time frame
3. Set the time frame and click OK.
Then you can click "Download Uploads in Time Range"
If the issue persists, please let us know.


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Everday Bravado Nursing Bra Designs
Every woman is different and their bodies will vary depending on how large or small or tall or short they might be. Bravado offers a wide variety of bras in a range of colors and designs that would suit any woman and her personal needs and wants. We have listed a few of their more popular designs that women an enjoy to wear no matter what they might be doing.
Supreme Nursing Bra
The supreme nursing bra is geared to those women who have fuller breasts than others. It is difficult for many of them to find a bra that is able to hold them comfortably and without squeezing them too tightly. The Bravado supreme nursing bra is able to provide women with the support they need – but without the need of an uncomfortable underwire.
This bra is able to distribute the weight of both breasts equally over the entire bra and upper body. It helps to relieve pressure and discomfort in the shoulders, back, and neck. It is made with lightweight fabric that is soft to the touch – but durable. It also has shoulder straps to help pad the shoulders and nursing clips that make it easy to open when you need to breastfeed. This bra costs around $50.
Lifestyle Microfiber Nursing Bra
The lifestyle microfiber nursing bra is so comfortable that women are able to wear it no matter what they are doing. This is a favorite for Julia Roberts who bought it while breastfeeding her baby and was photographed wearing it for Fit Mag.
This bra features a low neckline and molded cups that help to create a smooth and seamless shape. On the exterior of the bra is microfiber that helps to create the support that is needed and a shape that allows the woman to feel good about herself. On the inside of the bra is 100% cotton that allows the body to breathe and provides comfort at the same time. This costs around $40.


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$ 12.95
Whether you'll be going to the whiteboard for the first time or just need to brush up on your skills, these supporting materials for this webinar session will provide you with the groundwork needed to make a great first impression. Bundle includes the presentation slides, two whitepapers and code examples in Xcode Playground format.
Subscribe and receive expert tips on Swift, iOS development, technical interview preparation and computer science.


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The Memsie Stakes has attracted another big name player with trainer Kris Lees indicating that multiple Group 1 winner Le Romain will be heading to Caulfield for the $1 million weight for age contest over 1400m on September 2.
Le Romain pleased Lees with a handy barrier trial win over 1000m on his home track at Newcastle this morning and the rising five year old is on target to resume in the Group 2 $200,000 Missile Stakes (1200m) at Randwick on August 5.
"He's come back in good order," Lees told racing.com.
"He's pretty professional and did everything right and while he's still off fitness-wise, he'll be ready to go in a couple of weeks."
"We'll probably head to the Memsie second-up so if we start off first-up in the Missile in Sydney, we'll have four weeks to prepare for the race in Melbourne."
The Melbourne Racing Club has doubled the purse for the Memsie Stakes which is now a $1m race but Lees said that the weight for age race was also a part of Le Romain's Spring Carnival program even before the prizemoney increase was announced.
"To be honest, I thought it was the race to target before the extra prizemoney but that certainly confirmed things," Lees said.
Le Romain is the winner of three Group 1 races and claimed the scalp of the odds on favourite Press Statement when he won his first Group 1 trophy in the $1m Girvan Waugh Randwick Guineas (1600m) at Randwick during the 2016 Sydney Autumn Carnival.
Le Romain's next Group 1 victory came during the Melbourne Spring Carnival in the $1m Cantala Stakes (1600m) at Flemington and the four year old gelding was at his best again in the Autumn in Sydney this year when he beat Chautauqua in the Group 1 $500,000 BMW Canterbury Stakes (1300m) at Randwick on March 4.
Winx then proved too strong for Le Romain who led the rest of the field home, seven and a quarter lengths behind the champion mare, in the Group 1 $1m China Horse Club George Ryder Stakes (1500m) at Rosehill on March 18.
Le Romain was then never in the hunt when fifteenth to It's Somewhat in the Group 1 $3m The Star Doncaster Mile (1600m) at Randwick on April 1 but found some good form again at is next start when second to Tivaci in the Group 1 $600,000 Schweppes All Aged Stakes (1400m) at Randwick a fortnight later.
Le Romain has earned just below the $2.2 million mark from six wins and nine minor placings from eighteen starts.


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Jackson — When newly-elected Mayor Frank Melton lured former Mississippi Development Authority leader Jimmy Heidel to the post as the city's new economic development chief, he told him "you're totally in charge. Revamp it any way you want. Just let me know what's going on."
It was exactly what Heidel wanted to hear.
Even before Heidel officially began the post October 3, he was detailing action plans like a football coach readying for The Big Game. His 2006 calendar is already filled with dates around the country for trade shows, conferences and conventions, where he will be using his networking skills to aggressively market the city to prospects.
"I was eager to get started," he said. "It's good coming back to the metro area."
The Mississippi Business Journal asked Heidel, who recently moved into a home on East Northside Drive with his wife, Joanna, about his list of priorities, limitations, regional relationships and how he intends to revamp the department to better market the City of Jackson.
Mississippi Business Journal: You mentioned that you wanted to develop an inventory of available properties in Jackson. How is that process coming along?
Jimmy Heidel: We're probably about a quarter of the way through. We have a checklist and are going through each property, making sure there are soil borings on a property, for example, or that we have the dimensions of water lines, so it's going to take a good bit of time to put together.
So far, I have a listing of all my buildings, about 45 or 50, and we're going to put them all on a detailed Web site. We're hoping to have everything completed within the next five or six months.
MBJ: Now that you've been in place a few weeks, what other strategies and priorities have you identified, and in what order?
JH: Number one, projects that have been here for a while have never gotten off the ground. We're reworking all of those and as a result, several are going to get off the ground pretty quickly. We need to sit down with some people and go over with them in detail how to finance projects using all the different programs available, like the state bond, façade grant and Community Block Development Grant (CBDG) programs. These programs have never been explained to them in detail.
The second step is to deal with the increased interest people now have in the City of Jackson. The Southern Farm Bureau Classic brought in several investors, and we're putting together some packages in different areas for them. We're working on the convention center complex to get it off the ground.
At the next meeting, we'll take to the City Council a complete Tax Increment Financing (TIF) map of the City of Jackson. The TIF basically allows the city to issue bonds based on the tax revenue projected from those projects. It provides the infrastructure money that will give us a head start on bringing them back up to standard. It takes 60 to 90 days to do it in one area. We're marking sites in different areas of Jackson within the city limits — for example, Old Jackson Square and the Highway 80 corridor — and we're going to do one complete TIF for the entire city at once so we'll have everything set when a developer comes in.
MBJ: Jackson is somewhat limited in what it can do in the way of industrial development, so what types of economic development do you see as compatible for the city?
JH: Jackson has several industrial parks, but they're on the outer area of the city. One of the biggest potentials we have is developing downtown Jackson through existing buildings and properties, not only with retail but also with office space and housing. For example, we're looking at a whole section of town that Entergy owns around Hal & Mal's. We know people in the area are interested in working with developers to rejuvenate downtown with these developments and also the development of open spaces and bike trails.
MBJ: What's the status of the King Edward Hotel?
JH: We're taking a fresh approach to the King Edward. We have a grant to do the remediation of environmental problems, which has to be done whether it's developed or torn down, so we're not going to lose there. The loan monies that are going to be available are not what was initially projected, as far as the city's input, so we're looking at putting in the investors' money first, and bringing ours in on the tail end.
MBJ: You mentioned that you wanted to contact public leaders in the area to promote metro development. Have you begun that process and what have you encountered?
JH: Jackson's relationship with the Metro Economic Development Alliance (MEDA) was pretty well nonexistent in the past. Also, the city's relationship with MDA was not very good. The whole transition plan of the mayor greatly emphasizes a regional approach to economic development in the metro area. And that's my intent. My relationship with the chamber, MEDA and MDA has been strong for years and years. We've already begun working together on projects and I think you'll see those relationships strengthened a great deal. I'm enthusiastic about the potential for the City of Jackson and bringing together a coordinated effort to make it what it should be.
MBJ: You also said you wanted to make economic development a separate department from the planning department. Is that still the plan?
JH: I don't want to make it a separate department, but I do want to rename it. I've talked to everybody concerned, and they have no problems with it. It's under the city's Department of Planning and Development, which is more related to permitting and zoning. We need a point of contact for Corporate America.
MBJ: What size staff do you envision and have you already made hires or reassignments?
JH: There will be some changes in the staff. We don't have enough emphasis on economic development. We need to address that.
MBJ: How much image-building advertising will the city be able to do with the Metro Jackson Chamber of Commerce, Jackson Convention & Visitors Bureau or MEDA?
JH: We want to get the department's Web page up and running first. We'll be going to the City Council to ask for more marketing money and produce some publications in-house to market Jackson to prospects. Personally meeting with developers around the country to market the city is going to be the greatest influence we can exert.
MBJ: What attracted you to this position?
JH: To be honest with you, this is the Capital City, and the way Mississippi grows and develops is based in a lot of ways on what happens here. My wife and I prayed about it for a long time, and we kept coming back to, if I could help the city, it's what I needed to do. I'm really enthusiastic we can do a lot in the city if we all work together.
Contact MBJ contributing writer Lynne W. Jeter at firstname.lastname@example.com.
… we'd like to ask for your support. More people are reading the Mississippi Business Journal than ever before, but advertising revenues for all conventional media are falling fast. Unlike many, we do not use a pay wall, because we want to continue providing Mississippi's most comprehensive business news each and every day. But that takes time, money and hard work. We do it because it is important to us … and equally important to you, if you value the flow of trustworthy news and information which have always kept America strong and free for more than 200 years.
If those who read our content will help fund it, we can continue to bring you the very best in news and information. Please consider joining us as a valued member, or if you prefer, make a one-time contribution.Click for more info


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iPod Touch Wi-Fi Settings: How to connect & manage iPod Touch Wi-Fi network
iPod touch automatically locates wireless hotspots (WiFi), including protected networks within a given nearby range. If it's a secure network--the networks are listed with a padlock icon--the touch will prompt you for a password. This will be one of the keys listed on your router's wireless configuration pages.Show me more...
Once connected, verify that everything's working by using Safari to surf the Web. You can use the iPod touch WiFi to connect to the Internet using Safari web browser, YouTube, or iTunes WiFi Music Store.
How does the iPod Touch connect to the WiFiWhen you use the Safari, YouTube or the iTunes WiFi Music Store, the iPod touch automatically does the following in order until it is connected:
When iPod touch is connected to a WiFi network, the WiFi icon in the status bar at the top of the screen shows connection strength. The more bars you see, the stronger the connection.
How to change the iPod Touch Wi-Fi settings
Choose Settings > Wi-Fi and turn Wi-Fi ON or OFF.
If you enjoyed this post and want to stay in touch, make sure you subscribe to the RSS feed! or via email:
|©2012 appleipodtouch.blogspot.com. All rights reserved


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JQRacing Complete "Even Smoother" Gearing Set
|Low Price:
FREE Shipping Eligible!
This item is eligible for Super Saver Shipping Learn More
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|Low Price:
This is an optional JQ Products THE Even Smoother Gearing Set. This set includes the next evolution of JQ gearing, now called "THE Even Smoother Gearing Set." This gearing has even smoother and more controllable acceleration, less wheel spin and better overall throttle control. A must have for all racers! Developed to help drivers apply the power of their engine in the efficient way possible.
This product was added to our catalog on October 27, 2017


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Yamaha Grizzly Value
Yamaha Grizzly Value is free HD wallpaper. This wallpaper was upload at January 23, 2019 upload by admin in .You can download it in your computer by clicking resolution image in Download by size:. Don't forget to rate and comment if you interest with this wallpaper.
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- Title: Yamaha Grizzly Value
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616 Scenic Road, Unit 105
Price: $549,000
Status: For sale
Beautiful first floor living at the Bluegill Lodge. Unit 105 has the space for your guests and all the luxury for you and your family. 1,782 square feet of living space, plus a large deck and master suite overlooking the lake, this is just what you've been waiting for!
Basic Information
Style: 1st Floor
Year Built: 2015
Total Finish Sq Ft: 1,782 sq ft
Condo Fee: $525
Rooms: 5
Bedrooms: 3
Bathrooms: 2
Approximate Room Dimensions
Dining Rm: 9'-11" x 13'-1"
Kitchen: 13'-1" x 12'-8"
Living Rm: 15'-9" x 17'-3"
Master BR: 15'-7" x 13'-9"
2nd BR: 13'-2" x 16'-1"
3rd BR: 13'-6" x 12'-5"
Total Baths: 2
Full baths: 2
Half baths: 0
Amenities: Beach Rights, Cable, Ceiling Fans, Deck, Dining Area, Elevator Service, Exercise Center, Fireplace- Gas, Hard Wired Smoke Detector w/ Battery, Heated Underground Parking, Lake Views, Master BR with Bath, Mountain Views, Near Golf, Near Paths, Near Shopping, Near Ski, Near Snow Mobile Trails, Outdoor Pool, Pantry, Phone, Snow Removal, Sprinkler System, Under Ground Utilities
Construction: 2x6, Wood Frame
Driveway: Paved
Appliances: CO2 Detector, Dishwasher, Disposal, General Electric, Microwave, Stainless Steel, Range- Gas, Refrigerator, Smoke Detector
Heat Fuel: Propane
Fee Includes: Landscaping, Plowing, Recreation, Trash, Exterior/ Interior Insurance
Electric: 150 AMP, Underground
Exterior: Clapboard, Shingle
Heating/Cool: Central Air :Central Air, Hot Air, Multi Zone
Roof: Shingle- Architectural
Route 3 South, take a right onto Weirs Boulevard. Stay straight, Bluegill Lodge is on the left. From Route 3 North take a left onto Watson Road after the Tamarack Rest. Watson turns into Scenic. Bluegill Lodge is on your right.
Cheryl Leigh