I have as a definition of the closure of a set $E$ in a metric space $(S,d)$ that $E^-$ is the intersection of all closed sets containing E (Elementary analysis the theory of calculus by Kenneth Ross). I have as a hunch that the closure of $E = \{\frac{1}{n}: n \in \Bbb N\} $ is $[0,1]$ since this set seems to contain every element in $E$. How can I prove this to be true?
In $\mathbb{R}$ with standard topology the closure of the above set $M$ is $M \cup 0$. I fail showing this by purely topological means. I.e. not using the fact that this holds in the reals as a metric space and deduce it for the induced topology.  I assumed that there exist a closed subset $A$ containing $M$ but not 0. Hence, the complement of $A$ is an open subset containing 0. Does this already yield a contradiction or do I need a different approach?  Could you help me on this?
Does C# have extension properties? Find $n$ such that $n/2$ is a square, $n/3$ is a cube, and $n/5$ a fifth power MySQL Date Between Refer to the "name" of an equation, while a List of Equations is generated using these names Is Unicoins a April Fool’s Stunt? Is Unicoins a April Fool’s Stunt ? Opposite of "straight talk" How do I find packages to install via apt-get Proofs involving $ (A\setminus B) \cup (A \cap B)$ Give short arguments for each part below to complete the reasoning. Please do not use Venn diagrams.     a. Prove that $A = (A\setminus B) \cup (A \cap B)$.   (Hint: You should prove that $(A\setminus B)\cup (A\cap B) \subseteq A$ and vice-versa)      b. Prove that $(A \setminus B) \cap (A \cap B) = \emptyset$   (Hint: Try a proof by contradiction for this statement.).    I'm not too sure exactly where to start on this. Both statements make sense logically, but I don't know how exactly to prove them. Can anyone point me in the right direction? Number of permutations of $n$ elements where no number $i$ is in position $i$ I am trying to figure out how many permutations exist in a set where none of the numbers equal their own position in the set; for example, $3,1,5,2,4$ is an acceptable permutation where $3,1,2,4,5$ is not because 5 is in position 5.  I know that the number of total permutations is $n!$. Is there a formula for how many are acceptable given the case that no position holds its own number? Why does US Customs & Immigration ask the purpose of my trip when I'm returning home? As a US citizen, when I re-enter the US, I am often asked where I was, how long I was there, and the reason for my visit.  Why am I asked these things?  Am I required to answer? Don't I have a legal right to enter the US (as long as I'm not doing or carrying anything illegal)?  If I were to refuse to answer these questions, would I be permitted entry?  Note: I'm not really bothered by the questions (except when the agent asking them is especially rude), I'm just curious. Creating triangle commutative diagram with tikz-cd I've just started using TikZ and need to create a simple triangle commutative diagram like this one.    The closest I've been able to get is this.   How do I get the h arrow to go from A2 to A3? Everything I've tried just causes errors.  Here's my current code:  \begin{tikzcd}[column sep=small] A \arrow{r}{i}  \arrow{rd}{g}    &amp; A3 \arrow{d}{h} \\     &amp; A2 \end{tikzcd} Renaming hundreds of files at once for proper sorting "Literally" and "Decimate" misuse Recently I've heard American TV commentators say "[a person] was literally decimated" and "[a Senator] was literally thrown under the bus". In the first case I think the person was not actually 10% killed, but in the second, I believe they meant that 57 members of the US Senate carried #58 onto Constitution Avenue and threw him under a (hopefully moving) bus.  Are usages like these normal or acceptable now? I find them grating, myself. Terminal command 'login' fails with error message Coming from Mac OS X, there is a possibility for every ordinary user to open Terminal and type in 'login' to change to a different user account.  When I do this with Ubuntu, it provides the following error message:  login: Cannot possibly work without effective root   Is there any way to remedy this?  Which file do I need to edit and with what to enable the use of this command (even for a standard user)? How to set file input value when dropping file on page? As the title suggests, I'd like to prove that the sum $$ \sum_{n=2}^\infty (\ln n)^{- \ln n} $$ is finite.  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Example     Alice has 7 and 10   Bob has 7 and K   On the board are 7, 7, 2, 3, 4   In that case, both have a three of a kind. But because Bob has a king and Alice only a 10, he wins.  Most of poker websites explaining the rules clearly mention the role of the kicker applying to three of a kind and double pairs. But much fewer say something about colors, straights and full houses, and I found many contradicting answers. I also asked the question to some used online poker players and they also gave me contradicting answers.  I have tried to search for official tournament rules, but most of them only explain what happens with bad behaviors, bad deals, incorrect or confusing betting, showing cards when you shouldn't, acting when it's not your turn, etc. without mentioning subtleties about combinations at all.  To simplify my question, I will take three examples; I think it's better to start with examples before getting to the general answer if one exists. So, what's happening in the following 3 examples? Could you give a more generalized answer?  Example 1 - Flushes     Alice has 2&clubs; and 3&clubs;   Bob has 4&clubs; and 5&clubs;   On the board are 6&clubs;, 8&clubs;, 10&clubs;, 2&hearts;, 5&diams;   Contradicting answer 1: it's a tie, because the highest card included in the flush is the 10, which everybody chooses to include in their 5 showdown cards. Contradicting answer 2: Bob wins, because he has the greatest private card that is part of the flush  Example 2 - Straight     Alice has 6 and K   Bob has 6 and J   On the board are 4, 5, 7, 8, 10   Contradicting answer 1: it's a tie, because the greatest card in the straight is the 8 for both Alice and Bob Contradicting answer 2: Alice wins, because she owns an extra king, compared to the jack of Bob  Example 3 - Full house     Alice has 3 and 7   Bob has 3 and 6   On the board are 3, 3, 2, 2, 5   Contradicting answer 1: it's a tie, because one is supposed to choose only five cards to make a combination, and a full house is already five cards. There couldn't be any kicker, and thus their showdowns are strictly equals. Contradicting answer 2: Alice wins because of her extra 7, compared to the 6 of Bob  Thank you very much for your answers.  Note: I'm unable to post next to you, so I edit my own post; strange not be able to answer to an answer.  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My group is the set of all affine maps $x \rightarrow A.x+v$ from $R^n$ to $R^n$ under function composition.