Prove $(0,1)$ and $[0,1]$ have the same cardinality.    I've seen questions similar to this but I'm still having trouble.  I know that for $2$ sets to have the same cardinality there must exist a bijection function from one set to the other.  I think I can create a bijection function from $(0,1)$ to $[0,1]$, but I'm not sure how the opposite.  I'm having trouble creating a function that makes $[0,1]$ to $(0,1)$.  Best I can think of would be something like $x \over 2$.  Help would be great.
How to define a bijection between $(0,1)$ and $(0,1]$?   Or any other open and closed intervals?   If the intervals are both open like $(-1,2)\text{ and }(-5,4)$ I do a cheap trick (don't know if that's how you're supposed to do it): I make a function $f : (-1, 2)\rightarrow (-5, 4)$ of the form $f(x)=mx+b$ by \begin{align*} -5 = f(-1) &amp;= m(-1)+b \\ 4 = f(2) &amp;= m(2) + b \end{align*} Solving for $m$ and $b$ I find $m=3\text{ and }b=-2$ so then $f(x)=3x-2.$  Then I show that $f$ is a bijection by showing that it is injective and surjective.
Case insensitive searching in Oracle Prefer composition over inheritance? How can I run Debian stable but install some packages from testing? Say you're running a server and you don't want to upgrade to Testing (Squeeze) from Stable (Lenny) to just install a required package or two.  What's the best way of installing only certain packages from Testing? We all know that $\sqrt[3]{3}\not\in \mathbb{Q}({\sqrt[3]{2}})$ intuitively. We even know that $\sqrt[3]{p}\not\in \mathbb{Q}(\sqrt[3]{q})$ for two distinct primes $p, q$. However, I don't know how to prove these things rigorously. In case of $\sqrt{p}\not\in \mathbb{Q}(\sqrt{q})$, we can just set assume $\sqrt{p}=a+b\sqrt{q}$ for some $a, b\in\mathbb{Q}$ and deduce a contradiction. However, in cubic case, expansion of $(a+b\sqrt[3]{2}+c\sqrt[3]{4})^{3}$ is very complicated and even I expand this I couldn't get any contradiction by this.  To be specific, it gives us 3 diophantine equations  $$a^{3}+2b^{3}+4c^{3}+12abc=3$$ $$a^{2}b+2b^{2}c+2c^{2}a=0$$ $$ab^{2}+2bc^{2}+ca^{2}=0$$  and I don't know how to check solvability of this kind of diophantine equations (over $\mathbb{Q}$). Even if there exists a way to check solvability, I cannot convince that it can be used to prove $\sqrt[3]{p}\not\in \mathbb{Q}(\sqrt[3]{q})$.   So I tried to use Galois theory, but actually I don't know where to go. I assumed $\mathbb{Q}(\sqrt[3]{3})=\mathbb{Q}(\sqrt[3]{2})$ and take a Galois closure, i.e. $K=\mathbb{Q}(\sqrt[3]{3}, w)=\mathbb{Q}(\sqrt[3]{2}, w)$ where $w^{3}=1$ then tried to use field norm maps. We can check that $\beta=\sqrt[3]{3}\cdot\sigma(\sqrt[3]{3})\cdot\sigma^{2}(\sqrt[3]{3})\in \mathbb{Q}$ when $\sigma\in Gal(K/\mathbb{Q}), \sigma(\sqrt[3]{2})=\sqrt[3]{2}, \sigma(w)=w^{2}$. But I cannot get any information from this.   Edit : It would be great if there is a simple algorithm to check whether $\alpha\in K$ or not for a given algebraic number $\alpha$ and a number field $K$. Why it's true? $\arcsin(x) +\arccos(x) = \frac{\pi}{2}$ The following identity is true for any given $x \in [-1,1]$: $$\arcsin(x) + \arccos(x) = \frac{\pi}{2}$$  But I don't know how to explain it. I understand that the derivative of the equation is a truth clause, but why would the following be true, intuitively?  $$\int^{x}_{C1}\frac{1\cdot dx}{\sqrt{1-x^{2}}} + \int^{x}_{C2}\frac{-1 \cdot dx}{\sqrt{1-x^{2}}} =\\  \arcsin(x) - \arcsin(C1) + \arccos(x) - \arccos(C2) = 0 \\  \text{while } \arcsin(C1) + \arccos(C2) = \frac{\pi}{2}$$  I can't find the right words to explain why this is true?    Edit #1 (25 Jan, 20:10 UTC): The following is a truth clause: $$ \begin{array}{ll} \frac{d}{dx}(\arcsin(x) + \arccos(x)) = \frac{d}{dx}\frac{\pi}{2} \\ \\ \frac{1}{\sqrt{1-x^{2}}} + \frac{-1}{\sqrt{1-x^{2}}} = 0 \end{array} $$  By integrating the last equation, using the limits $k$ (a constant) and $x$ (variable), I get the following:  $$ \begin{array}{ll} \int^x_k\frac{1}{\sqrt{1-x^{2}}}dx + \int^x_k\frac{-1}{\sqrt{1-x^{2}}}dx = \int^x_k0 \\ \\ \arcsin(x) - \arcsin(k) + \arccos(x) - \arccos(k) = m  \text{ (m is a constant)}\\ \\ \arcsin(x) + \arccos(x) = m + \arcsin(k) + \arccos(k) \\ \\ \text{Assuming that } A = m + \arcsin(k) + \arccos(k) = \frac{\pi}{2} \text{ ,for } x \in [-1,1] \end{array} $$ Using Calculus, why is that true for every $x \in [-1,1]$?    Edit #2:  A big mistake of mine was to think that $\int^x_k0 = m \text{ (m is const.)}$, but that isn't true for definite integrals.  Thus the equations from "Edit #1" should be as follows: $$ \begin{array}{ll} \int^x_k\frac{1}{\sqrt{1-x^{2}}}dx + \int^x_k\frac{-1}{\sqrt{1-x^{2}}}dx = \int^x_k0 \\ \\ \arcsin(x) - \arcsin(k) + \arccos(x) - \arccos(k) = 0\\ \\ \arcsin(x) + \arccos(x) = \arcsin(k) + \arccos(k) \\ \\ A = \arcsin(k) + \arccos(k) = \frac{\pi}{2} \text{ ,for } x \in [-1,1] \end{array} $$ Is there a way to do custom /say commands? I've looked it up, and apparently there is not a way to. But, there's an NPC in my map where I have it "talking" with a /say command. So, is there a way to make it say something like [(NPC Name)] (Text) without mods? How much to tip in the United States? I have an excel spreadsheet with header and survey drill data. Header data contains hole ID, and location coordinates, and the survey data contains related downhole survey with Distance, Azimuth and Dip values.   Since I know the hole location and surface elevation, I would like to be able to convert the survey table to XYZ coordinates as well.  Does anyone has a function / procedure / example?  (VB &amp; ArcObjects)  Header Data:    Survey Data: I'm new to regular expressions and would appreciate your help.  I'm trying to put together an expression that will split the example string using all spaces that are not surrounded by single or double quotes. My last attempt looks like this: (?!") and isn't quite working. It's splitting on the space before the quote.  Example input:   This is a string that "will be" highlighted when your 'regular expression' matches something.   Desired output:  This is a string that will be highlighted when your regular expression matches something.   Note that "will be" and 'regular expression' retain the space between the words. How should I deal with becoming discouraged as a graduate student? it VS. this Vs. that Why does the question's closed status disappear when clicking to load a new edit? What is the term for the grandson of my aunt? How are classical optics phenomena explained in QED (Snell's law)? How is the following classical optics phenomenon explained in quantum electrodynamics?   Reflection and Refraction   Are they simply due to photons being absorbed and re-emitted? How do we get to Snell's law, for example, in that case?  Split by request: See the other part of this question . MapServer (MS4W) test page from shapefile shows only blank white square? I am trying to create a test page with mapserver and shapefile. At the moment I just want to get the map shown on my test page. I have gotten the EXTENT coordinates by opening shapefiles with QGIS - I am sure those are correct. I have followed this  as an example.  I am accessing my test page via this link:  http://localhost:1025/cgi-bin/mapserv.exe?map=C:\MS4W\ms4w\Apache\htdocs\mapfile.map   Problem I have - map is not loaded - I only see a blank white square.  When I look at contents of ms4w\Apache\htdocs\output , I see that images are actually getting generated but all of them are white (empty), for example:  lat_pmts_region13941906093724.png lat_pmts_region13941910095144.png lat_pmts_region13941913405500.png ...   Shapefiles are located at this location: C:\MS4W\ms4w\Apache\htdocs\shapefiles.  An example of what is logged to error file:  [Fri Apr 11 11:19:32 2014].883000 CGI Request 1 on process 7888 msWCSDispatch20(): wrong service (none) msWCSParseRequest(): request is KVP. msWCSDispatch(): SERVICE and REQUEST not included msDrawMap(): rendering using outputformat named png (AGG/PNG). msDrawMap(): WMS/WFS set-up and query, 0.000s msDrawMap(): Layer 0 (lat_pmts_region), 0.001s msDrawMap(): Drawing Label Cache, 0.000s msDrawMap() total time: 0.002s msSaveImage(C:/MS4W/ms4w/Apache/htdocs/output/lat_pmts_region13972043727888.png) total time: 0.138s [Fri Apr 11 11:19:33 2014].24000 mapserv request processing time (loadmap not incl.): 0.141s [Fri Apr 11 11:19:33 2014].24000 msFreeMap(): freeing map at 0246EFD8.   ms4w\Apache\htdocs\template.html:  &lt;html&gt; &lt;head&gt;&lt;title&gt;Mapserver Template HTML Document&lt;/title&gt;&lt;/head&gt; &lt;body bgcolor=#AAAAAA&gt; &lt;h2&gt;Template&lt;/h2&gt; &lt;img src="[img]" border=1&gt; &lt;/body&gt; &lt;/html&gt;   It looks like this:   C:\MS4W\ms4w\Apache\htdocs\mapfile.map:  MAP     CONFIG "MS_ERRORFILE" "C:/MS4W/ms4w/tmp/ms_error.txt"     DEBUG 5     NAME lat_pmts_region     STATUS ON     SIZE 800 600      EXTENT 286878.53 142087.25 799084.10 481123.75     #   lat_pmts_region      UNITS METERS     SHAPEPATH "shapefiles"     IMAGECOLOR 155 155 155      WEB     TEMPLATE "template.html"     IMAGEPATH "C:/MS4W/ms4w/Apache/htdocs/output/"     IMAGEURL "/output/"     END      LAYER     NAME "lat_pmts_region"     DATA lat_pmts_region     STATUS DEFAULT     TYPE POLYGON     END END   I will gladly provide more information if it's asked. How do I learn where the source file for a given Python module is installed? Is the method different on Windows than on Linux?  I'm trying to look for the source of the datetime module in particular, but I'm interested in a more general answer as well. How can I get the DateTime for the start of the week? There is a nice property of normal spaces, namely, closed disjoint subsets can be separated by continuous functions into $\mathbb R$.   Then you ask yourself, what about Hausdorff spaces?, are all Hausdorff spaces completely Hausdorff?   Well, if we choose our separating space $X$, that is, the space such that for any Hausdorff $Y$ and any distinct $a,b\in Y$ there is some continuous $f:Y\rightarrow X$ with $f(a)\neq f(b)$, to be completely regular and Hausdorff, this cannot happen by an example of E. Hewitt, Hewitt’s Condensed Corkscrew, call it $\mathbb H$; as any continuous function from $\mathbb H$ into $\mathbb R$ is constant, and any Hausdorff completely regular space $X$ can be embedded into a product of closed intervals $\prod_{i\in I}[0,1]$, and thus any continuous $f:\mathbb H\rightarrow X\subseteq \prod_{i\in I}[0,1]$ is constant, as each $\pi_i\circ f$ is constant, so our candidate to be the separating space $X$ cannot even be $[0,1]$.  My question is, does there exist a Hausdorff space $X$ such that for all Hausdorff spaces $Y$, and any distinct $a,b\in Y$ there exists a continuous $f:Y\to X$ such that $f(a)\neq f(b)$?  Thanks. Who is the Community user? I see the  on the user page.  It seems to edit posts sometimes.  What is it? And why has it downvoted over 1800 times? See also: I pressed F2 to take a few Minecraft screenshots. Later I couldn't find them. Where are they stored on my computer?