I read the Jech, Set theory, and saw following proposition.  (☆) If S is a finite family of nonempty sets, existence of choice function of S can be proved without axiom of choice.  I tried to prove the proposition using axioms in ZF... But how can I pick the element of X in S explicitly despite I don't know information about X in S?  Edit: There seems to be confusions on what exactly I am asking.  I think ZFC Axioms instruct when we call a object set.(If object is not set, the object is called urelement or proper class) So I think we must show that nonemptyset has a element that is set. For example, if S is singleton, US is element of S and is set by axiom of union. That is, we can find a element of S that is set. Then using other axioms in ZF, we can construct a set we call choice function of S.  But, If S has two elements, how I can know the element of S is set??
Can anyone explain how we choose one sock from each of finitely many pairs without the axiom of choice?  I mean : To choose one sock from each of infinitely many pairs of socks requires the Axiom of Choice, but for shoes the Axiom is not needed.  The idea is that the two socks in a pair are identical in appearance, and so we must make an arbitrary choice if we wish to choose one of them. For shoes, we can use an explicit algorithm -- e.g., "always choose the left shoe." Why does Russell's statement mention infinitely many pairs? Well, if we only have finitely many pairs of socks, then AC is not needed -- we can choose one member of each pair using the definition of "nonempty," and we can repeat an operation finitely many times using the rules of formal logic (not discussed here).
I have to prove that:     $$\sum_{k=0}^{n} {{m+k} \choose{m}} = { m+n+1 \choose m+1 }$$   I tried to open up the right side with Pascal's definition that: $$ { n \choose k} = {n-1 \choose {k}} + {n-1 \choose {k-1}}$$  Here is what I came up with, and I am sure it is wrong because it does not equal the left side:  $$ {m+n+1 \choose m+1} = {m+n \choose m+1} + {m+n \choose m} = ... ={m+n \choose m+1} + {m+n-1 \choose m} + {m+n-2 \choose m-1} + ... + {m \choose m+1} = \sum_{k=0}^{n} {m+k \choose m+k-n+1 }  $$  Which, again, probably is wrong because it is not equal $\sum_{k=0}^{n} { m+k \choose m}$. Any help is appreciated When needing to ask a professor about something important and potentially delicate, what is an appropriate way to phrase the question or write the email?  Examples include:   Requesting a recommendation letter Asking for exceptions to policy (e.g., ignoring a course prerequisite, rescheduling an exam) Asking if you can do research with them, join a group, get into a program Scheduling a meeting, exam, or defense Letting them know about a change of plans Letting them know about a life event affecting your work or studies (e.g., wedding, funeral, sickness, vacation) are the uses of 'bare infinitive' & 'participle' same here? I saw him sing     I saw him singing    In the first example, the verb sing is used as a bare infinitive and in the following one it is exemplified as a present participle. My question is  if both of them are expressing the same meaning? How can I make a plane's edges taper inwards? How can I deform one edge or multiple edges of a plane or cube so they are tapered in and rounded equally? Prove $\frac{a+b+c}{abc} \leq \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}$. So I have to prove $$ \frac{a+b+c}{abc} \leq \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}.$$  I rearranged it   $$ a^2bc + ab^2c + abc^2 \leq b^2c^2 + a^2c^2 + a^2b^2 .$$  My idea from there is somehow using the AM-GM inequality. Not sure how though. Any ideas?  Thanks QGIS apply style to shapefile Can every non-simple group $G$ be written as a semidirect product? When people want to classify a group with certain (small) order, they seem to find a normal subgroup $H$ and a subgroup $K$, and then consider how many distinct $H \rtimes_\varphi K$ are there.  My question is: For any group $G$, does it always exist normal subgroup $H$, subgroup $K$ and $\phi : K \rightarrow Aut(H)$ s.t. $G\cong H \rtimes_\varphi K $?  Sorry if this question is too basic...  Thanks the comment and the answers. Then if the $G$ is not simple, can we have can $G\cong H\rtimes_\varphi K$ for proper subgroup $H$, $K$, with $H$ normal?  EDIT:  If we have a group $G$, that is not simple (then $G$ has at least one normal subgroup), are there always a non-trivial normal subgroup $H$ and a proper subgroup $K$ s.t. $G \cong H \rtimes K$. (Note that if $G$ is simple this is clearly not possible).  But the quaternion group and $Z_4$ given in answers have shown that we cannot always have it. Change style of labels of Tikz nodes of a given style How to use the keyboard to drag and drop? How to input a specific Pantone color number in Illustrator? In email applications should the text of the cover letter be stated directly in the email or be attached in a separate file? Creating System-Wide Keyboard Shortcut To Launch Applications I recently switched from Ubuntu, and I was feeling very comfortable with using Ctrl + Alt + T to launch terminal. Can I assign Shift + Cmd + T to do the same thing in Yosemite? Specifically, I want a new terminal window to be opened. We're planning to upgrade an existing farm from 2007 to 2013. I learned that the best way and maybe the only way to achieve this is by migrating in two steps. 2007 to 2010 and 2010 to 2013. The actual farm is a sql-server(2008r2 64 bit) and a sperate sharepoint-server(web-frontend, etc....) I want to keep the sql-server as it is, if possible and we want to install a new 64bit host for sharepoint 2013. We will design a new branding for 2013 afterwards, so we need to keep the content, but not the layout. There's so much about migration available on the web, but Can I find something like best practises? How do I show that $a^p + b^p > (a + b)^p$ for $a,b>0$ and $0<p<1$? A question on my math homework asks us to show that if $0 &lt; p &lt; 1$ and $a, b &gt; 0$, then $a^p + b^p &gt; (a + b)^p$. I have no idea how to do this, any pointers? What is the sum of $$\sum_{n=0}^{\infty}\frac{1}{2^n(n+1)}$$    I've spent an insane amount of time on this problem. I checked on Wolfram and it gives $ln(4)$, which I assume you get from $2\ln(2)$.   I know that the integral of $\frac{1}{n+1}$ is $\ln(n+1)$. What am I missing?  Thank you! Could anyone give me an explanation of the meaning of "get one's finger in the air" and the usage of it? Word for seeing both sides of an argument raw_input in python without pressing enter MVC DateTime binding with incorrect date format How do simulacra behave?