Assume we have a non-empty  $X \equiv \{ A_i \}_{i \in I}$. Then the  asserts that there exists a choice function $F: X \rightarrow \cup X$ such that $F(A_i) \in A_i$. I claim that it is possible to build the choice function using just the . Of course this cannot be true because choice is independent of $ZF$. So where am I going wrong? Constructing the choice function $F$ using $X \equiv \{ A_i \}_{i \in I}$ and powerset axiom  Given an indexed family of sets $X \equiv \{ A_i \}_{i \in I}$, first make this family unique, by creating a new collection of elements $Y \equiv \cup_{i \in I}\{i\} \times A_i$. This  Build the powerset $P = 2^Y$.  Filter subsets of the powerset that have a single element $(i, \text{some element in }A_i)$ for each $i \in I$. Formally, create the set:   \begin{align*} &amp;\operatorname{ChoiceFns} \equiv \{ p \in P : \forall i \in I, \text{there exists a unique $a$ such that $(i, a) \in p$}\} \\ &amp;\operatorname{ChoiceFns} \equiv \{ p \in P : \forall i \in I, (\exists a \in A_i, ((i, a) \in p) \land (\lnot \exists a' \in A_i, a' \neq a \land (i, a') \in p))  \} \end{align*} This now gives us a collection of choice functions, since each element $p \in P$ maps  the index $i \in I$ to a unique element $a \in p$? What am I missing?  How am I able to build a set of choice functions ?
I have been thinking about the following formulation of the Axiom of Choice.:     For any relation $R$, there is a function $H$, which is a subset of $R$, and such that $\operatorname{dmn}H=\operatorname{dmn}R$.   At first, it seemed surprising to me that this did not follow from Comprehension (i.e., from the schema  $\forall x_1, x_2, ..., x_n \forall C\exists B \forall x ( x \in B \Leftrightarrow x \in C \land \phi )$, where $B$ does not occur free in $\phi$).    I thought that for any relation, it seems that we should be able to define a subset of it with the same domain meeting a "functional condition".  But when I tried to do this, I couldn't.  However, I felt like I was getting close when I let $B$ be free in $\phi$, i.e., when I let myself say that $w$ is in $R$ and also anything in $H$ meets such and such conditions, where $H$ is the set (function) whose existence I'm trying to prove.    But my understanding is that Choice does not even follow from Naive Comprehension.  So, I wondered if someone could explain the following precisely.   Why does the above formulation of Choice does not follow from regular Comprehension? Why does it not even follow from Naive Comprehension (assuming that it does not)?
Is it really possible to break the speed of light by flicking your wrist with a laser pointer? I'm planning to travel to France a few times in the next few months (tourist). Since I need a Schengen visa, I just don't know how to fill out the forms (which only have questions about travel dates of ONE single trip).  How do I report that I am traveling a few times, because if I don't they might only give me a single entry visa (or a visa not valid long enough for my next trips). I can only see that I'm requesting a multiple entry visa, nothing else about how long I need it to be valid.  Do I need to provide proof of hotel/flight reservations and travel insurance for all the trips or would the first one suffice?  How do I approach this issue so that I get a Schengen visa for at least 4.5 months? I am applying in the UK at their London application centre. If $\mathfrak p \subset R$ is a prime ideal, prove that for every nilpotent $r \in R$ it follows that $r \in \mathfrak p$.   The only hint that my tutor gave me was to use induction. Can someone explain what he means by this?  Thanks for the help! I'm looking for the title of a short story by Ursula Le Guin about a scientist (?) who is given a drug that slows the timescale of his perception enormously, so that he can perceive that an alien (?) forest is actually behaving as a sentient being, reacting to current events, but at an enormously slow (from a normal human point of view) timescale. It might be in , but I don't remember the title (and it doesn't jump out at me from the list of titles) and don't have access to that collection right now. It is not &quot;The Word for World is Forest&quot;, nor &quot;Direction of the Road&quot; (from ), nor &quot;Vaster than Empires and More Slow&quot;, the latter two of which look at similar plant perception/time-scale issues. Update: it turns out that the description was right, but I had most of the details wrong: not Le Guin, not alien, not a scientist ... Fedora 14 uses vi by default when one runs visudo. Is there a way to change this to something else? I want to propose for a Persian version of Super User (I will explain why later) so I want to create a proposal in Area 51, but this means non-English question and answers, Is this OK? P.S: This may be a question for other non-English proposals too. Update: The main question is: How can non-English site proposals pass the Commitment phase? How can the English community support non-English proposals? You can find some information in  about the Stack Exchange 2.0 Site Creation Process (specially Commitment phase):  How does this thermometer work? Intuitively, if Jon Skeet says that he’ll participate in the Sock Puppet Stack Exchange, that commitment is a better sign that the site will succeed than if we get a commitment from a random Internet user who has never participated in Stack Overflow. Sure, they’re both wonderful people, I’m sure, but Jon Skeet has proven that he likes to participate in Stack Overflow so it’s a good bet that he’ll participate in SockExchange, too. .... we’re going to require a selection of existing users with certain badges and reputation that proves that they’ll participate. For instance (and I’m making these numbers up), we might require that a site get at least 100 commitments from people with the Teacher badge, at least 20 from people with the Enthusiast badge, and at least 50 from people with a reputation of 1000 or more on some of our sites. Suppose $K/F$ is an extension field, and $E_1$ and $E_2$ are two intermediate subfields of finite degree. Let $E$ be the subfields of $K$ generated by $E_1$ and $E_2$. I'm trying to prove that $$ [E:F]\leq[E_1:F][E_2:F].$$  Since $E_1$ and $E_2$ are finite extensions, I suppose they have bases $\{a_1,\dots,a_n\}$ and $\{b_1,\dots,b_m\}$, respectively. If $E_1=F$ or $E_2=F$, then the inequality is actually equality, so I suppose both are proper extension fields. I think $E=F(a_1,\dots,a_n,b_1,\dots,b_m)$. Since $\{a_1,\dots,a_n,b_1,\dots,b_m\}$ is a spanning set for $E$ over $F$,  $$[E:F]\leq n+m\leq nm=[E_1:F][E_2:F]$$ since $m,n&gt;1$.  Is this sufficient? I'm weirded out since the problem did not ask to show $[E:F]\leq [E_1:F]+[E_2:F]$ which I feel will generally be a better upper bound. Content Type field on NewForm.aspx I have a custom list and I am using content types to add items to the list. I am using OOTB forms for new and edit items. One field that I see on the EditForm.aspx is the one that I hope I get on the NewForm.aspx also and it is the Content Type field. It is shown in the following screenshot. Do you know how to add that field on the NewForm.aspx page also?    Thanks. how can i forward requests to my own IP address to a local IP address in my network (possibly using named)? My setup  I have a linux server running in my basement. I currently have my router looking to that server for DNS lookup, so I can use named to spoof a bunch of domain names to point to my local server's local address, 192.168.0.111. This works great.  I also am running Apache and PHP on this server, and have forwarded port 80 on my router to this machine, so my public IP xx.xxx.x.xx is sent to the basement server.  This is a great development set up for me, because I can develop websites in my local network and pull them up on numerous devices with a *.mylocal.net address. I can also quickly share a website on my public IP with others that are outside of my network.  I have a static IP, so that makes this even easier.  The problem  Some applications need to have an absolute URL. My spoofed DNS doesn't work outside my network, and neither does my internal IP address.  I need to be able to access the sites at the same IP that they are accessed publicly, which is my public IP: xx.xxx.x.xx    For some reason, I can't access my own public IP address from within my own network.  My router's public IP is xx.xxx.x.xx, and when I hit xx.xxx.x.xx from outside my network I see exactly what I want -- my development website. But, when I hit xx.xxx.x.xx from inside my network, the connection times out.    I assume my ISP has a firewall that blocks requests from xx.xxx.x.xx to xx.xxx.x.xx where the origin and the destination are the same IP (or maybe this is impossible on any network?)  So how can I set up my network to forward the internal request to my public IP to the local IP of my basement server?   I would like to do this within the local DNS server I set up, or in my router, so the change can be applied to all devices in my network, and not just the basement server or the windows tablet/laptop, etc individually. I have this code:  DECLARE @MyTable AS TABLE (    [Month] INT,    Salary INT );  INSERT INTO @MyTable VALUES (1,2000), (1,3100);  SELECT [Month], Salary FROM @MyTable;   Output:    I want to concat the Salary (grouping by month) so that it will be NVARCHAR like this: '2000,3100'  How would I do this efficiently? The app forgets that I voted when I scroll the question How to watch live youtube stream using vlc? Is it possible for now to watch live youtube video streams using vlc? I just don't want to use flash player. I know that vlc can play youtube videos using link for the page where that video is located, but how about live youtube videos? How do I disable the profiles button in Chrome? Since Chrome version 38, there's a pretty ugly button in the title bar of the Chrome window that allows you to switch profiles quickly.  Since I'm the only one that uses my computer, this is not really useful to me, and I'd rather reclaim the space for more open tabs.  I've looked in the settings but don't see a way to disable it.  How can I get rid of the button? Applying Dupli-Objects Is it possible to bake dupliverts into one mesh for the BGE? (I have a large ground landscape that I use as the parent to duplicate trees, and the BGE doesn't support most modifiers.) In Illustrator is there a way to precisely scale to a pivot point or "reference point" that is off the screen? I mean when you want to zoom in somewhere far away from where you set that point and then scale precisely to align something.   Here's :     To scale relative to a different reference point , click where you   want the reference point to be in the document window, move the   pointer away from the reference point, and then drag until the object   is the desired size.   Well, I tried using this click-drag scaling method as well, but the level of precision is too low to be usable. It overshoots or undershoots the guide I'm aiming for.  The Scale tool's pop-up dialog isn't useful either. It only offers percentages. And moving/zooming are disabled while the dialog window is up, meaning I cannot navigate to the guide I need to see.  For these reasons, I need to be zoomed in on the guide, tweaking the measurements from the top Control Panel.  Some images to help illustrate what I mean:    ▲ The circle marks where I would like the center of my scale transformation to be.    ▲ Now zoomed in somewhere else, I am trying to get the black border of this bitmap image to align perfectly with the vector guide by scaling it.    ▲ I nudge the numbers in the Control Panel until the image is perfectly aligned.  One would think all would be perfect, but then I look and my reference point has vanished!   It seems the scaling has been executed relative to one of these limited nine reference point choices  instead of the reference point I specified.  There must be a way to scale precisely to a specific reference point in illustrator. Can somebody please enlighten me? How do I match any character across multiple lines in a regular expression? The equivalent electric field of a magnetic field How to get WordPress' hooks/actions run sequence? In what order add_action hooks execute? i.e. init wp_head wp_footer after_theme_setup  etc... ??? ??? ???     EDIT: I've also posted my solution. Finding the sum of series $\sum_{n=0}^∞ \frac{2^n + 3^n}{6^n}$ Finding $\lim\limits_{x\to1^-}\Bigl(\prod\limits_{n=0}^{\infty}\Bigl(\frac{1+x^{n+1}}{1+x^n}\Bigr)^{x^n}\Bigr)$ Compute : $$L=\lim_{x\to 1^-} \left(\prod_{n=0}^{\infty} \left(\frac {1+x^{n+1}}{1+x^n}\right)^{x^n}\right)$$   My try:   Writing out the first few terms I noticed that the limit can be expressed as $$L=\lim_{x\to 1^-} \frac 12\left(\prod_{n=1}^{\infty} (1+x^n)^{x^{n-1} -x^n}\right)$$ $$L=\frac 12\lim_{x\to 1^-} \left(\prod_{n=1}^{\infty} (1+x^n)^{x^{n-1}}\right)^{1-x}$$  Converting to exponential form I get $$L=\frac 12 \exp {\left(\lim_{x\to 1^-} (1-x)\sum_{n=1}^{\infty} \left(x^{n-1}\ln(1+x^n)\right)\right)}$$  And got stuck here. I did try to use approximation $\ln x\sim x$ hereon to get final answer as $\frac {\sqrt e}{2}$ but I think it was improper to use approximation and hence I believe that the answer I got is flawed. I also noticed that the form I obtained is quite similar to be used for a Riemann Sum but I didn't get a way to tackle the problem using that way.  Can someone please help me with this problem....