I know there is a way to view all my questions in reverse chronological order from the PC by going to my profile. But I would like to do the same from the mobile app for Android. On Android in the left upper hand corner I can only see the latest answers to my posts, but I would be able to view all my posts.
I want to see all my Questions and Answers in a particular SE site -  - profile page through Stack Exchange Android App. How can I see that? When I tried to see that I found only Top Posts(5) and Top Tags.
A doubt from Milnor's "Topology from a Differentiable Viewpoint". This is a doubt from Milnor's "Topology from a Differentiable Viewpoint".     For a smooth $f:M\to N$, with $M$ compact, and a regular value $y\in N$, we define $n(f^{-1}(y))$ to be the number of points in $f^{-1}(y)$. It is easy to see that $n(f^{-1}(y))$ is finite. It can also be seen that $n(f^{-1}(y))$ is locally constant; i.e. there is a neighbourhood of $y$ such that for every point $y'$ in that neighbourhood, $n(f^{-1}(y'))=n(f^{-1}(y))$.   The proof for the statement in bold is the following:     Let $x_1,x_2,\dots,x_n$ be the points of $f^{-1}(y)$, and choose disjoint neighbourhoods $U_1,U_2,\dots,U_n$ of these points, which are mapped diffeomorphically into neighbourhoods $V_1,V_2,\dots,V_n$ of $y$. The required neighbourhood is $(V_1\cap V_2\cap\dots V_n)-f(M-U_1-U_2-\dots U_n)$.   I don't understand how $(V_1\cap V_2\cap\dots V_n)-f(M-U_1-U_2-\dots U_n)$ is open. I know that $(V_1\cap V_2\cap\dots V_n)$ is open. However, how can one prove that $f(M-U_1-U_2-\dots U_n)$ is closed? I can see this working if $f$ is surjective, but not otherwise. I need to execute a directory copy upon a user action, but the directories are quite large, so I would like to be able to perform such an action without the user being aware of the time it takes for the copy to complete.  Any suggestions would be much appreciated. Age and Graduate school How does $e^{i x}$ produce rotation around the imaginary unit circle? Euler’s formula states that $e^{i  x} = \cos(x) + i \sin(x)$.  I can see from the MacLaurin Expansion that this is indeed true; however, I don’t intuitively understand how raising $e$ to the power of $ix$ produces rotation.  Can anyone give me an intuitive understanding? When we canonically quantize the scalar field in QFT, we use a Lorentz invariant integration measure given by $$\widetilde{dk} \equiv \frac{d^3k}{(2\pi)^3 2\omega(\textbf{k})}.$$  How can I show that it is Lorentz invariant? I'm fairly new to DMing and so is my group, so we are all on a learning curve together. We're playing D&amp;D 5e.  Last week a fighter in our group tamed a wolf with Animal Handling and then requested that he keep the wolf. I allowed this, but now I'm not sure about how to play the wolf.  Specifically, how can he train the wolf to perform certain actions such as attacking?  What other things might I need to be aware of with a PC owning a "pet"? Regex to strip lat/long from a string Anyone have a regex to strip lat/long from a string? such as:  ID: 39.825 -86.88333 No non-trivial clopen sets in $\mathbb{R}$? How to give a direct proof? How to give a direct proof of the following result?     Let $A$ be a subset of $\mathbb{R}$ such that $A$ is both open and closed. Then $A$ is either empty or all of $\mathbb{R}$?   My work:   If $A$ is not empty, let $u \in A$. Then there is some open interval $(a,b)$ such that $$u \in (a,b) \subset A.$$  Now if $\mathbb{R} - A$ is not empty either, then let $v \in \mathbb{R} - A$. Then there is some open interval $(c,d)$ such that  $$v \in (c,d) \subset \mathbb{R} - A.$$  Suppose that $u &lt; v$.   Since $$\emptyset \subset (a,b) \cap (c,d) \subset A \cap (\mathbb{R} - A) = \emptyset,$$ we must have $$(a,b) \cap (c,d) = \emptyset.$$ So we can conlclude that $$b \leq c.$$ But $b \in A$ and $c \in \mathbb{R} - A$. So we must have $$b &lt; c.$$  What next? Can anybody here please help complete the proof from here on?  An edit based on a comment by Marc Paul:   Let us assume that the set $A$ is a non-trivial clopen set in $\mathbb{R}$. Let us define a function $f \colon \mathbb{R} \to \mathbb{R}$ as follows:  $$f(x) \colon= \begin{cases} 1 \ \mbox{ for } \ x \in A; \\ 0 \ \mbox{ for } \ x \in \mathbb{R} - A. \end{cases}$$   Let $V$ be an open set in the range space $\mathbb{R}$. We show that the inverse image set $f^{-1}(V)$ is open in the domain space $\mathbb{R}$.   The following cases arise:   If $0, 1 \in V$, then $f^{-1}(V) = \mathbb{R}$.   If $0 \in V$ but $1 \not\in V$, then $f^{-1}(V) = \mathbb{R} - A$.   If $1 \in V$ but $0 \not\in V$, then $f^{-1}(V) = A$.   And, if $0 \not\in V$ and $1 \not\in V$, then $f^{-1}(V) = \emptyset$.   Thus, $f^{-1}(V)$ is open in the domain space $\mathbb{R}$.   Hence the function $f$ is continuous.   What next? How does this lead to our desired conclusion?   [Yet another edit, again based on valuable comments from Marc Paul: ]  So if both $A$ and $\mathbb{R} - A$ were non-empty, then let's suppose $a \in A$ and $b \in \mathbb{R} - A$, and we can assume without any loss of generality that $a &lt; b$.   Then $f(a) = 1$ and $f(b) = 0$. So by the intermediate value theorem there is a real number $c \in (a,b)$ such that $f(c) = 1/2$, which is a contradiction because the image set of $f$ does not contain $1/2$. How do I detect unsigned integer multiply overflow? I was writing a program in C++ to find all solutions of ab = c, where a, b and c together use all the digits 0-9 exactly once. The program looped over values of a and b, and it ran a digit-counting routine each time on a, b and ab to check if the digits condition was satisfied.  However, spurious solutions can be generated when ab overflows the integer limit. I ended up checking for this using code like:  unsigned long b, c, c_test; ... c_test=c*b;         // Possible overflow if (c_test/b != c) {/* There has been an overflow*/} else c=c_test;      // No overflow   Is there a better way of testing for overflow? I know that some chips have an internal flag that is set when overflow occurs, but I've never seen it accessed through C or C++.    Beware that signed int overflow is undefined behaviour in C and C++, and thus you have to detect it without actually causing it. For signed int overflow before addition, see . Sequence of a product of functions in $L^p. L^q$ with $p,q$ conjugate The question is: if $f_i$ is a sequence of functions in $L^p$ converging to $f$ and $g_i$ a sequence in $L^q$ converging to $g$ show that $f_ig_i$ converges to $fg$ in $L^1$ for $p,q$ finite and $\frac{1}{p}+\frac{1}{q}=1$. Does this result hold if $p=1, q=\infty$?  So I think I showed the first part. $||f_ig_i-fg||_1=||f_ig_i-f_ig+f_ig-fg||_1\leq||f_ig_i-f_ig||_1+||f_ig-fg||_1 = ||f_i(g_i-g)||_1+||g(f_i-f)||_1\leq||f_i||_p||g_i-g||_q+||g||_q||f_i-f||_p$   (by Holder's inequality) which goes to $0$ as $i\rightarrow\infty$.  However the second part of the question throws me off because I don't see why this proof doesn't work just as well for $p=1,q=\infty$.  Am I missing something? Does the second part also hold? I want to know if my server is running Subversion 1.5.  How can I find that out?  Also would be nice to know my SVN client version number. svn help hasn't been helpful.  Note: I don't want my project's revision number, etc. This question is about the Subversion software itself. Implementing a PIVOT query In bash, read after a pipe is not setting values If neutrinos are disfavoured as DM candidates why aren't axions? How can I get a complete list of fields for a standard object? wrapfig: frame around image and caption I'm currently the proud owner of Frost the horse. What I want to know is, what will happen to him if I ride another horse?  Obviously nothing will happen as long as I have him in sight, e.g. I'll be able to ride Frost to make him my horse again.  Scenarios  But what will happen if, before I get back on Frost, I fast travel somewhere? Will he return to where I originally stole him, will he disappear forever, or can I own two horses simultaneously?  Ownership?  If he goes back to where I originally stole him, will I have to steal him again, and therefore never be his true owner again?  Two Horses?  The reason I'm asking is because I want to get Shadowmere (the Dark Brotherhood horse) while keeping Frost. Is it possible to do this, or will riding Shadowmere make me lose Frost forever? Setting NTP on Windows 2008 Domain Number Theory: Prove that $m | \phi(a^m - 1)$ Text highlighting in revisions is wrong For example, when watching revision 2 in  in the side-by-side mode, word "answer" marked as deleted on the left hand side and as added again on the right hand side:        If we switch to the side-by-side markdown mode no "answer" word is highlighted at all (correctly):