Is it true that $M=(M^{\perp})^{\perp}$ if $M$ is a closed subspace of a PRE hilbert space (a space with a scalar product, but that is not complete)?  The proof of the analog fact for hilbert spaces uses the projection on the subspaces $M^{\perp}$ and $(M^{\perp})^{\perp}$, but if the space is not complete that projection may not exist.
I am looking for an example such that in a pre-Hilbert space $H$ we have for a subspace $U$ that  (i) $\bar{U} \oplus U^\perp \neq H$  (ii) $ \bar{U} \neq U^{\perp \perp}$  Since finite and closed subspaces will probably not do it, I am not good at inventing nice examples. Does anybody here have a few at hand?
I understand that there is an upper limit. But why do you have to type at least 15 characters to post a comment? It is easy to work around it anyway (like adding dots or other useless things). Chrome: remove URL from autocomplete which doesn't show up in history? I accidentally visited gist.gihub.com (note the missing t), and now that URL auto completes each time I start typing gist.…. How can I remove this URL?  I've tried removing it from my history, but that doesn't help:     (note the empty search results) What's the limit of the sequence $\lim\limits_{n \to\infty} \frac{n!}{n^n}$? A generic list of anonymous class Object not visible in render I have a model which you can see here (camera perspective).    In the second image, you can see how the render turned out.    I'm at my wit's end trying to figure our why it's not showing up in the render. If anyone needs more information, just ask and I'll get it to you. I've already checked the normals and they're not the cause. $X_1, X_2, \ldots, X_n$ are $n$ i.i.d. uniform random variables. Let $Y = \min(X_1, X_2,\ldots, X_n)$. Then, what's the expectation of $Y$(i.e., $E(Y)$)?  I have conducted some simulations by Matlab, and the results show that $E(Y)$ may equal to $\frac{1}{n+1}$. Can anyone give a rigorous proof or some hints? Thanks! Why is python ordering my dictionary like so? Why is pressure in a liquid the same in all directions? I'd like answers both in the more intuitive side an on the more precise side.   Thinking of water as "cubes" of water, for example, would allow pressure in the z axis to be independent of the y or x axis. Choosing other "shapes", like a  triangular prism, yield different results. Will this fact, then, be dependent on the format of the molecules of the liquid ?  I've heard of the rotational symmetry of liquids. What is its precise statement ? Why is it true ? How to prove that the set $A = \left\{ {p:{p^2} < 2,p \in {\Bbb Q^+}} \right\}$ has no greatest element? More specifically, I have to find a $q$ in $A$ such that $q$ is larger than any $p$ in $A$.  The only thing I can think of is using (${2 - {p^2}}$) somehow, or ${(p+x)^2}&lt;2$, but other than that I don't know how to proceed. This exercise is taken from the very first chapter of Walter Rudin's Principles of Mathematical Analysis. Quoting the exercise:     To do this, we associate with each rational $p &gt; 0$ the number:     $$q = p - \frac{{{p^2} - 2}}{{p + 2}} = \frac{{2p + 2}}{{p + 2}}$$   But he didn't explain how he came up with that number.   The only thing I've thought of is the following procedure:  $$\eqalign{   &amp; {p^2} + (2 - {p^2}) = 2 \Rightarrow {p^2} + (\frac{{2 - {p^2}}}{2}) &lt; 2,{\text{ and since }}2 - {p^2}{\text{ is always positive,}}  \cr    &amp; {p^2} &lt; \left[ {{p^2} + (\frac{{2 - {p^2}}}{2})} \right] &lt; 2  \cr    &amp; {\text{Therefore, let }}q = \sqrt {{p^2} + (\frac{{2 - {p^2}}}{2})} ,{\text{ then:}}  \cr    &amp; p &lt; q &lt; \sqrt 2  \cr} $$  However, my result doesn't necessarily belong to ${\Bbb Q}$. I wonder how did he obtain the value of $q$ in the book.  EDIT: Feel free to point out if there are more appropriate tags for this question. Is my ESTA valid for a second trip? Given $a>b>2$ both positive integers, which of $a^b$ and $b^a$ is larger? Given $a&gt;b&gt;2$ both positive integers, which of $a^b$ and $b^a$ is larger?  I tried an induction approach. First I showed that if $b=3$ then any $a \geq4$ satisfied $a^b&lt;b^a$.  Then using that as my base case I tried to show that given any pair of positive integers $a,b$ satisfying $a&gt;b&gt;2$ and $a^b&lt;b^a$, then $(a+1)^{b+1}&lt;(b+1)^{a+1}$ - but that is where I got stuck.  Any help would be appreciated. This problem is quite easily solved by using logarithms and derivative and forming the function $f(x,y) = x^y - y^x$, however there are assertions that this problem can be solved without using either of the two. I can not see how one would proceed to solve this problem in the absence of logarithmic manipulation and differentiation. Can anyone shed some light on this? What's the difference between !, %, and @ in replacement tokens? Hamiltonian Cycle Problem At the moment I'm trying to prove the statement:     $K_n$ is an edge disjoint union of Hamiltonian cycles when $n$ is odd.   ($K_n$ is the complete graph with $n$ vertices)  So far, I think I've come up with a proof. We know the total number of edges in $K_n$ is $n(n-1)/2$ (or $n \choose 2$) and we can split our graph into individual Hamiltonian cycles of degree 2.   We also know that for n vertices all having degree 2, there must consequently be $n$ edges. Thus we write $n(n-1)/2 = n + n + ... n$ (here I'm just splitting $K_n$'s edges into some number of distinct Hamiltonian paths) and the deduction that $n$ must be odd follows easily.  However, the assumption I made - that we can always split $K_n$ into Hamiltonian paths of degree 2 if $K_n$ can be written as a disjoint union described above - I'm having trouble proving. I've only relied on trying different values for $n$ trials and it hasn't faltered yet. So, I'm asking:     If it is true, how do you prove that if $K_n$ can be split into distinct Hamiltonian cycles, it can be split in such a way that each Hamiltonian cycle is of degree 2? Is there an analytic function $f$ on $B(0,1)$ the open ball with radius 1 such that $f(1/n)=e^{-n}$ for $n=2,3,4,...$? Is there an analytic function $f$ on $B(0,1)\subset\mathbb{C}$ such that $f(1/n)=e^{-n}$ for $n=2,3,4,...$? I know the following doesn't work:  Let $g(z)=\exp(-1/z)$. Then, $f=g$ on a sequence with a limit point in $B(0,1)$ and so $f=g$ on $B(0,1)$. Since $g$ is not $\mathbb{C}$-differentiable at $0$, neither is $f$ and so such a function cannot exist.  This is not the solution because you cannot use the identity principle with a non analytic function like $\exp(-1/z)$ is not analytic at $z=0$. Any help using the identity principle another way? How can I install a Nautilus script? A simple algebra that is not semisimple Force JavaScript exception/error when reading an undefined object property? How to make a bib entry appear also in the main document I have always had a problem installing and removing compiled software, so I have decided I would like to build software from source into a .deb package for easier installation/removal.  I would like to know of an easy and short way to build source into a .deb package, as an end user.  I have tried:   , which did not work for me This  guide, but I found it too long