@Ilya yes. A subspace of a Polish space admits a complete metric if and only if it is a $G_\delta$. This is a theorem of Alexandroff, Kuratowski et al.
@tb so the metric completeness does depend on the metric: even if $d_1,d_2$ define the same topology, $(X,d_1)$ may be complete and $(X,d_2)$ can be non-complete?
Of course, $\mathbb{R}$ is complete $(-\pi/2, \pi/2)$ is not complete (with the usual metric from the reals) but they are homeomorphic via the $\arctan$
@JonasTeuwen, i'm trying to make work what we discussed yesterday and the implementation wasn't going as i'd thought. so i want to show that $\int G_t(y)(1-\phi)f(x-y)\ dy$ goes to zero with $t$. So what I thought of doing was letting $V$, the support of $\phi$, be the unit ball of radius $R$, and then this gives us the task of estimating $\int_{\mathbb{R}\setminus V} G_t(y)(1-\phi)f(x-y)\ dy$
@HenningMakholm The Hilbert cube is the product space $[0,1]^\mathbb{N}$ which is compact metrizable. It is realized as a subspace of $\ell^2$ e.g. as the set of sequences with $|x_n| \leq \frac{1}{2^n}$, for example.
@tb, maybe you can help. Let $G_t(x)=(4\pi t)^{-n/2} e^{-|x|^2/4t}$, and $f$ locally integrable on $\mathbb{R}^n$ such that $|f(x)|\le C_\epsilon e^{\epsilon|x|^2}$. I want to show that $\lim_{t\to 0} G_t* f=f(x)$. my idea is above
@tb ok. Let me state it in another way: I mean $\mathcal H = \prod\limits_{k=1}^\infty [0,1]$ with a product topology when I say "the Hilbert cube". My book suggests it, so these are not my words. Is $\mathcal H$ compact?
@Ilya yes. A space is Polish if and only if it is homeomorphic to a $G_\delta$ in that $\mathcal{H}$. That one is compact in the product topology. As I said, I give a detailed argument here
(but the product topology is not the $\ell^\infty$-norm topology, it's rather the weak$^\ast$-topology.)
@EricGregor Hence we can find a nice dominating function... Like the exponential times the supremum of your continuous monkey. So you can apply LDC by replacing $t$ by some sequence tending to $0$.
(and then note that this holds for all such sequences and you're done)
@JonasTeuwen one thing i'm confused about. when you send $t\to 0$, $\epsilon$ is fixed. so the exponential is not decaying since in the limit you just have $C_\epsilon e^{-|x-y|^2(\epsilon)}$
that $x-y$ factor is because we're doing the convolution
@JonasTeuwen, so i've shown that $\int G_t(x-y)(1-\phi)f(y)\ dy$ goes to $0$ with $t$. Now i show that because of this $u(x,t)-f(x)=\int G_t(x-y)(\phi f(y)-f(x))dy=0$, right? and this is when i should be using dominated convergence? i don't know why this problem is frustrating me so much
@EricGregor Yes. Write down everything step by step. If you're not sure about a step send me the document. I can type it up for you but 1) I don't have much time 2) It will not help you.
so we have $\int G_t(x-y)(\phi f(y)-f(x))dy\le(4\pi t)^{-n/2}\int e^{-|x|^2/4t}(\text{(1)})\ dy$, where i want to take (1) to be the uniform norm of $f(x)-f(y)$ over all $y$ in $V$
@HenningMakholm got it. Don't worry about. I try to do not spam. I never had asked something to you directly. I usually ask here and got answers or discussion or hints. I ask you with this aim. I think that's good. I try to do the work myself. However sometimes I got stuck. The point is: I don't want to be a silly boy. If that's what it feel, let me know please
@JM Apparently not. Experimentally, it's not possible to submit the question without filling in the title field. But it's a great mystery to me how so many askers think things like "Let $V$ be a finite-dimensional vector space" are appropriate question titles.
@JonasTeuwen, it seems like we should be done earlier in fact since if we can show that $[(1-\phi)f]*G_t\to 0$ on $V$, then we have shown that $f*G_t\to f$ on $V$. Since this is true for all bounded open sets, we should be able to conclude that $\lim_{t\to 0} G_t*f=f$
@Henning: Would you happen to know where I might find a full proof that Ackermann is not primitive recursive? I can't make sense of one step in my lecture notes...
@ZhenLin Hmm. I don't actually think I've ever seen that proof written out. I just saw it claimed and sort of convinced myself that I could probably prove that $x\to A(n,x)$ grows faster than any p.r. function with fewer than $n$ levels of recursion if I put my mind to it ...
It looks like there are two approaches... one is to show that Ackermann dominates w.r.t. to the max "norm", and one is to show that Ackermann dominates w.r.t. the $\ell^1$ "norm". I can prove half of the claim for each of those approaches, so I only have to show that the two "norms" are equivalent...
"I think the ordering by elegance and the ordering by simplicity are partial orders, not total orders. I'm sure there are plenty of maximal elements in the corresponding set, but I'm not sure there is a maximum (even if each of the orders is total, the product order is not...)" - oh Arturo... :D
This might seem like a strange question, but a younger friend of mine asked me about the equation $log_10(x^3)+log_10(10/x) = 2$, are the solutions to this one $x=\pm10$ or $x=10$. I would say the latter, but I am quite unsure.
I'm not even considering branch cuts. If you take the principal branch of the logarithm of a negative real, you have a result with nonzero imaginary part, yes?
Well, his argument is not more convincing. It seems to boil down to: prove Cauchy-Schwarz, and define the angle between two vectors as $\theta = \arccos{\frac{a \cdot b}{|a|\,|b|}}$.
let $f:\mathbb{R}\to\mathbb{R}$ continuous and differentiable. why is easy to see that if $t\in\mathbb{R}$ then there exist an $\alpha\gt 0$ such that $f'$ does not change the sign on $]t,t+\alpha[$?
Quick sanity check before I post an answer. @tb If $S\subseteq R$, the kernel of the map $A\otimes_S B\to A\otimes_RB$ generated by $u\otimes v\mapsto u\otimes v$ is $\cong A\otimes_{R/S}B$, right?
There may be nothing I can say that will convince André to remove the downvote. That makes the hypothesis that the downvote is competition or spite more feasible.
Yes. I've used it a time or 2 in Project Euler problems. I think there was one problem I solved where I used that method and someone else gave another method of generating coprime numbers. Trying to see if I can find it. If only I can remember which problem it was...
Hmm... Somehow managed to find the topic. Looks like it was the same Wikipedia link I gave before. And that gave me a segmentation fault. Looks like I'm going with the Farey sequence.
"Assuming that ZF is consistent, adding the Axiom of choice wont make it inconsistent". Can we say the same about any other axioms of ZF. For example, Assuming ZF without regularity axiom is consistent, can we deduce that ZF is consistent?
I'll get from 2 to 3 Master students this year. And I'm not allowed to do any measure theory with them, only applications. My supervisor regards me as a bit of a crazy mathematician, but you guys know that it's not true!
@FortuonPaendrag there are quite a few such considerations. I suggest that you look e.g. in Moschowakis's book on set theory which is a very pleasant read. If you have a model for $ZF$ without regularity you can build the von Neumann hierarchy inside that model and you get a model of ZF. This doesn't work for all the axioms but for regularity it does.
General (pointset) topology is to set theory like parsley to Greek food: some of it gets in almost every dish, but there are no great “parsley recipes” that the good Greek cook needs to know.
Keeping my fingers crossed on this program. Wikipedia algorithm for generating coprime numbers caused a segmentation fault. Using the Farey sequence like I did in the past caused another segmentation fault. I'm desperately trying the Wikipedia algorithm again, but with a different order by making the branch that should terminate quickest go first. It's been running a while so far. No segmentation faults yet...
