@mo-_- yeah sure. That's easier

although it doesn't work to show that the sum of real solutions is $-8$

it only shows that the sum of complex roots of the polynomial $(z-1)(z-3)(z+5)(z+7)-297$ is $-8$

so not even solutions of that equation

The real solutions are solutions to $v = 36$ i.e. $u = \pm 6$

that is $z = -2\pm 6$ so their sum is $-4$, and not $-8$

Consider $\{(x,x):x\in\mathbb R\}$. I've managed to show this set is closed in $\mathbb R^2$. Now using what I've already showed, how can I argue that $\{(x,x):x\in\overline{\mathbb R}\}$ is closed in $\overline{\mathbb R}^2$?

psie the details of anything like this would depend sensitively on how you define overline{R} and the topology on it but if you somehow know or believe that R is a closed subset of overline{R} and that singletons in overline{R} are closed in overline{R} and that a finite union of closed sets is closed then yes you can use that kind of reasoning

i'm not sure you can say much about how closedness in R relates to closedness in overline{R} (or R^2 and overline{R}^2) without definitions of the topology, and are you approaching this via topological axioms or is this coming from some metric or what

what a lot of these kinds of questions have in common is (a) sensitive dependence on definitions (not that the truth of the result is in question, but that what constitutes a proof of the result depends on what you already know/assume/allow yourself to use vs. what you don't) and (b) no definitions provided

up above i should maybe have said that if you somehow know or believe that the "diagonal" in RxR is closed in overline{R}^2 as one of the postulates, and not so much that R is closed in overline{R}, similarly singletons in overline{R}^2 not in overline{R}

although all of these things are again related, what relates them? definitions

however you define 'closed,' finite unions of closed sets being closed should be very close to that definition, and the argument you're suggesting fits into that framework provided you understand the three things that you're unioning together to be closed already (whether by definition or from some work you have done)

"singletons are closed" is not a general property of arbitrary topological spaces, but it is a property enjoyed by the one you're talking about

@leslietownes ok 👍 hmm, I don't know what to say anymore. overline{R} has the order topology I would say. Not sure if that makes the diagonal in R^2 as a subset of overline{R} x overline{R} closed, nor if the singletons are closed in overline{R}.

maybe @Jakobian wants to add something to the discussion. As said above, I'd like to show $\{(x,x):x\in\overline{\mathbb R}\}$ is closed in $\overline{\mathbb R}^2$ using the fact that $\{(x,x):x\in\mathbb R\}$ is closed in $\mathbb R^2$. I know of the theorem/proposition the diagonal is closed iff the space is Hausdorff, but I don't want to use this as I don't really know what Hausdorff is nor how to prove that theorem/proposition.

Hi everyone. I just realized that the Hadamard product (element-wise product) of a square matrix $A$ and the cofactor matrix of $A$ forms a magic square (all rows and columns add up to $\operatorname{det}{A}$). Are there any references (preferably online) which discuss this fact?

@psie any LOTS is a $T_1$, in fact a $T_5$ space

moreover we can give $\overline{\mathbb{R}}$ a specific metric to show its topology coincides with its order topology

so it is in fact metrizable

and as such every singleton is closed, every countable product is metrizable, and each subspace of those products is metrizable

that is $\overline{\mathbb{R}}\times \overline{\mathbb{R}}$ is metrizable, so every singleton of it is closed

in your argument what's straight up not true, is that the diagonal of $\mathbb{R}^2$ is closed in $\overline{\mathbb{R}}^2$

its in fact, not closed in it

if $\Delta_X$ denotes the diagonal of $X\times X$ for any given topological space $X$, then $\overline{\Delta_{\mathbb{R}}} = \Delta_{\overline{\mathbb{R}}}$

the property you are asking about, that $\Delta_X$ is closed in $X\times X$, is equivalent to $X$ being Hausdorff

this is known for both $\mathbb{R}$ and $\overline{\mathbb{R}}$

@Jakobian here closure in $\overline{\mathbb{R}}$

But lets start somewhere

I don't know what you want, you have to tell me what you want

if you know what a topological space is and have agreed on your definitions, great

first lets show that $X$ is Hausdorff iff $\Delta_X$ is closed in $X\times X$, and what Hausdorff is

then lets prove that both of those spaces are Hausdorff - we can do so by explicit argument if you want

alternatively, we could explicitly show the diagonal is closed

you need to tell me your background and how much do you know

I have no idea how much have you learned about topological spaces or if you are doing some weird restriction to metric spaces or something

@Huibong Hi

you seem to be new here - my name is modularmindset

math level please?

oh c'mon

don't ping lurkers

...and demand they reveal information about themselves

Sorry I was joking I know Huibong @BenSteffan

ah, ok :)

@ModularMindset yeah me too

I totally know who they are

$$\int a_{\vec k_1}a_{\vec k_2}a_{\vec k_3}a_{\vec k_4}e^{-ix(\vec k_1 + \vec k_2 + \vec k_3 +\vec k_4)}\frac{d^3k_1}{(2\pi)^3 2\omega_{k_1}} \frac{d^3k_2}{(2\pi)^3 2\omega_{k_2}}\frac{d^3k_3}{(2\pi)^3 2\omega_{k_3}}\frac{d^3k_4}{(2\pi)^3 2\omega_{k_4}}$$

How does one solve such cases?

Where you integrate over n variables and the delta depends over those n variables?

would for example: $\vec k_4=-(\vec k_1 + \vec k_2 + \vec k_3)$ be a solution that

would help me get rid of the delta and one integration variable ?

@imbAF are you asking for integral of the kind $\int \delta(f(x)) dx$?

of the kinda were you have several variables you integrate over

like $x=(x_1,\dots,x_n)$ ?

Like $\int f(x_1,x_2,...x_n) \delta(x_1 + x_2...+x_n)dx_1dx_2...dx_2$

the general formula is $\int \delta(f(x)) \varphi(x) dx =\int_{\{f=0\}}\frac{\varphi(x)}{|\nabla f(x)|}d \sigma $

@Thorgott did you see this answer math.stackexchange.com/a/5010703/681666

it's so simple, it's always so simple

for $f:\mathbb R^n\to \mathbb R$

Does what I wrote fall into that category ?

@imbAF I struggle to understand your notation

If you have $\int e^{i(k_1-k_2)}\delta(k_1+k_2)dk_1dk_2=\int e^{2ik_1}dk_1$

Now consider more than 2 variables

@imbAF where did you find this formula?

@BenSteffan duh

I was looking at projective spaces and hoping to get a contradiction via cup product, but the product was too obscure a concept to occur to me...

highly advanced constructions such as products,

I tried for a while to see whether I could show that $[\Sigma^\infty \mathbb{R}\mathrm{P}^n, \Sigma^\infty \mathbb{R}\mathrm{P}^m] = 0$ for some combination of $n, m > 1$ but that's probably a foolish endeavor

(which would imply that there is no map $\mathbb{R}\mathrm{P}^n \to \mathbb{R}\mathrm{P}^m$ realizing an isomorphism on fundamental groups)

@Jakobian thank you for the comments :) the thing is, I'm reading about the following result. Let $(f_n)$ be a sequence of measurable functions from $E$ into $\mathbb R$. Then the set $A$ of all $x\in E$ for which $f_n(x)$ converges in $\mathbb R$ as $n\to\infty$ is measurable. The proof goes like this (I'm paraphrasing):

> Define $G:E\to\overline{\mathbb R}^2$ by $G(x)=(\liminf f_n(x),\limsup f_n(x))$. Then $G$ is measurable and if $D=\{(x,x):x\in\mathbb R\}$, then $$A=\{x\in E:-\infty<\liminf f_n(x)=\limsup f_n(x)<\infty\}=G^{-1}(D).$$So the measurability of $A$ follows since $D$ is measurable subset of $\overline{\mathbb R}^2$ (note that $\{(x,x):x\in\overline{\mathbb R}\}$ is measurable as a closed subset of $\overline{\mathbb R}^2$).

I just wanted to understand the part in parenthesis. There is this, but I feel very reluctant to just read some separate theorem in topology from this site. I'd rather read a book instead on the subject then, but currently I'm reading another book :) I feel torn.

In fact, I have started on a topology book and reached a couple of chapters into it, but it's still quite a few pages until they talk about separation axioms :(

My current "understanding" is simply intuitional; the diagonal is like a line (or perhaps is in fact a line). So there's no way to fit a ball into this line. Maybe this "understanding" is just bonkers.

@BenSteffan that (for $n>m$, I assume) is just a cup product argument, no?

well now that you say it

but the statement about stable homotopy classes is interesting in its own right

I also didn't realize that the question was about endomorphisms $X \to X$

@Jakobian I was glancing at this, and maybe one can replicate this to show also $\{(x,x):x\in\overline{\mathbb R}\}$ is closed in $\overline{\mathbb R}^2$. I actually don't see why it wouldn't work. What do you think?

@SineoftheTime you encounter this in physics

@BenSteffan yeah, if you wanted automorphisms, it becomes a lot easier

(I certainly believe that the presentation complex of $\langle a\vert a^3\rangle$ does not have an automorphism realizing the non-trivial automorphism of $\mathbb{Z}/3\mathbb{Z}$)

@psie nah, I think this doesn't work. The answerer uses the 2-norm there to derive a contradiction, but $\overline{\mathbb R}$ can't be equipped with the 2-norm. Hmm.

actually, if $n = m + 1$, then I only need to show that $(\Sigma^\infty \mathbb{R}\mathrm{P}^m)^{m + 2}(\mathbb{R}\mathrm{P}^m) = 0$ I think

maybe I should try a little harder

I have a question about this video

(It is my video by the way)

full disclosure

I wonder why this works?

Is there a better explanation out there?

@ModularMindset I have no interest in watching a video in order to answer your question. Perhaps if you actually ask the question here, I will have more interest (no promises, but I'm not watching the video).

um, the first rule of the torus nobody talks about is that nobody talks about the torus nobody talks about

This might sound complex, but you take the foliational completion ($\mathcal F$-completion) of the unit square $X^2=[0,1]^2$ i.e. the $\mathcal F$-completion, $CX^2_{V},$ where $\partial X^2 = X^2-(0,1)^2.$ Then you extend $CX^2_{V}$ in two different ways: 1. By generalizing the leaves of the $\mathcal F$-completion and simultaneously generalizing the boundary, and 2. Building a particular quotient space from $X^2.$ These two ways turn out to be homeomorphic in the category $\mathrm{Top}$

@psie Prove: If $X$ is a metric space then $\Delta_X = \{(x, x) : x\in X\}$ is closed in $X\times X$

something simple for the non-topological people

@XanderHenderson And I will define $CX^2_{V}$ now

If $d$ is a metric on $X$, pick your favourite metric on $X\times X$, then a sequence $(x_n)$ such that $(x_n, x_n)\to (x, y)$. Prove that $(x, y)\in \Delta_X$

@ModularMindset Anything involving "foliations" is already outside my area of interest / expertise. I cannot help you.

You don't need anything fancy here... topological knowledge would make this trivial, but its not necessary

@ModularMindset are you basically using the homeomorphism between a unit square modulo some points and a torus to obtain two subsets of a torus that are homeomorphic

@Jakobian I'm obtaining the torus by the standard quotient space construction (identifying edges of a square) and then I'm also obtaining a torus by building out an $\mathcal F$-completion on $X^2$ and generalizing that definition to $X^3$

they are only equivalent topologically

but homeomorphic implies diffeomorphic in dimension $< 4$?

I'm working with a self intersecting shape so not a manifold. More like a stratified space

hi All, I am working on uniform convergence of $x^n$ in $(-1,1)$. Some answers in the math forum said $(1 - 1/n)^n = 1/e$ in the limit. Then I tried $(0.999 - 1/n)^n$ and $(0.9999 - 1/n)^n$ and these all go to $0$. Any intuitive way to understand why $1$ is special.

well $(0.999 - 1/n)^n \leq (0.999)^n \xrightarrow{n \to \infty} 0$. For 1 this doesn't happen.

damn. That is good Ben. Thank you!