maybe it's clopen :)

half closed half open

It's closed.

@BalarkaSen we haven't done closed and open sets yet so I'm not familiar with them.

and the definition of the closed set seems to be a bit confusing at the moment

If $S \subset \Bbb R^n$, then $S$ is said to be closed if any convergent sequence of points in $S$ converges in $S$.

That is, there is no point outside $S$ which can be approximated arbitrarily well by points inside $S$.

To see how it fits with the interval thing in the 1 dimensional setting, $[a, b]$ is closed because no matter whatever point $x$ you choose outside of $[a, b]$, regardless of how close $x$ is to $a$ and $b$, you can never approximate $x$ arbitrarily well by points inside $[a, b]$ (because you're bounded by $a$ and $b$!)

This definition essentially says that if for any $x_0$ point in $\Bbb R^n$ if $B_r(x_0)$ (the ball of radius $r$ around $x_0$, defined by $\{x \in \Bbb R^n : ||x - x_0|| < r\}$) contains a point of $S$ for any positive real $r$, then $x_0 \in S$.

Hey, what should one learn after point-set topology?

(I don't know algebraic topology. Maybe that?)

In the one-dimensional setting, $(a, b)$ is not closed because you an approximate $b$ arbitrarily well by points in $(a, b)$. Take any small interval around $b$. Now matter how small it is, it will intersect $(a, b)$, thus will contain a point of it.

In $(-\infty, 1] = \{x \in \Bbb R : x \leq 1\}$, no matter whatever point $y > 1$ you choose, regardless of how close to $1$, you can never get points in $(-\infty, 1]$ to approximate $y$ arbitrarily well (well, you're bounded by $1$ above!)

So $(-\infty, 1]$ is closed.

@AkivaWeinberger Fair warning : algebraic topology would require some amount of algebra.

I know a bit of Galois theory, and abstract algebra.

Not the Sylow stuff though.

(Regarding closed sets: a related notion is compact, which — for subsets of $\Bbb R$ — means closed and bounded. A compact interval is of the form $[a,b]$.) @Paradox101

@BalarkaSen but if we take the -infinity part in this interval how is that closed?

I'd recommend you spending time on algebra first. Groups, rings, fields, modules.

@Paradox101 Eh? $-\infty$ is not a real number.

$(-\infty, 1]$ is just a notation for $\{x \in \Bbb R : x \leq 1\}$.

Nor is it a complex number.

@BalarkaSen The only word there I was unfamiliar with was modules.

Even being familiar with all of the words above, I am afraid, would not be enough of a prerequisite.

:P

ok @Ted I think I got it. Let $w_1...w_n,v_1...v_k$ be the base of $V$ (where $w_1...w_n$ is the base of $W$). I want to show that $v_j+W=[0]=W$ implies all the coefficent=0, I write a linear combination as $a_1(v_1+W)+...a_n(v_n+W)$ and I rewrite that as $(a_1v_1+...a_nb_n)+W=W$ using the operations I have on $V/W$. So I get the independence from the fact that the $v_i$ are independent

By spending time, I mean studying so that you can be able to do computations with them.

@BalarkaSen ok. But if for instance i take [1, $\infty$) this time will it be open or closed?

@Paradox101 What do you think?

I gave you the definition of a closed set. Tell me if it is one.

@AkivaWeinberger How much of group theory are you familiar with?

I read A Book of Abstract Algebra, which went over groups, rings, and fields. I also read Stewart's Galois Theory.

OK, but I have not read the book, so I don't know how much group theory it covers. That is why I was asking.

Whoops

Didn't mean to post that

@BalarkaSen it's closed?

Yes. Why?

I think I'd try algebraic topology, and if I find I don't know enough prerequisites I'll stop.

Bad idea, to be honest.

But you are free to do what you want.

@Jasper How would you even know how old I am?

@Alessandro Close. You mean $\sum a_j(v_j+W) = 0+W$ implies all the $a_j=0$. You actually need to use more at the end, though. Can you see where? Grazie :)

hi @Jasper @Balarka (again)

What did you have as lunch, @Ted?

:)

Just a sandwich

@BalarkaSen because if we take any convergent sequence in this interval it will converge to a value in this interval?

@Paradox101 Well, you have to prove that.

Almost, @Jasper. Haven't you noticed?

@TedShifrin Huh. For me, it's usually just a biscuit.

Hm, I'm not sure @Ted, can't I just rewrite it as $W+\sum a_jv_j$ and use the independence of the $v_j$?

No, @Alessandro. What if some linear combination of the $v_j$ happened to land in $W$?

You don't have $\sum a_jv_j = 0$, after all.

if $\sum a_jv_j=w\in W$ isn't $w+W$ just $W$ again?

@Balarka: You still claiming to be sick? :D

@Jasper: We're over 100 miles apart, so that isn't too likely.

And our knowledge bases are quite different.

Ted's old. Mike's impatient. Definitely not the same user.

But go ahead and amuse yourself.

glares @Balarka

@BalarkaSen if we are proving it formally then one example wouldn't be enough right?

Nope, @Paradox101

:D @TedShifrin

Balarka is Mike?

Totally not.

What a useless guess.

@BalarkaSen or if I prove that it's complement is open then would that be a sufficient proof?

Who is Sarah?

(also aren't the elements of $W$ uniquely written as a linear combination as $\sum a_iw_i+\sum b_jv_j$ where all of the $b_j=0$?) @Ted

Where you use "user" generically.

Yes @Alessandro. So you need to use this.

I don't get axiom of choice. Does the difficulty come when given sets in some family, one doesn't know if a particular element is in one of those sets?

ok, so $\sum a_jv_j$ can't land in $W$ and I can use the same reasoning as before, right @Ted?

@Paradox101 Have you already proven that the complement of open sets are closed?

I am so sad that he's so depressed that he lashes out at everyone, @Jasper, in particular, at me.

@Paradox101 I gave you 2 definitions of closed set, none being the complement-of-open-set one.

Certainly all of them are equivalent, but I'd like you to do it with those 2.

No, @Alessandro, not quite. What does it mean to say $v+W = W$?

He's just a troll.

@idonutunderstand Consider $\cal P(\Bbb R)$, the set of subsets of the real line.

that $v\in W$, but that's possible only if $\sum a_jv_j=0$, I think @Ted

From interactions a year ago, @Jasper, I think it's probably depression plus being in over his head mathematically.

But you need to use linear independence of $\{w_i,v_j\}$ in the first place to get there, @Alessandro.

@Balarka: Did you write up the continuous bijection thing?

And don't forget to do spanning, too.

Goodnight, @MikeM.

@idonutunderstand Is there a function $f:\cal P(\Bbb R)\to\Bbb R$ that takes each subset of $\Bbb R$ to one of its elements?

You might try making $f$ select the smallest element of the subset, but not all subsets of $\Bbb R$ have a smallest element.

Oh no. I forgot. Let me think.

there's plenty of such functions, but not a lot of injective ones(?)

@AkivaWeinberger I think it depends on how one defines the function?

Can you define such a function?

Does it exist?

The only way, it seems, to construct $f$ is to go through each subset $S\subseteq\Bbb R$, one-by-one, and choose an element $s\in S$ and let $f(S)=s$.

I think it depends on how one defines the notion of ''subset''?

@BalarkaSen ok

A function mapping every element of $\mathcal{P}(\mathbb{R})$ to $5$?

But that involves making infinitely many arbitrary choices.

anyway, back to linear algebra, I'm not sure I'm sure I see what's missing @Ted

@Alessandro We want the function to map it to one of its elements.

The axiom of choice says that we're allowed to make infinitely many arbitrary choices like that to make a function.

Ah, ok, I read the "its" in "one of its elements" as referring to $\mathbb{R}$, not to the subset, sorry

I'm being pedantic, @Alessandro, but I'm trying to get you to put the pieces together in one or two sentences. You have a proof.

We want $f(S)\in S$ for all $S\subseteq\Bbb R$.

The axiom of choice guarantees that such a function exists.

It seems to me there's only one function $f$ and no axiom of choice is needed. @Akiva

How so? $f(\Bbb R)$ can be any real.

This has to hold for every subset $S$?

@AkivaWeinberger that's really interesting.

Sure, $f(\{a\})$ has to be $a$, but that's the only thing that's forced.

I thought $f\: \Bbb R\to\Bbb R$?

Or is the domain of $f$ the power set?

No, $f:\cal P(\Bbb R)\to\Bbb R$

Yeah

ah, ok ...

Serves me right for interrupting.

I think this is the function involved in proving that $\Bbb R$ has a well-ordering.

@MikeMiller when $M$ and $N$ has the same dimension, then $f : M \to N$ has to be chartwise one-to-one, so once can just invariance of domain to get $f$ is an open map, right? so that case can be dealt with easily.

ok, so, I write the linear combinations as $W+\sum a_jv_j$. I know that $w+W=W$ iff $w\in W$ but $\sum a_jv_j$ can't be nonzero and in $W$ since the elements of $W$ can be written as a linear combination of $w_i$ and $\{w_i,v_j\}$ is linearly independent, so the only possibility is $\sum a_jv_j=0$ but using the independence of $v_j$ that's possible iff $a_j=0$ for every $j$ @Ted

First, I recommend using letters that suggest where things live, so don't write $w$ for a general element of $V$. @Alessandro. Fewer words, more tight concepts. Suppose $\sum a_jv_j \in W$. This means that $\sum a_jv_j = \sum b_iw_i$ for some scalars $b_i$. Now finish.

@AkivaWeinberger what was your comment about question of divisibility per 2^j aiming to ?

Sure, @BalarkaSen,though I'm not the one who wanted to see the proof.

is it some lame sarcasm ?

oh, right. let me deal with the inconsistent dimensional cases, then.

@Agawa001 What are you referring to?

I'll ping it to Alec when I have it.

When did I say that?

btw, nice to see you again, Columbus8 :)

One is harder than the other.

That's not possible unless all of the $a_j$ and all of the $b_j$ are $0$, since $\sum a_jv_j-\sum b_iw_i=0$ and $\{w_i,v_j\}$ are independent in $V$

Right, @Alessandro. Clean and precise and done.

@TedShifrin You too! (Did you remember only now that I'm columbus8myhw?)

@Akiva, yes. :)

Huh?!

@Alessandro: I'm very pleased with how well you're learning so much stuff :)

You're trying to prove $f^{-1}$ is continuous. That's the whole hard part!

3:27 AM, again.

Of course I don't know if $f^{-1}$ is continuous :P

purposely has no idea of what Mike and Balarka are babbling

thanks @Ted now give me a moment to think about the spanning part

@TedShifrin: Here's a question I posed earlier, which led to bickering about a different question, which we're now talking about.

@Agawa001 Dominic said something involving the last $j$ digits of a number. You protested that not all numbers are big enough to have a "last $j$ digits," and I responded that they do if you allow zeroes in front.

Suppose $f: M \to N$ is a smooth bijection between manifolds without boundary, not necessarily compact (though I don't know how to resolve the compact case). Are $M$ and $N$ still diffeomorphic, even though $f$ needn't be a diffeomorphism?

The different question is that a continuous bijection between topological manifolds (without boundary) is automatically a homeomorphism. To prove this when the dimensions needn't be the same needs some care.

(So in particular, my $M$ and $N$ are homeomorphic - just possibly with exotic smooth structures.)

@Agawa001 I guess I might've misunderstood your comment.

I believe $\text{dim} M > \text{dim} N$ should be easier.

Correct.

I guess even in the smooth case, it's not obvious that $\dim M\le \dim N$, @MikeM.

@TedShifrin: Well, just invoke the continuous theorem above...

I don't know how to do the smooth question, so I'm not looking for elegance of solution.

@AlecTeal no

ok, isn't the spanning part basically already done? Since I can write every $v\in V$ as $\sum a_iv_i$ I can get every class $v+W$ (and I get the class $[0]=W$ when all of the $a_i=0$) @Ted

I've never thought about this question. I'm suspicious, @MikeM.

Of what?

I suspect the answer might be no, @MikeM.

Sure. Yeah, I'm agnostic as to the solution. If I know more about exotic $\Bbb R^4$s that might be the place to look.

@Alessandro: Better style is to start with an arbitrary element of $V/W$ and show how to express it as a linear combination of our vectors.

Note that the critical points form a closed set. If they were isolated it would be easy to resolve. But there's no reason for that to be true.

There is the strengthened version of Sard's Theorem that I once pointed you to, I think, @MikeM, which appears in Federer (which I no longer possess).

wondering: in our situ, do the critical points still necessarily have measure 0? Not very helpful even if so, but worthj knoting

@AkivaWeinberger it is meant to be less than j digits for big and small numbers

Sorry, Federer's what?

The huge GMT tome.

He has the Hausdorff-measure strongest theorem.

So all we want to do is to prove that there is no continuous injection $\Bbb R^n \to \Bbb R^m$ for $n > m$. Let's look at $n = 3, m = 2$. Erm, why not look at the image of $S^1 \in \Bbb R^3$ in $\Bbb R^2$? Invoke Jordan curve theorem to the image to conclude discontinuity.

Hm, does this generalize to higher dimension?

@AkivaWeinberger i couldnt find time to complete my research but i found it less than first j digits

No, because you have Alexander's horned spheres in higher dimensions, @Balarka.

Not that I'm paying any attention to anything here.

I start with $[v]=v+W\in V/W$, $v=\sum a_jv_j$ so $v+W=\sum a_j(v_j+W)$ which is a linear combination of the equivalence classes in the basis @Ted

They still disconnect, @Ted. This is part of how one normally proves invariance of domain.

Not so, @Alessandro. $v = \sum a_jv_j + \sum b_iw_i$ !!

Cute argument, not what I had in mind. In particular, the local theorem you want to prove is not for $\Bbb R^m$. The codomain can be $\Bbb R^n$ if you like, but the domain needs to be an open subset of $\Bbb R^m$.

Yeah, the ball it bounds is just not simply connected.

Who cares about that?

(I was worried about the Alexander horned sphere too)

About what?

Of course, you would need to tell me why it's true that they do still separate. ;)

About one of the connected components not being simply connected.

@MikeMiller This is in Hatcher 2.B, right?

Sure, fine.

There is a much cleaner answer with invariance of domain itself.

$\sum b_iw_i=w\in W$ so $v+W=\sum a_jv_j+(w+W)=\sum a_jv_j+W$ @Ted?

Restrict to a subspace of $\Bbb R^m$ homeomorphic to $\Bbb R^n$?

Still continuous + injective.

Suppose $U \to \Bbb R^n$ is a continuous bijection, where $U$ is an open subset of $\Bbb R^m$, $m>n$. Now compose with a linear inclusion into $\Bbb R^m$. The image of $U$ is obviously not open.

finally :D @Ted!

"Oops!"

Well, at your stage of mathematics, @Alessandro, it's important to learn to say things carefully and correctly. I was being tough on you.

I know, my professor does that all the time too @Ted

Well, that's good for you!

@TedShifrin: I still don't know the name of the book you're talking about and can't find it.

@MikeMiller What are the scare quotes for this time?

That's a contradiction.

@MikeM: Federer wrote an 800-page book (Springer, I think) called Geometric Measure Theory. It's the unreadable bible.

The oops is spoken by the poor soul to have supposed there's such a continuous bijection.

ok, funny way to say it. :P

I believe Oops appears in some of my videos, @Balarka :P

Damn, why do I have the physics chat open in my tab?

I think you must have pointed this out to someone else, or at least I didn't see your message about this book and its version of Sard.

Good Q.

oh, ok, sorry, then, @MikeM ....

He has Hausdorff dimension of the set of critical values.

No need to apologize... just explaining my confusion

OK, I believe that. I still am not sure it helps. :P

I actually had to use this in my second published paper.

thanks a lot for your help @Ted, I think I had enough math for today and I'm going to sleep now!

Buono notte, @Alessandro :)

@TedShifrin Lurie's HA is also 800 pages :P

(or so Alex informed me)

I was just guestimating, @Balarka. I no longer have the book to check.

@MikeM: Check this out!

@TedShifrin: Do you think one can make sense of codimension of nasty sets in an inf-dim space?

I dunno. Check for Smale's generalization of Sard to infinite dimensional manifolds.

Hello @TedShifrin !! Could you take a look at my question math.stackexchange.com/questions/1481950/centre-of-the-circle/… and explain to me why the unit normal vector $n(s)$ points towards the circle of the center? ?

I would really quite like a theorem like "The critical set has 'codimension' blah"

I know & am a big fan of that theorem... dunno if these fancy types have thought about it.

I dunno, @MikeM.

@MaryStar: I am not going to read through all of that. The right way to do it is to use Taylor's Theorem and the local canonical form for space curves.

Nice little survey.

lol. google translates eau de toillette to toilet water.

ok, back to incosistent dimensional work.

eau @Balarka

Federer's book seems nice. And with such a strong tennis career, too!

It's so technical as to be unreadable, but there's tons of important, deep stuff in there, @MikeM.

Yes, Roger has had to work hard to overcome his name of Herbert.

Only 672pp.

@TedShifrin Yes... My question is how we know at which direction the unit normal vector points. To the inner of the cirle or outwards.

Always into the osculating circle.

John Lott has some notes titled "Notes on Perelman's papers", or something similar. This brief set of notes, intended to clarify some of the more technical details of the papers, comes out to a little under 400pp.

Why? @TedShifrin

Is that all? @MikeM

Perhaps they were being brief.

Because the unit normal points in the direction in which the curve is bending at that moment, @MaryStar.

On a question about an introduction to symplectic geometry, I say McDuff-Salamon is the canonical source.

"Do you mean "Introduction to Symplectic Topology" or "J-holomorphic Curves and Symplectic Topology"?"

Hmmm... I wonder.

Still not an unfair question, come on.

You should have known about Federer's GMT bible, but you didn't.

Actually, I am not at all sure what should I do for $\dim M < \dim N$. Lots of continuous surjections $M \to N$, but that none of them is injective looks tough.

You should go to sleep, @Balarka.

@BalarkaSen: Ultimately, you're going to invoke the Baire category theorem.

@MikeMiller Uh-oh.

Nighty night :)

Do I really have to invoke BCT? shifts uneasily

Fabulous theorem!

@BalarkaSen: Try to think how you'd do it if the domain were $\Bbb R^n$.

It's a powerful tool that often replaces measure theory ideas.

Yes, but I thought there was a purely algebro-topological proof.

@MikeMiller Ohh. Right.

I never said that.

I recall why space-filling curves are not injective.

Why?

I asked you about that the other day, @Balarka? How un-injective must they be?

@MikeMiller Take a space-filling curve in $\Bbb R^2$. Chop off pieces (images of closed intervals which cover the domain $\Bbb R$. None of these pieces are dense anywhere.

That does not sound to me like a proof.

Why are the pieces not dense anywhere?

I'm not sure that's correct.

In fact, I'm pretty sure it can't be.

I don't agree with that assertion, @Ted.

Well, fine, then :D

I was gonna disappear anyhow.

Well, if the piece was dense somewhere, then you could chuck out points so that the piece would be connected, no?

(I am trying to recall the gritty details, give me some time)

I think you should probably write a proof instead of whatever this is. :P

I think I'll ask my students tomorrow about that smooth bijection question, @Ted, and offer $20 if they can figure it out.

Out of interest, may I ask how to bound $x^{k}$ from above over $[0,1]$ if $k$ is positive?

From above by what?

By another function s.t. $|x^{k}|$ is less than that function, @MikeMiller.

(Forgot that important detail!)

Strictly less?

Nope, $\leq$ @MikeMiller.

$x^k$ seems to work pretty well.

$(1-x)^k$ is an upper bound for that. I think I'm pretty good at this :)

OK, $\{\mathcal{V}_i\}$ be a covering of $\Bbb R$ by closed intervals. $f(\mathcal{V}_i)$ is then a closed subset of $\Bbb R^n$ since $f$ is an injection implies this piece is homeomorphic to $\mathcal{V}_i$ (domain compact, codomain Hausdorff, $f$ is by hypothesis an injection).

@Khallil: Why don't you actually just integrate it? I'm confused.

Oh, it's a product of those terms but with different powers that I'm looking at, @MikeMiller.

Go on.

Oh, oh, I recall it. If it's somewhere dense, take it's closure.

That must contains a disk.

But as it's closed in the first place, it itself contain a disk. Clearly a contradiction, that cannot be homeomorphic to a closed interval

You probably should flesh out the details, but that's the gist, yes. Now at some point you may want to write up this proof where the domain is no longer $\Bbb R$ and send it to the person who wanted to see it.

OK, $\text{im} f$ is union of $\aleph_0$-many nowhere dense sets, so it has to have empty interior -- hell of a contradiction.

@MikeMiller Alright, I'll try it out.

Now find someone who can answer the question I originally asked 'cuz I wanna know the answer.

I can ask my prof.

So the question was that whether there exists nondiffeomorphic $M$ and $N$ such that $f : M \to N$ is a smooth bijection, right?

Yeah.

That seems very related. But I don't see a yes or no for the general question.

It turns out that "google it" would have been a good first avenue of approach.

hehe.

Hey everyone!

So how long have you guys been here?

Greg's answer literally answers it: there is such a map for every pair of exotic spheres in dimension $n>4$.

I just saw an easy question and thought I'd answer it :)

Indeed, for any pair of PL isomorphic smooth manifolds.

Aw.

That's pretty cool though.

Oh, well IDK

I was just exploring the site a bit..

Wow

This is my first chat here