Mathematics

SteamyRoot

7:00 PM

o.O

First this chat goes wild about fixed points, and now about periodic points

Semiclassical

Now we just need to start talking about period 3 points and we'll be bound for chaos.

Akiva Weinberger

@SteamyRoot Whoah I think I have a counterexample to the fixed set thingy from this morning

Or at least the set-theoretic version

SteamyRoot

To what part of the fixed set thingy exactly?

Akiva Weinberger

There exists a set $X$ and a function $f:X\to X$ such that $f(\bigcap_n f^n(X))\ne\bigcap_n f^n(X)$

That is, $\bigcap_nf^n(X)$ not necessarily fixed by $f$

SteamyRoot

Okay, so what's the example?

Akiva Weinberger

7:07 PM

Trying to think of the best way to describe it

Secret

$\{\forall a,b \in \omega_2,(-\infty,b),(a,\infty),(a,b)\}$
check convergence on $(0,0)$:

$\forall a \in \omega_2, (0,0)\subseteq(0,a)$ thus $(0,a)$ is a nbhd of $(0,0)$

Thus $(0,0)\subseteq (0,0)$ thus $(0,0)$ is a nbhd

Any countable sequence $\{(a,b)\}$, with $a,b$ both $\to 0$ converges to $(0,0)$

Uncountable nets, need to check its definition again...

SteamyRoot

o.O

Not sure what $\omega_2$ is.

Akiva Weinberger

$X=\Bbb Z_{\le0}\cup\{1'\}\cup\{1'',2''\}\cup\{1''',2''',3'''\}\cup\dotsb$

SteamyRoot

But - you're sure this space is Hausdorff and compact?

Akiva Weinberger

where $\Bbb Z_{\le0}=\{\dots,-2,-1,0\}$.

Secret

7:08 PM

$\omega_2$ second uncountable ordinal.

I am not sure, I just try to work out its topological properties. The set is $\omega_2 \times [0,1)$ btw

Akiva Weinberger

$f$ subtracts $1$ from each element, keeping the same amount of tick marks if possibly

SteamyRoot

Oh, nevermind, for a second I thought that was Akiva's counterexample >.<

Akiva Weinberger

so $1^{\prime\prime\dots\prime}\mapsto0$ no matter how many tick marks there are, for example

Semiclassical

Ugh, I thought I'd left my days of thinking about steepest descent approximations behind.

SteamyRoot

Right... And this set has the discrete topology?

Semiclassical

7:10 PM

But they just keep dragging me back in :P

Akiva Weinberger

@SteamyRoot I'm doing the set-theoretic version right now, so no topology

In any case, $\bigcap_nf^n(X)=\Bbb Z_{\le0}$, but that's not fixed under $f$.

Secret

nope, because it is basically the closed long ray with $\omega_1$ (first uncountable ordinal) replaced by $\omega_2$ it should have the natural topology the order topology

SteamyRoot

@AkivaWeinberger What do you mean with "set-theoretic version" ?

Akiva Weinberger

5 mins ago, by Akiva Weinberger

There exists a set $X$ and a function $f:X\to X$ such that $f(\bigcap_n f^n(X))\ne\bigcap_n f^n(X)$

SteamyRoot

Oh.

Akiva Weinberger

7:11 PM

Dealing with sets rather than topological spaces

SteamyRoot

So no continuity conditions on $f$

Akiva Weinberger

However, it may be possible to embed this into a topological space

(Into a compact and Hausdorff one)

But does the example make sense?

For any $n$, there's a sequence of length $n$ that kind of "feeds into" $0$

but there's no infinite sequence

Secret

Hmm.... given an increasing countable sequence $\{(a,b)\}$ with $a,b \to \infty$, it does not seemed clear to me which $\omega_n$ it will converge to. I need to think about this later...

Mike Miller

@PVAL-inactive This is equivalently asking if we can start with S^n and add finitely many cells of degree n+1 or higher so that the resulting space is different but has the same homotopy groups. This seems very unlikely but also completely intractable.

SteamyRoot

Wait, why is $\cap_n f^n(X) = \mathbb{Z}_{\leq 0}$ ?

Akiva Weinberger

7:15 PM

@SteamyRoot Well, first off, do you see why $0\in\bigcap_nf^n(X)$?

SteamyRoot

For any $k$ and any $n$, the element $f^n (k+n) = k$

Akiva Weinberger

@SteamyRoot And do you see why $1'\notin\bigcap_nf^n(X)$?

SteamyRoot

Oh, wait

Akiva Weinberger

Note that there is no element called $2'$

SteamyRoot

Yeah, those ticks.

PVAL-inactive

7:16 PM

@MikeMiller Why do you know you only have 1 n-cell (up to homotopy type)

Leaky Nun

@Secret be specific

SteamyRoot

Right, yes, that works as a space.

Akiva Weinberger

@SteamyRoot I'm thinking maybe there should be another way to describe the set

but yeah, there are $n$ elements with $n$ ticks

PVAL-inactive

I mean you have the map $S^n \to X$ gives an isomorphism on $\pi_n$ so there's something you get.

Akiva Weinberger

so $f^n(n^{\prime\prime\dotsb\prime})=0$, yeah?

Secret

7:17 PM

@LeakyNun well, it seemed to heavily depend on the terms in the sequence for example $\{(n,n)\}$ $n\to \Bbb{N}$, will it converge to $\omega_1$ or $\omega_2$?

Mike Miller

@PVAL We know its nth homology and you can make a space out of as few cells as its homology demands.

Akiva Weinberger

But there's nothing that'll give you $n^{\prime\prime\dotsb\prime}$

SteamyRoot

Yup.

Akiva Weinberger

(if there are $n$ ticks there)

Leaky Nun

@Secret which topology? the ultra-long line?

Akiva Weinberger

7:18 PM

So we agree the intersection of repeated applications of $f$ is $\Bbb Z_{\le0}$ @SteamyRoot

But then $f(\Bbb Z_{\le0})=\Bbb Z_{\le-1}$

Mike Miller

Ya I know all that. There's only one map the equivalence could be. That's why you can talk about building it up from the sphere.

Secret

@LeakyNun It should be the order topology since it is basically constructed by replacing $\omega_1$ of the long line with $\omega_2$

SteamyRoot

Yeah, don't worry, I already understand the entire thing now :P

Mike Miller

I thought about it for a bit before responding I just made no progress.

SteamyRoot

It's indeed a counterexample to the "set-theoretic" version

Secret

7:19 PM

I am trying to use it to understand higher cardinals that is not just about their sizes

Akiva Weinberger

@SteamyRoot I knew there was something bothering me this morning when some claimed to prove it using just set-theoretic techniques!

Leaky Nun

@Secret what is the long line?

Secret

$\omega_1 \times [0,1)$

Mike Miller

The picture I have is that the first level the map isn't an iso you add a cell and in a dimension above that add some cells to kill off your original pi_n+k and remake it with new relator cells.

Akiva Weinberger

@Secret Other way 'round I think

Leaky Nun

7:20 PM

@Secret then what the hell is $(n,n)$?

SteamyRoot

Well, yeah, it would be very suspicious that the question asks it for a Hausdorff compact space and a continuous map; and you wouldn't need those at all :P

Leaky Nun

@AkivaWeinberger I don't think so

Long line (topology)

In topology, the long line (or Alexandroff line) is a topological space somewhat similar to the real line, but in a certain way "longer". It behaves locally just like the real line, but has different large-scale properties (e.g., it is neither Lindelöf nor separable). Therefore, it serves as one of the basic counterexamples of topology. Intuitively, the usual real-number line consists of a countable number of line segments [0, 1) laid end-to-end, whereas the long line is constructed from an uncountable number of such segments. == Definition == The closed long ray L is defined as the carte...

Akiva Weinberger

Well, the continuous and compactness stuff was used to make sure the intersection is nonempty

@SteamyRoot

Mike Miller

But this so chaotically and drastically changes higher homotopy groups you have no reason to guess some number of cells will magically end up changing all the higher ones back.

Secret

Since long line (and also the "ultralong line") has the order topology, its points are of the form $(a,b)$

PVAL-inactive

7:21 PM

@Mike I think you can do that, and there should be a proof that something like that never terminates analogous to the infinite dimensional ness of Eilenberg MacLane spaces.

Secret

so $\{(n,n)\},n\in \Bbb{N}$ is the sequence $\{(0,0),(1,1),(2,2,),...\}$

Leaky Nun

@Secret no, (1,1) isn't in the topology. $(a,b)$ where $a \in \omega_2$ and $b \in [0,1)$.

Mike Miller

You can do that but I don't expect a proof it can't terminate

Akiva Weinberger

@Secret What's the ultralong line, $\omega_2\times[0,1)$? I thought that wasn't locally homeomorphic to $\Bbb R$

You get problems at like $\langle\omega_1,0\rangle$

PVAL-inactive

I suspect most sequences of homotopy groups can't be realized by a f.d. complex.

Secret

7:23 PM

I am not currently interested with $\Bbb{R}$, I want to understand the bahaviour of higher cardinal sets and I am pretty sure it is not just about their size

Akiva Weinberger

Oh OK sure

Paul Plummer

He has the lexographical order topology on it

Akiva Weinberger

@SteamyRoot So the question is… can we embed that into a Hausdorff compact space

SteamyRoot

No.

Because the theorem holds for a Hausdorff compact space :P

Secret

uh wait a sec.. I am confused temporarily...

Akiva Weinberger

7:24 PM

Oh, wait, I think the reason is that we can consider the sequence $(1',2'',3''',\dots)$

Axoren

Why can't all curves be straight lines?

Akiva Weinberger

It's a compact space so it must have a convergent subsequences and that messes it up somehow

@Axoren What?

Leaky Nun

@AkivaWeinberger it boggles me that $\omega_1 \times [0,1)$ must have cardinality $\mathfrak c$...

Secret

@LeakyNun Order topology is generated by the base $\{(-\infty,b),(a,b),(a,\infty)\}$, right?

Akiva Weinberger

@LeakyNun Is that true??

Danu

7:25 PM

But what's the historical reason behind "arithmetic" @MikeMiller?

Akiva Weinberger

Oh, in ZFC probably

Not necessarily in ZF

I think

Leaky Nun

sure

Axoren

@AkivaWeinberger I'm just kind of frustrated that some formulas I'm working on are complicated by the fact that I'm not guaranteed straight lines.

Akiva Weinberger

What subject?

PVAL-inactive

@Mike for instance if the map wasn't surjective for exactly one of the $\pi_k$ you could kill off the original sphere and then get a f.d. Eilenberg MacLane space.

Leaky Nun

7:26 PM

@Secret yes

but beware that $b$ is also a tuple

Secret

ah ok

Axoren

@AkivaWeinberger I guess Differential Geometry, but it's more like most subjects of math all at once. I'm trying to get a simple formulation for gradient descent of a specific cost function.

SteamyRoot

@AkivaWeinberger That's sequentially compact, though :P

Axoren

And the integral curve doesn't end up being a geodesic, so I need to calculate it's curve rather than just using the exponential map.

Secret

The cool thing about higher cardinals is that you can no longer count or rely on $\Bbb{R}$ thus working with them force you to think with infinities

(paraphasing thinking with portals)

PVAL-inactive

7:28 PM

actually nevermind this probably messes with higher homotopy groups.

Paul Plummer

When do you think about $\mathbb{R}$ when thinking about cardinals?

Are you talking about $\omega$ and everything else is higher

Secret

$\aleph_1$ is the cardinality of $\Bbb{R}$ if continuum hypothesis holds

Paul Plummer

Really?

Akiva Weinberger

@SteamyRoot Compact implies sequentially compact

@Secret Speedy set goes in, speedy set comes out

Leaky Nun

@PaulPlummer I just imagine a very large amount.

Paul Plummer

7:29 PM

If it is false $\aleph_2$ could be the cardinality of $\mathbb{R}$,

Secret

yup

SteamyRoot

@AkivaWeinberger wut

Secret

So my idea of higher cardinals is $\aleph_n,n > 1$

Paul Plummer

It might be helpful for countable cardinals, since you get embedding results (but you can just think about the rationals

Akiva Weinberger

@SteamyRoot Doesn't it? It's the reverse that fails

SteamyRoot

7:30 PM

Both fail in general.

Akiva Weinberger

Every sequence in a compact set has a convergent subsequence

SteamyRoot

That's just the definition on sequential compactness :/

Akiva Weinberger

@SteamyRoot Really??

Do we need compact and Hausdorff?

Paul Plummer

Not really seeing where $\mathbb{R}$ comes in. Do you have an example where you rely on $\mathbb{R}$ to get results about $\aleph_1$ (that isn't about reals) @Secret

Akiva Weinberger

I thought for sure this was true…

Secret

7:32 PM

I am working under ZFC where GCH is true, so $|\Bbb{R}|=\aleph_1=2^{\aleph_0}$

SteamyRoot

Time to check Counterexamples in Topology

Secret

uh, wait a sec.. I might have misunderstood you question, in that case I am not sure except for the cardinality

SteamyRoot

$I^I$ is a compact Hausdorff space that is not sequentially compact

Paul Plummer

I am no expert, but pretty sure most people interested in cardinals find GCH makes the theory quite boring. I don't really think the reals give a good intuition about $\aleph$'s in either case anyways (you don't have an explicit bijection or way to think about such an ordering of the reals).

I could be mistaken though

Axoren

"Erik Christopher Zeeman tried for 7 years to prove that one cannot untie a knot on a 4-sphere. Then one day he decided to try to prove the opposite, and succeeded in a few hours."

9

Akiva Weinberger

7:39 PM

@SteamyRoot What's the sequence

Axoren

"A "theorem" of Jan-Erik Roos in 1961 stated that in an [AB4*] abelian category, lim1 vanishes on Mittag-Leffler sequences. This "theorem" was used by many people since then, but it was disproved by counterexample in 2002 by Amnon Neeman."

Akiva Weinberger

@Axoren What, a regular one-dimensional knot?

Isn't that well-known?

Axoren

@Akiva let me check the specifics. I'm just looking at disproven mathematical hypotheses right now. Many of them are well-known now, such as irrational numbers.

SteamyRoot

According to Counterexamples: "$I^I$ is not sequentially compact since the sequence of functions $\alpha_n \in I^I$ defined by $\alpha_n(x) = $ the $n$th digit in the binary expansion of $x$ has no convergent subsequence. "

Akiva Weinberger

Oh god

Wow

SteamyRoot

7:43 PM

The Stone-Cech compactification of the integers is supposedly another counterexample, but I'm not even going to try understand that one.

Mike Miller

@Danu I am wrong. Hirzebruch introduced it.

Just win baby

\o @MikeMiller

Secret

$\omega_2 \times [0,1)$

Base: $\{(a,b)\},a \in \omega_2, b \in [0,1)$

Check $(0,0)$ convergence

Since $(0,0) \subseteq (0,0)$ thus $(0,0)$ is nbhd of itself. Now $(0,0) \subseteq (0,b), b \in [0,1)$ therefore sequences and nets that converge to $(0,0)$ must be of the form $\{(a,b),a,b \to 0\}$. For example $\{(0,n)\}$ where n are the dyadic rationals $\frac{1}{2^m},m\in \Bbb{N}$ arranged in increasing powers of $\frac{1}{2}$

Check $(\omega_1,0)$ convergence

Ok it's getting late (early?) 5:44, will continue this tmr...

Axoren

@AkivaWeinberger I can't find the specifics of what type of knot, but what might be well-known now might not have been back when he proved it.

The article wikipedia uses for reference is behind a paywall, and I can't find any other mention of it

SteamyRoot

The proof for $I^I$ is actually really doable. If $(\alpha_n)_n$ has a convergent subsequence $(\alpha_{n_k})_k$, say it converges to $\alpha$. Then for any $x \in I$, $\alpha_{n_k}(x)$ converges to $\alpha(x)$.

Now let $p \in I$ such that $\alpha_{n_k}(p) = (n_k \mod 2)$. Then the sequence $\alpha_{n_k}(p)$ is just $0, 1, 0, 1, \dots$, which does not converge.

Just win baby

7:48 PM

@Axoren by "opposite" do they mean converse?

PVAL-inactive

@Axoren This quote stinks of ignorance on whoever was commenting.

SteamyRoot

@Axoren The way this is written, it sounds like he could've skipped all those years and proven the counterexample right away.

PVAL-inactive

I mean the other quote.

I don't think Zeeman or really anyone after Reidmeister thought you couldn't unknot a circle in 4-space.

I suspect the theorem Zeeman proved was quite a bit better of a result than that.

Daminark

Hey everyone!

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