Mathematics - 2024-03-11 (page 2 of 2)

Balarka Sen

19:00

$\|\cdot\|$ agrees with a linear functional on $V$ restricted to the subspace

Alessandro Codenotti

So the optimistic statement here would be that the space on which the norm is linear is convex

Jakobian

@Thorgott Before when we were talking about Alexandroff extension functor, I had this idea. You mentioned we can just set whole space to be the only neighbourhood of infinity. My point is that I want Alexandroff extension of locally compact Hausdorff space to be Hausdorff, and for this whole construction to be a functor. Can we then prove some kind of uniqueness?

Xander Henderson

@Pizza I don't understand.

Jakobian

Then we can proudly say, yes, this is THE one point compactification, rightfully called so

Thorgott

@EE18 indeed

@AlessandroCodenotti the issue, I think, is that there's no unique such space

Pizza

19:06

for example if I have g(x)=47 for every x in A[0,1]

Balarka Sen

maybe its not true

im not 100% sure

Xander Henderson

The general idea of an extension is that you start with a function $g$, defined on some domain, and then seek to expand the domain to some larger set. So, for example, I could start with the function $g : [0,1] \to \mathbb{R} : x \mapsto 47$, and ask "How can I extend this function so that it is defined on all of $\mathbb{R}$?"

EE18

@Thorgott This is kinda (a lot) shocking but I guess it's true...anyway, back to it :)

Xander Henderson

"Extending the function", in this case, means to find a new function which agrees with $g$ on the set $[0,1]$, but which is defined on all of $\mathbb{R}$. That is, the goal is to find a function $f : \mathbb{R} \to \mathbb{R}$ such that $f(x) = 47$ whenever $x \in [0,1]$.

Pizza

an extension of g on R would be a function f(x) different from g(x) which takes the same values as g in A

?

EE18

19:08

Actually one last question Thorgott...should it be obvious to me how I can construct a function such that the claim is false without AC?

Xander Henderson

@Pizza What do you mean "different from"? $f$ and $g$ are defined on different domains, so they are necessarily different functions.

Thorgott

@Jakobian if $X_+$ is CH and contains $X$ as a subspace with one-point complement, then $X$ is LCH

EE18

That is, I follow in the proof how AC is sufficient to complete the proof...but is there a way to see it's necessary? (I think demonstrating a function F s.t. no such H exists would do this?)

Jakobian

@Thorgott I know, what do you mean?

Pizza

@XanderHenderson yes, but calculated on A they take on the same value, if g(x)=47 in A, and f(x)=47 in A, but outside of A it is different from 47, isn't it an extension of g on f?

Thorgott

19:10

nothing is obvious in the absence of AC and I really suggest not worrying about it

that said, it shouldn't be too bad to show this claim is equivalent to AC

@Jakobian I don't think that leaves any other choice for the topology than the usual one

Balarka Sen

@Thorgott @AlessandroCodenotti Suppose $x, y, z \in V$ are three points such that $|\cdot|$ is linear on $[x, y], [x, z], [y, z]$ *and* $|x + y + z| = |x| + |y| + |z|$. Then
$$|x + y + z| \leq |x + ty + sz +(1 - t)y+ (1 - s)z| \leq |x + ty + sz| + (1-t)|y| + (1 - s)|z| \leq |x| + |y|+|z|$$
Equality forces $|x + ty + sz| = |x| + t |y| +s|z|$. So $\|\cdot\|$ is linear on the simplex $[x, y, z]$.

Thorgott

or am I misunderstanding what kind of uniqueness you want

Jakobian

One point compactification of an LCH space has unique Hausdorff topology, yes

but my concern is, do we need this exact construction for the non-LCH spaces?

can we get away with something else?

or does this force us to use this construction in some way

Balarka Sen

It seems important to have $|x + y + z|= |x| + |y| + |z|$, having barycenter as a reference point

Jakobian

in order for it to be canonical

Balarka Sen

19:12

In fact I don't seem to need linearity on $[x, y], [y, z], [x, z]$ lol

EE18

@Thorgott I will definitely take this advice. I see that later in the book I'm meant to prove equivalence of AC to the form of AC used here so that will be interesting. I just wasn't sure how quick or easily one could show AC was necessary for a given claim (rather than just being sufficient, like when I did a stupid proof yesterday of the existence of a basis for a vector spanned by a finite subset of vectors)

Thorgott

@Jakobian so you're asking if there's another functor from spaces to compact spaces that agrees with Alexandroff on the lch spaces?

Jakobian

Yes, a functor that would be a "compactification by one point"

Thorgott

with one-point remainder, I should add

actually, this is nonsense

cause Alexandroff extension isn't functorial

only wrt "proper" maps

(in the sense of preimage of compact is compact)

Jakobian

I think thats a fine thing to assume

that we aren't taking whole Top category

But I'm not sure if proper maps would be the right choice, because if we use them then it sounds like we're forcing it to be Alexandroff extension

Pizza

19:20

from what you wrote perhaps I understood; I start from a function g(x)
defined on a subset A which takes on certain values; at this point I decide to extend it to a larger domain. I have to find a function f that is defined on a larger set, but which in A always takes on the same values as g. Ex: \begin{cases}
f_1(x) = 47, & \text{if } x \in A \\
f_1(x) = 420, & \text{if } x \notin A
\end{cases} in this case from g(x)=47 I extended g over the whole of R, now g is f

Thorgott

idk, perhaps you can still build something stupid, but I'm not sure

Jakobian

yeah no worries. Its a stupid question

If $X$ is a topological space, and $Z$ its compact extension by one point, then $X\to Z$ factors through $\alpha X$, that is $X\hookrightarrow \alpha X\to Z$, here $\alpha X$ is Alexandroff extension

because Alexandroff extension is actually the largest topology which is a compact extension by one point

so its canonical in this sense at least

Xander Henderson

19:35

@Pizza You should not say "now $g$ is $f$". They are not the same function.

Also, what is $f$? You've defined an $f_1$, but not an $f$...

Pizza

ah

f1=f

Thorgott

@Jakobian oh yeah, nice observation, I think this is probably the "right" answer

Xander Henderson

Assuming that you meant to write $f_1$, then $f_1$ is an extension of $g$, yes.

Pizza

nice

thank you

Xander Henderson

But the question was "are extensions unique?"

Can you answer the question?

Pizza

19:37

no

extensions is not unique

because I could also define the function on a subset of R, in this case the subsets that can be formed from R are infinite. Therefore the extension is not unique

@XanderHenderson right?

Xander Henderson

@Pizza Extensions *are not unique.

That is correct.

But your reasoning is not correct.

Pizza

ah...

Xander Henderson

19:52

It isn't about the subsets on which the functions are defined (though that is certainly a place where non-uniqueness arises). It is about the fact that you can define an extension of $g : A \to B$ however you like for $x \not\in A$.

Pizza

ah

like

\begin{cases}
f_1(x) = 47, & \text{se } x \in A \\
f_1(x) = 420, & \text{se } x \notin A
\end{cases}

\begin{cases}
f_1(x) = 47, & \text{se } x \in A \\
f_1(x) = 69, & \text{se } x \notin A
\end{cases}

Xander Henderson

For example, two possible extensions of $g : [0,1] \to \mathbb{R}$ are $$ f_1(x) = \begin{cases} g(x) & \text{if $x \in [0,1]$, and} \\ x^2 + 17 & \text{otherwise,} \end{cases} $$ and $$ f_2(x) = \begin{cases} g(x) & \text{if $x \in [0,1]$, and} \\ \sin(x) + \mathrm{e}^x & \text{otherwise.} \end{cases} $$

@Pizza Yes.

sharon_puthuparambil

Can someone help me prove a property about lower semicontinuous functions‘

‘

?

Pizza

@XanderHenderson ok , i understand thank you so much

Xander Henderson

Now, what often happens is that you want to extend a function to a larger domain while retaining some useful property. For example, the function $f : (0,1) \to \mathbb{R}: x \mapsto x^2$ can be extended to $[0,1]$ in any number of ways, but only one of those extensions is continuous.

Jakobian

20:06

@sharon_puthuparambil Like I said, you get one inequality for free. The other one you need to show from definition of lower semicontinuity

Pizza

\begin{cases}
f(x) = x^2 & \text{if } x \in (0,1) \\
f(x) = x, & \text{if } x \notin (0,1)
\end{cases}

with this extension it remains continuous in [0,1]???

Xander Henderson

@Pizza I wouldn't write it that way.

Either define the new function by $f(x) = x^2$ on $[0,1]$, or write $$ f(x) = \begin{cases} x^2 & \text{if $x\in[0,1]$,} \\ 0 & \text{if $x=0$, and} \\ 1 & \text{if $x=1$.}\end{cases}$$

sharon_puthuparambil

But how can I construct a sequence x_n such that liminf x_n = sup_r inf_B(x,r) f(y)?

Xander Henderson

But I don't think that the piecewise definition is terribly useful.

Jakobian

@sharon_puthuparambil I don't understand what you're asking

Xander Henderson

20:11

The formula $f(x) = x^2$ is probably the most "intuitive" presentation.

Jakobian

if you want to talk about your question, you need to provide your definition of lower semicontinuity first

also see LaTeX in chat in description of this chat

right upper corner

you just need to add a bookmark

Pizza

@XanderHenderson thanks!

Jakobian

@XanderHenderson someone seems to have a question for the question you closed in CURED

Xander Henderson

@Jakobian I don't understand. If they have a question about something in CURED, they can ping me there...

Jakobian

I thought you only want to be pinged there if it requires moderator attention?

I thought it wasn't a big deal to tell you since you just finished helping Pizza, and are already here

sharon_puthuparambil

20:19

I want to prove that if f is sequentially lower semicontinuous, then f(x) = sup_{r>0}inf_{y in B(x,r)} f(y).
As you said, one inequality if for free, namely the >=.
So I need to prove that f(x) <= sup_{r>0}inf_{y in B(x,r)} f(y).
If I can construct a sequence x_n such that x_n converges to x and liminf f(x_n) = sup_{r>0}inf_{y in B(x,r)} f(y), then the proof is finished since liminf f(x_n) >= f(x)

Xander Henderson

If moderator attention (or my attention, in particular) is not required, then yes, I don't want to be pinged about stuff happening in CURED. But if there is something happening in CURED, and you think that I should know about it, ping me there, not here.

I have no context here for whatever it is you are talking about in CURED.

But if someone has a question about a moderator action, it seems to me that this is something which requires moderator attention...

Jakobian

oh. I see, I didn't take that into consideration in my definition of things requiring moderator attention

I'll consider it next time

@sharon_puthuparambil do you know how to use LaTeX?

which definition of lower semicontinuity are you using? (asking again)

Seems to me that you know how to use LaTeX but perhaps not how to use it in this chat

sharon_puthuparambil

I want to prove that if $f$ is sequentially lower semicontinuous, then $f(x) = \sup_{r>0}\inf_{y\in B(x,r)} f(y)$.
As you said, one inequality if for free, namely the $\geq$.
So I need to prove that $f(x) \leq \sup_{r>0}\inf_{y\in B(x,r)} f(y)$.
If I can construct a sequence $(x_n)_n$ such that $x_n$ converges to $x$ and $\liminf_{n\to\infty} f(x_n) = \sup_{r>0}\inf_{y \in B(x,r)} f(y)$, then the proof is finished since $\liminf_{n\to\infty} f(x_n) \geq f(x)$

Jakobian

I think this is the definition that $f$ is lower semcontinuous at $x$ if for any sequence $x_n\to x$ we have $\liminf_{n\to \infty} f(x_n)\geq f(x)$?

Random Variable

@robjohn Is the MathJax preview working OK in Firefox on your Mac? It's still broken in Firefox on my Windows desktop. A clean install of Firefox didn't fix it.

Jakobian

20:34

Why not for each $n$ take $x_n\in B(x, 1/n)$ such that $f(x_n)-\inf_{y\in B(x, 1/n)} f(y) \leq 1/n$?

assuming the infimum is finite for large $n$

If its infinite for all $n$ then we can easily obtain a contradiction by taking $x_n$ to be such that $f(x_n)\leq -n$ for all $n$

unless, of course, $f$ can take infinite values, which I assume it doesn't

Either way we should have $\lim_{n\to\infty} f(x_n) = \sup_{r > 0} \inf_{y\in B(x, r)} f(y)$

@sharon_puthuparambil

sharon_puthuparambil

Thanks. A costruction based on a similar idea works!!

Someone responded my original post

Have a nice evening

Jakobian

well, my idea is pretty precise and definitely works

its also concise

EE18

23:18

My text asks me to "Show that a finite field cannot be ordered."

They mean that it can't be made into an ordered field right?

cause of course i can define any order on it?

robjohn

@RandomVariable It is okay on my computer, but I have an update of Firefox pending. I will try updating Firefox and see if it still works.

I installed the update, and ChatJax still works.

Oh, you're looking at the MathJax preview. Let me look.

Thorgott

@EE18 yes

ordering a field means making it into an ordered field

EE18

Merci Thorgott :)

oh is that common phrasing?

leslie townes

i have seen it before, i would not adopt it as a model in your own life

great example of math textbooks kinda modeling shitty writing, pardon my french

Thorgott

moreso it's not uncommon, and it's the only thing that makes sense

EE18

23:22

That's very fair and very true

Another question which just came to mind. I know about the ring of formal power series induced by a given ring. Suppose that ring is in fact a field $K$. The corresponding induced ring $K[[X ]]$ is still just a ring in general. I'm sure there is some extension that gets us into a field somehow? What is said extension?

I guess these are the rational functions or something like that?

Joe

One often formally defines a field as a triple $(F,+,\cdot)$ satisfying various axioms, and an ordered field as a quadruple $(F,+,\cdot,\le)$, also satisfying various axioms. To say that a field $(F,+,\cdot)$ can be ordered means that there exists a relation $\le$ on $F$ such that $(F,+,\cdot,\le)$ is an ordered field.

leslie townes

there's some general theory around when you can embed rings in fields

EE18

Got it, so nothing special in terms of the ring of formal power series

OK I will keep an eye out for said theory

leslie townes

you can associate formal series to many rational functions but you will not get all formal series that way

the coefficients of a formal series that come from a rational functions will satisfy relations that are not going to hold for all formal series (which can be defined for any sequences of coefficients whatsoever)

if you work this out over C (i don't know about more generally), when a formal laurent series over C arises from a rational function can be characterized in terms of the rank of an infinite "hankel matrix" associated to the function in a natural way, see en.wikipedia.org/wiki/Hankel_matrix

ooh there is even a MO thread on this

19

Q: When a formal power series is a rational function in disguise

Given a formal power series $f \in k[[X]]$, where $k$ is a commutative field, is there any good way to tell whether or not $f\in k(X)$? Edit: To clarify, "good way to tell" means "computable algorithm to tell". Edit 2: I really screwed up this question, so I am recusing myself from accepting an...

ac.commutative-algebra

robjohn

@RandomVariable the preview window when editing a post works for me.

Jakobian

23:35

@EE18 I love ordered fields

@leslietownes its theory? Doesn't you just need to be an integral domain? I think its a theory for non-commutative fields i.e. division rings, with the Ore condition playing the main role

@EE18 isn't this a field?

Oh no you need to have $a_0\neq 0$ as first term

leslie townes

jakobian: yes, i am engaging in the practice of simplifying what may ultimately become a complicated story (see e.g. "the ore condition") for someone who is self studying and might be encountering the first chunk of that story in a page or two

Jakobian

I'm not really knowledgeable about Ore conditions, if you want you can tell me what it is about

all I know is what I already told you

leslie townes

i don't know anything, i just expect that it's more complicated than "oh, the algebra 1 textbook theorem about embedding integral domains in fields is the final word that anyone could say on this"

Jakobian

ah okay. So we are the same on this one

leslie townes

the way i might say "there are theorems that generalize the textbook FTC" without like, wandering into the nightmare world of those integrals that you like

yes, exactly

Jakobian

23:42

there is something I might add though, is that, from what I know, there is no known criteria for when a ring can be embedded into a division ring

i.e. Ore condition is not needed for it, its just sufficient

I have a question on that note though, functional analysis is often about non-commutative stuff. Do you ever embed rings into division rings there?

leslie townes

when i was in grad school, i really wanted to take a class on the general theory of noncommutative rings, but it was only offered occasionally and never at the right time for me

i also suspected that after lecture 1 it just becomes a trip through dante's inferno with prof. lam in place of virgil

Xander Henderson

@EE18 Typically, when an author writes "a [foo] cannot be made into a [bar]", they mean that a [foo] cannot be made into a [bar] while preserving the relevant structures of the original [foo] so that they are compatible with [bar]. So when an author writes that a finite field cannot be ordered, the implicit (but unwritten) clause is "in a way that is compatible with the field structure / in a way that produces an ordered field".

It is not the best kind of exposition, and I would not encourage you to write things like that yourself, but it is pretty common.

leslie townes

@Jakobian maybe incidentally (it wouldn't surprise me) but i don't know of, like, a general reason why you'd consider that desirable

Xander Henderson

@leslietownes Who plays Beatrice in that story?

Jakobian

I see

leslie townes

23:46

i dunno, hartshorne? eisenbud? some commutative algebra guy that you don't meet until "paradiso"? isn't part of the point of inferno that beatrice isn't there

Xander Henderson

@leslietownes Yeah, but Virgil is still pining for her the entire time.

Jakobian

No one needs non-commutative rings, but... yeah no never mind

Xander Henderson

But I'm pretty sure that Hartshorne is somewhere around the 7th circle.

Thorgott

@EE18 The thing you're thinking of is not the ring $K(X)$ of rational functions, but the ring $K((X))$ of formal Laurent series.

It is the quotient field of $K[[X]]$, which has the very nice property of being what's called a DVR.

Jakobian

You just need $1/x$ I'm pretty sure

Xander Henderson

23:51

@Thorgott So, like, you are save your videos on it?

Thorgott

This, roughly speaking, allows you to write $K((X))=\bigcup_{n\ge0}\frac{1}{X^n}K[[X]]$, the latter being closer to how you typically think of Laurent series.

Jakobian

i.e. the field which contains $K[[x]]$ is obtained by adding an element $y$ such that $xy = 1$

Thorgott

@XanderHenderson it's very handy!

Jakobian

oh okay. Laurent series, right. From right and left

Thorgott

@Jakobian yes, it's local with maximal ideal principally generated by $x$

Jakobian

23:54

$(x)$ is a maximal ideal?

Thorgott

yeah

Jakobian

I mean yeah of course

$K[[x]]/(x) \cong K$ by obvious map

Thorgott

yeah, evaluation at other points isn't well-defined, but evaluation at $0$ is

Xander Henderson

$$\operatorname{obvious} : K[[x]]/(x) \to K : t \mapsto \operatorname{derp}(t). $$

Where $\operatorname{derp}(t) = \operatorname{NoDoi}(t)$.

Jakobian

@Thorgott why does it matter that its a maximal ideal?

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