Mathematics - 2024-10-23

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Thomas Finley

04:31

0

Q: Let $(Y,d_Y) $ be a subspace of a metric space $(X,d).$ If $Y$ is complete then show that $Y$ is closed.

Let $(Y,d_Y) $ be a subspace of a metric space $(X,d).$ If $Y$ is complete then show that $Y$ is closed. My solution goes like this: Let $y$ be a limit point of $Y.$ This means $\exists$ a convergent sequence $(y_n)$ such that $y_n\neq y,\forall n\in \Bbb N$ satisfying $\lim y_n=y\in X.$ This mea...

sequences-and-series solution-verification convergence-divergence metric-spaces complete-spaces

Need some help with this :?)

It's a solution verification question.

2 hours later…

Sahaj

06:25

Hey

does anybody know if I can somehow find a criterion on $k$ so that a prime of form $p\equiv 3\pmod 4$ divides $(4k+3)^5-1$?

Sine of the Time

07:08

user image

these quotes are getting out of hand

Soumik Mukherjee

I once wrote $m^1$ but my $1$ was so small that it looked like $m'$

@Sahaj you have to do case by case probably, for $p=3$, $k\equiv 1\pmod 3$ satisfies it

Debug

08:05

Is this topologists' anonymous?

0

Q: Is it true or not whether the inverse image of a compact set under a *uniformly continuous* map is again compact?

Let $f : X \to Y$ be a uniformly continuous map, where $X,Y$ are topological spaces, each a Hausdorff metric space if that is needed. Let $K \subset Y$ be compact and let $(x_n)_{n\geq 1}$ be a sequence of points in $f^{-1}(K)$. Clearly this means that $f(x_n) \in K$ fora all $n\geq 1$. As $K$ i...

sequences-and-series general-topology continuity compactness uniform-continuity

Rusurano

08:47

Anybody good with probability theory? I've got a quick question I cannot come up on how to solve.

Vladimir Lysikov

08:58

@Rusurano Don't ask to ask, just ask the question

Ryder Rude

09:17

hi

2 hours later…

Madder

11:22

in the definition of a module over an associative Algebra A, it said it is a vectorspace V with a bilinear mapping s.t $A\times V\rightarrow V , (a,v)\mapsto av$.
My question: i do not understand, why the element $av \in V$. As far as i understand, the associative algebra is a ring $R$. And this vectorspace might be defined over some field $k$ that do not share elements with that ring. With $av$ they mean multiplication in this vectorspace right?

Jakobian

11:54

@Rusurano probability theory can mean various things... and people may be good at it to various degrees

Rusurano

@Jakobian math.stackexchange.com/questions/4988559

I've made a table and got an answer of 0.1004...

Jakobian

@Rusurano the probability that a given iPhone breaks is clearly 1/k

Madder

I am sincerly confused. Chatgpt is telling me that av is not scalarmultiplication of the vectorspace, rather something difernet, why not use a*v then to show it is not the same as $\lambda . v$ for scalars over hte field $k$ where this vector space V is defined.

Jakobian

There is no (real-valued) random variables in this problem but you can set something up if you want

Its then possible to phrase it as a sum of random variables with values in {0, 1} so it becomes binomial distribution

Sine of the Time

The real question is: why are you using chatgpt?

Madder

12:06

Well its sometimes helpfull.

think_meaning_builds

chatGPT likes to argue.

With a good textbook, you fill in the arguments.

think_meaning_builds

12:45

> you do not merely read a math textbook – you work through it! The information has to be dug out, not just skimmed from the surface. It is a slow business, and the only good way to understand what the math text is trying to tell you

GPTs skim their answers from the surface of their searches

Jakobian

@think_meaning_builds well... chatGPT is made to seem like it has a lot to say, more like that

but its in fact very shallow

think_meaning_builds

Yup, the variety of voice inflections makes them interesting to listen to.

Jakobian

13:07

@AlessandroCodenotti hey, want to think about something?

Alessandro Codenotti

It depends on what it is

Jakobian

general topology of course

question is like this

Pick a Tychonoff space $X$. I assume that you know that if $X$ is compact then $X$ is totally disconnected iff $\dim X = 0$ where $\dim$ is the Lebesgue covering dimension (i.e. $X$ is strongly zero-dimensional)

Alessandro Codenotti

Sure. Is tychonoff really needed? Hausdorff may be enough iirc

Oh well hausdorff compact is tychonoff, nevermind

Jakobian

well its compact so whatever. I'm just saying this to point out the topic of focus

Alessandro Codenotti

I haven't had much coffee today

Anyway I'm familiar with that, go on

Jakobian

13:20

You might know that $X$ has a zero-dimensional compactification iff $X$ is zero-dimensional as well

(i.e. has basis of clopen sets)

and you might know that $\dim \beta X = \dim X$

Alessandro Codenotti

Yes, all good so far

Jakobian

so clearly if all compactifications of $X$ are zero-dimensional then $\dim X = 0$, but I am wondering, does the converse hold?

if $\text{ind} X = 0$ but $\dim X > 0$ then clearly not, but maybe $\dim X = 0$ is enough

Alessandro Codenotti

Embed $\mathbb Q$ as a dense subset of $[0,1]^2$?

Jakobian

ah right I forgot sorry

this is clearly not enough

Alessandro Codenotti

Or of $[0,1]$ actually

Maybe with some extra conditions?

Jakobian

13:22

but either way, the issue is when will all compactifications of $X$ be zero-dimensional

Alessandro Codenotti

Hmm do you have an example of such a space which is not finite? Does it hold for all discrete spaces?

Jakobian

I was thinking maybe Mrówka spaces?

I think $\mathbb{N}$ is dense in those and they can be made arbitrary dimension

Alessandro Codenotti

Oof I haven't thought about those in a very long time. They are the ones built from almost disjoint families right?

Jakobian

and we can just take Stone-Cech compactification of those

@AlessandroCodenotti yeah

Alessandro Codenotti

Yeah that makes sense

Jakobian

13:29

actually, maybe I'm wrong about this, but doesn't $\dim X < \infty$, at least when using the non-modified dimensions, imply that $X$ is normal?

Alessandro Codenotti

That seems suspicious to me

Jakobian

well, at least for $\dim X = 0$ since this is the same as $\text{Ind} X = 0$

Alessandro Codenotti

But I'm not so familiar with dimension theory beyond metric spaces to be able to confirm or deny on the spot. I'd have to go looking into Pears's book

Jakobian

so its equivalent to, for each disjoint closed $A, B$ where is clopen $U$ with $A\subseteq U, B\subseteq U^c$

so if you take a space $X$, embedd it as a dense subset of a non-normal space $Y$ and take $\beta Y$ then it will be a compactification of $X$ that's not zero-dimensional since $\dim \beta Y = \dim Y > 0$

so such spaces can only embedd densely into normal spaces

@AlessandroCodenotti sure, watch out for different definitions of Lebesgue covering dimension though

Alessandro Codenotti

This is getting to the point where I'd need to spend some time thinking but I have to teach a class soon so I need to prepare for that

Jakobian

13:35

or rather, if $X$ embedds densely in $Y$, then we must have $\dim Y = 0$

ah alright, bye bye

But it does look like progress

@AlessandroCodenotti $\omega_1$

that's for one that isn't compact, any compact zero-dimensional space will do

Cantor set

Tychonoff plank is an example as well

$\Omega = (\omega_1+1)^2\setminus \{(\omega_1, \omega_1)\}$ is another example

maybe the appropriate condition is $X$ is almost compact and $\dim X = 0$

Jakobian

14:04

that's not based on anything other than its a sufficient condition

Vladimir Lysikov

@Madder
There is a lot of pieces and a lot of multiplications here
I assume this is the definition of a module over $A$ where $A$ is an algebra over a field. All vector spaces in this context are over the same base field $k$.
An algebra is both a vector space and a ring. More specifically, we have a bilinear associative multiplication $A \times A \to A$ and an element $1 \in A$ such that $1x = x1 = x$ for all $x$.
A module $M$ is another vector space on which the algebra acts. This means that we can multiply an element of $A$ by an element of $M$ and get another element of $M$. So now we ha…

Thomas Finley

Let $(X,d)$ be a metric space and $S_1,S_2\subseteq X$ such that $S_1\cap S_2=\emptyset.$ Prove that $diam(A\cup B)\leq diam (A)+diam (B).$

I have no idea how to show this.

Thorgott

did $S_1$ and $S_2$ turn into $A$ and $B$

Xander Henderson

@ThomasFinley Well, it isn't true. Consider $A = \{0\}$ and $B = \{1\}$ as subsets of $\mathbb{R}$ with the usual metric. Then $\operatorname{diam}(A) = \operatorname{diam}(B) = 0$, but $\operatorname{diam}(A\cup B)=1$.

Thomas Finley

@Thorgott yes... my bad...

Xander Henderson

14:16

Perhaps you mean $A\cap B \ne \varnothing$?

Thorgott

I agree with Xander

Thomas Finley

@XanderHenderson yes...fixing it... ugh... I made so many typos...

The editing time is over...

Can't edit it now...

Let $(X,d)$ be a metric space and $A,B\subseteq X$ such that $A\cap B\neq\emptyset.$ Prove that $diam(A\cup B)\leq diam (A)+diam (B).$

Xander Henderson

Well, do you have thoughts?

Thomas Finley

@XanderHenderson I think somewhere I should use that $n(A\cup B)=n(A)+n(B)-n(A\cap B)$. But this is just a relation of set cardinalities and in the long run I don't think it's at all related to it...

Xander Henderson

What does this have to do with cardinality?

You need to be measuring distances, not cardinalities. What is the definition of $\DeclareMathOperator{diam}{diam}\diam(X)$?

Thomas Finley

14:23

@XanderHenderson $diam (X)=\sup\{d(x,y): x,y\in X\}$

This is the definition I am using.

My definition also says that a set is bounded if its diameter is finite.

In this case, I think the problem assumes that A,B are bounded, i.e. have a finite diameter.

Xander Henderson

Right. Use that.

What are you trying to show?

Thomas Finley

@XanderHenderson $diam(A\cup B)\leq diam(A)+diam(B)$

Xander Henderson

@ThomasFinley Honestly, that isn't even a hurdle. If either is unbounded, then $A \cup B$ will also be unbounded, and so $\infty = \diam(A\cup B) \le \diam(A) + \diam(B) = \infty$. No worries.

@ThomasFinley No, that is the ultimate statement that you are trying to prove. But what are you actually trying to show? In order for that statement to be true, what do you actually need to demonstrate?

Break it down into smaller steps.

Thomas Finley

@XanderHenderson I just don't have an idea

Maybe assume it's the opposite

Then use the method of contradiction?

Xander Henderson

@ThomasFinley Well, I suggested that you recall the definition of the diameter. So perhaps the first thing you should do is write down the conclusion of the theorem in terms of those definitions.

@ThomasFinley No, don't start by thinking about what method of proof you are going to use. Clearly write down what it is that you actually need to show. Strip back the definitions and break the problem down into smaller pieces. You trying to attack this problem from 10,000 feet up, but you need to actually get down into the weeds and figure out what you are doing, first.

Thomas Finley

14:30

@XanderHenderson $\sup\{d(a,b):a,b\in A\cup B\} \leq \sup\{d(a,a'):a,a'\in A\}+\sup\{d(b,b'):b,b'\in B\}$

Xander Henderson

Okay! That's a start.

So, maybe start with some $a,b\in A\cup B$. Can you bound the distance between them in any way?

Hint 1: the fact that $A \cap B \ne \varnothing$ is important here.

Hint 2: Triangle Inequality.

Thomas Finley

@XanderHenderson $d(a,b)\leq d(a,i)+d(i,b)$ where $i\in A\cap B.$

Xander Henderson

Okay. Then what?

Thomas Finley

@XanderHenderson $d(a,i)+d(i,b)\leq diam A+diam B$

Xander Henderson

Are you sure? Aren't you making some additional assumptions about $a$ and $b$ here?

Thorgott

14:35

@XanderHenderson the secret to every analysis problem

Thomas Finley

@XanderHenderson It seems that I am making no additional assumptions till now. I think I am wrong tho.

Xander Henderson

@Thorgott Indeed.

@ThomasFinley You definitely are. Why are you able to bound the right side by the left side?

I believe that you are assuming that $d(a,i) \le \diam(A)$. Why is this true?

Thomas Finley

@XanderHenderson Yes, $d(a,i)\leq diam (A)$ as diam(A) is just the (least ) upper bound of all the d(a,a')s where $a,a'\in A$. Isn't it?

and $i\in A$ as well

Xander Henderson

Well, you are making an assumption about $a$, no?

In particular, you are assuming that $a \in A$. But the only thing you said initially is that $a, b \in A\cup B$. Why do you get to assume that $a \in A$?

Thomas Finley

@XanderHenderson Ohh.. I see... yes, I assumed $a\in A.$

@XanderHenderson yeah... I totally missed it.

Xander Henderson

14:38

So you actually have three cases to cover: $a, b \in A$, $a,b\in B$, and $a\in A$, $b\in B$.

The first two are kind of dull, but you still need to do something with them.

Thomas Finley

@XanderHenderson of which the last case is solved just now, as it seems.

If $a,b\in A$ then $d(a,i)\leq diam (A).$ But then how to show that $d(a,i)+d(b,i)\leq diam(A)+diam(B).$

I think this is the most challenging part

Thorgott

(also, why is the case $b\in A,a\in B$ redundant?)

Xander Henderson

Not quite. If $a,b\in A$, then $d(a,b) \le \diam(A) \le \diam(A) + \diam(B)$ (since the diameter of a set is always going to be nonnegative).

When you introduce $i$, as you have, the best you can say is that $d(a,b) \le d(a,i) + d(i,b) \le \diam(A) + \diam(A) = 2 \diam(A)$.

Why introduce an extra point if you don't have to? Simplify your life...

Thomas Finley

@XanderHenderson good gracious! That was such a smart trick.

@XanderHenderson I made things too complicated.

Xander Henderson

@ThomasFinley There are two very important tricks that every analyst should know: add zero, and multiply by one. Nearly every analysis proof ultimately devolves to at least one of those and the triangle inequality.

Thomas Finley

14:45

@XanderHenderson You know I just wrote it in my notepad so that I don't forget it. Yeah, I have experienced these are often very very handy...

The case for $a\in A,b\in B$ is just what I was doing earlier, i.e. $d(a,b)\leq d(a,i)+d(i,b)\leq diam(A)+diam(B).$

So in all the cases, I have, $d(a,b)\leq diam(A)+diam(B)$

Since, $a,b\in A\cup B$ are arbitrary elements so, $diam(A\cup B)\leq diam(A)+diam(B).$

Thanks, @XanderHenderson ! I remain grateful to you, as always.

Xander Henderson

@ThomasFinley Yes. If $a \in A$ and $b\in B$ are arbitrary, then choose some $x \in A\cap B$. Then, by the triangle inequality, $d(a,b) \le d(a,x) + d(x,b) \le \diam(A) + \diam(B)$. As this is true for any such $a$ and $b$, it follows that $\sup\{ d(a,b) : a\in A, b\in B \} \le \diam(A) + \diam(B)$.

Then, with the two other cases, you are done.

Thomas Finley

15:01

@XanderHenderson I get it. Thanks again!

psie

15:41

I'm reading a passage in Introduction to Topology by Gamelin and Greene. They say a linear operator $T$ is bounded if $Tx$ has uniformly bounded norm as $x$ varies over the unit ball in $X$. What does the word "uniformly" mean here? I know the definition of a bounded linear operator is one for which $\sup\{ \|Tx\|:x\in X,\|x\|\leq 1\}<\infty$, but I don't understand the usage of the word uniformly here. What does it mean?

Sine of the Time

I think this means that the constant you use to bound $\|Tx\|$ does not depend on $x$

psie

@SineoftheTime yeah, that would be reminiscent of uniform convergence, where we find $N$ independent of $x$. Then that word is used in a similar fashion here.

Jakobian

16:03

@psie the word "uniformly" in this case doesn't add anything

psie

yeah, I agree, it feels stupid.

Xander Henderson

16:19

@SineoftheTime Yes, that is what it means.

For any particular $x$, you would expect $Tx$ to be bounded ($Tx$ is some element of $X$, and each element of $X$ has a well-defined norm, and so is, individually, bounded---simply saying $Tx$ is bounded is, possibly, a little ambiguous).

Hence "uniformly" here is acting as a bit of an intensifier to emphasize that the bound should not depend on the choice(s) for $x$.

psie

makes sense 👍

Jakobian

it would also make sense to simply say that $T$ restricted to the unit ball of $X$ is a bounded function

psie

17:29

I'm reading a lemma that all bounded linear operators from $X$ to $Y$ form a normed space, where $X,Y$ are vector spaces. I've seen this in a couple of proofs, where the authors only check closure under scalar multiplication and addition. I understand if these hold for a subset of a vector space, then that makes the subset into a vector space. But in this case, no information of the overlying vector space has been given. I find this sloppy. I'd rather see the axioms being checked.

The norm for the normed space in question is the one I gave above.

Ben Steffan

There's an obvious """overlying""" vector space

psie

I see. I need to read up on it in some other book then. Meeh.

Thorgott

17:44

this is basic linear algebra

Sine of the Time

everything in linear algebra is basic :D

Xander Henderson

@BenSteffan "obvious" X(

(though you are correct)

@SineoftheTime ur basic!

Sine of the Time

No I'm acid

Xander Henderson

Yo mama so basic, she dissolves human flesh!

psie

18:11

@BenSteffan ok, the obvious space is the space of linear transformation from $X$ to $Y$. I'm checking the axioms now one by one, but I can't spot anywhere where we require $X$ to be a vector space. Do you know if this assumption can be omitted? I.e. only require $Y$ to be a vector space?

Pizza

👋

Ben Steffan

well you can't talk about linear maps if $X$ is not a vector space :)

but in general the set of functions $S \to Y$ where $S$ is any set will form a vector space, yes

psie

ah ok 👍 right, a linear transformation is a function between two vector spaces, silly of me

Ben Steffan

your proof will use that if $f, g\colon X \to Y$ are two linear maps, then $(f + g)(v) := f(v) + g(v)$ defines a linear map, and this uses the vector space structure of $X$ (in some sense)

psie

@BenSteffan hmm, in what sense? For example, to verify associativity of addition, we have $(f+g)(v)=f(v)+g(v)=g(v)+f(v)=(g+f)(v)$. I feel like I only used the vector space structure of $Y$ here in the second equality.

Ben Steffan

18:22

that is orthogonal to showing $f + g$ is linear

you need to show e.g. that $(f + g)(v + w) = (f + g)(v) + (f + g)(w)$

for all $v, w \in X$

this is basically trivial, but it uses the structure on $X$

put the other way around, you can find a family $F$ of functions $X \to Y$ such that there are $f, g \in F$ with $f + g \notin F$

psie

@BenSteffan ok, I see. Yeah, of course, closure under addition and scalar multiplication. I was only staring at the (other) axioms and forgot about those two.

2 hours later…

Martin Brandenburg

20:26

also have basic linear algebra question ... if does $\Lambda^2 K^I$ embed into the $\Lambda^2 K^E$, where $E$ runs through all finite subsets of the infinite set $I$?

I already checked injectivity on the pure wedges

Ben Steffan

20:40

What do you mean by $\Lambda^2$?

ModularMindset

I'm assuming it's the wedge product of $K$

Ben Steffan

Then I don't understand the problem

$|I| = |E|$, so $K^I \cong K^E$, so $\Lambda^2 K^I \cong \Lambda^2 K^E$

Martin Brandenburg

$I$ is an infinite set. $E$ runs through all finite subsets of $I$.

I consider the canonical map $\Lambda^2 K^I \to \prod_{E} \Lambda^2 K^E$. Is it injective?

$\Lambda^2 V$ is the 2nd exterior power of $V$.

Notice that $K^I$ is an infinite-dimensional vector space.

(Forget to find a basis of it ... let alone of the exterior power)

Ok I think it's true

But that entirely obvious

NOT entirely obvious - sry

ok my proof just broke. no idea

Ben Steffan

21:09

it is surprisingly unclear for a statement in vector spaces

whether it is true or not, that is

Martin Brandenburg

maybe I can show my proof for pure wedges

Ben Steffan

isn't it rather clear for pure wedges?

Martin Brandenburg

say we have $v,w \in K^I$ such that $v|_E \wedge w|_E = 0$ for all finite subsets $E$. we claim $v \wedge w = 0$. if $v = 0$ or $w = 0$, that's clear. otherwise, consider only finite subsets that contain indices where they are non-zero (they are cofinal so that will be enough anyway). now, $v|_E \wedge w|_E = 0$ means that the two vectors are linearly dependent, and non-zero as mentioned,

so $v|_E = \lambda_E w|_E$ for a unique (!) scalar $\lambda \in K^{\times}$. if we make $ E$ larger, then the scalar must be the same. so we cover a cofinal set of finite subset and hence all indices and find a scalar $\lambda$ with $v = \lambda w$. Hence, $v \wedge w = 0$.

Ben Steffan

21:28

somebody remind me what the deal with categorical fibrations is

what does being a categorical fibration gain me over being an inner fibration?

Thorgott

an inner fibration between $\infty$-categories is a categorical fibration iff the induced functor on homotopy categories is an isofibration (you can lift isomorphisms along either source or target)

categorical fibrations between general simplicial sets are awkward

Ben Steffan

I only care about $\infty$-categories

ModularMindset

aren't categorical fibrations more useful in ordinary category theory, for ex. fibered categories while inner fibrations extend that concept more naturally to higher category theory which makes it easier to deal with higher dim. morphisms and possibly quasicategories?

Ben Steffan

I'm not interested in 1-categories :^)

Categorial fibrations appear to be of importance in $\infty$-category land

ModularMindset

Oh gotcha

Thorgott

21:36

they are the fibrations in the Joyal model structure on sSet

Ben Steffan

yeah

that much I know, but I'm still not sure what this gives me in terms of theory

Thorgott

categorical fibration + categorical equivalence => trivial fibration

I use this all the time

Ben Steffan

A fibration of $\infty$-operads, say, is a map of $\infty$-operads that's also a categorical fibration, but right now it is mysterious to me why it should be a categorical fibration, as opposed to some other kind of fibration

@Thorgott ah, so a trivial fibration is a trivial fibration in the Joyal model structure

I somehow didn't realize

Thorgott

yup, as well as in the Kan model structure

and they also have the same cofibrations, which means (I think) that one is a Bousfield localization of the other or something

Ben Steffan

according to nlab the Kan model structure is the Bousfield localization of the Joyal one at the outer horn inclusions

which intuitively should make a lot of sense

Thorgott

21:46

yeah that sounds about right

psie

> Two norms $\|\cdot\|_a$ and $\|\cdot\|_b$ on a vector space $X$ are equivalent if there exists a $c>0$ such that $$\frac1c\|x\|_b\leq\|x\|_a\leq c\|x\|_b.$$ Show that the norms $\|\cdot\|_a$ and $\|\cdot\|_b$ are equivalent iff the identity map of $X$ is bicontinuous between the $\|\cdot\|_a$-topology of $X$ and the $\|\cdot\|_b$-topology of $X$.

This exercise confuses me. In particular, which identity map are they talking about, i.e. what does it input and output? Since they mention different topologies, I'm confused about the identity map they are talking about.

Ben Steffan

They are talking about the map $\mathrm{id}_X\colon X \to X$

considered as a map of sets

psie

@BenSteffan ah ok, so it does not take vectors as inputs

Ben Steffan

???

It takes elements of $X$ and maps them to themselves

The elements of $X$ happen to be vectors here

psie

what does map of sets mean?

Ben Steffan

21:52

function

just a good ol' function :)

Sine of the Time

everything is a vector, if you have enough fantasy

psie

@BenSteffan so I need to show, for $\mathrm{id}_X: (X,\|\cdot\|_a) \to (X,\|\cdot\|_b)$, that $\mathrm{id}_X(A)$ is open for any open $A\in (X,\|\cdot\|_a)$ and likewise that $\mathrm{id}_X^{-1}(B)$ is open where $B\in (X,\|\cdot\|_b)$, or?

Ben Steffan

yes

psie

ok 👍

Ben Steffan

Phrased differently, you need to show that $(X, \lVert \cdot\rVert_a)$ and $(X, \lVert \cdot\rVert_b)$ have the same open sets

Sine of the Time

22:05

equivalent norms induce the same topology

Ben Steffan

@SineoftheTime that is in fact what is to be shown here, yes :)

Sine of the Time

yep :)

psie

isn't the identity map a bounded linear operator hence it is always continuous, and the function equals its inverse, so the inverse is also continuous?

Ben Steffan

in a word, no

at least not a priori

Sine of the Time

the identity is not always continuous

Ben Steffan

22:14

nor is it always a bounded operator

although in this case... etc.

psie

@BenSteffan isn't $\|I\|=\sup\{\|Ix\|:x\in X,\|x\|\leq 1\}$ and since $Ix=x$ and $\|x\|\leq 1$, the $\sup$ is no greater than $1$?

Ben Steffan

There's no $\lVert \cdot\rVert$

Claudio

guys, has it ever occurred to you that the mathjax chrome extension stopped working all of a sudden? Because I can't seem to make it work anymore

Ben Steffan

@psie This is true for $I\colon (X, \lVert \cdot\lVert) \to (X, \lVert \cdot\lVert)$. That's not the situation you're in.

psie

ok, hmm

Sine of the Time

22:19

@Claudio never happened

psie

@BenSteffan I guess we are in the situation $\|I\|=\sup\{\|Ix\|_b:x\in X,\|x\|_a \leq 1\}$.

Ben Steffan

yes

Ryder Rude

hi

psie

OK. So if the norms are equivalent, i.e. $\frac1c\|x\|_b\leq\|x\|_a\leq c\|x\|_b$, then the first inequality yields $\|x\|_b\leq c\|x\|_a$ and taking the $\sup$ over all $\|x\|_a\leq 1$ gives that $\|I\|\leq c<\infty$, hence $I$ is continuous as a bounded operator. The inequality $\|x\|_a\leq c\|x\|_b$ likewise shows that $\|I^{-1}\|=\sup\{\|I^{-1}x\|_a:x\in X,\|x\|_b \leq 1\}<\infty$. So I've proved one direction of the exercise (I think). Phew.

ModularMindset

@RyderRude hi

Ryder Rude

22:34

@ModularMindset hi

psie

22:47

@SineoftheTime but $I=I^{-1}$, right? Maybe it's not correct to say that the two are equal since they have different domains and codomains, but can we at least say they have the same operator norm?

Actually wait.

They do have the same domain and codomain.

Ben Steffan

They don't

Not as normed spaces

And they don't necessarily have the same operator norm

but you're not far away from a proof now

hint: show that equivalence of norms is an equivalence relation

Sine of the Time

23:02

@psie consider $C([0,1],\|\cdot\|_{\infty})$ and $C([0,1],\|\cdot\|_1)$. $I:C([0,1],\|\cdot\|_{\infty}) \to C([0,1],\|\cdot\|_1)$ is bounded but $I^{-1}$ is not bounded

ModularMindset

Is it possible for a connection to exhibit a multi-valued nature, where sections change sign as they are transported around a loop?

maybe a piece-wise defined connection?

psie

@SineoftheTime Ok.

I feel though my solution above to one direction is ok. If $I$ and $I^{-1}$ are continuous equivalently bounded, we can use the inequality for bounded operators that states that $\|Tx\|_u\leq \|T\|\|x\|_v$. Then $\|x\|_b=\|Ix\|_b\leq \|I\| \|x\|_a$ and $\|x\|_a=\|I^{-1}x\|_a\leq \|I^{-1}\| \|x\|_b$. But I'm not sure which number of $\|I\|$ and $\|I^{-1}\|$ I put as $c$? I guess the maximum will do.

Ryder Rude

23:18

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