Mathematics - 2024-09-17

Ben Steffan

12:01 AM

:(

I assume "here" does not refer to Math SE

Sine of the Time

Violence is not the answer, it's the question and the answer is "yes"

Xander Henderson

@BenSteffan I don't work at SE. :)

Ben Steffan

no, but being a mod is like a job something something

Jakobian

Hmm... every open cover has an open refinement of size $\leq \kappa$ iff every open cover has a subcover of size $\leq \kappa$

never thought about this

This also shows how paracompactness is closer to compactness than one thinks

Xander Henderson

Oh, this isn't related to my position on the disciplinary committee... This is related to my role as chair of the Instructional Council. Darn I actually have to organize this shindig... :( :( :(

Jakobian

12:20 AM

And if every open cover has a locally finite subcover, then say $(U_i)$ is an open cover, pick some non-empty $U_j$ and let $V_i = U_i\cup U_j$ then if we take a locally finite subcover of $(V_i)$ then at some $x\in U_j$ the local finiteness shows that the subcover needs to actually be finite. So this is actually the same as compactness

psie

> Given a bounded function $f:[a,b]\to\mathbb R$, let $$H(x)=\lim_{\delta\to0}\sup_{|y-x|\leq\delta}f(y),\quad h(x)=\lim_{\delta\to0}\inf_{|y-x|\leq\delta}f(y).$$Show that $H(x)=h(x)$ iff $f$ is continuous at $x$.

This is an exercise from Folland's book in proving the Lebesgue criterion. I think I know a solution, but I'm more concerned about elementary things. Are $h,H$ both functions from $[a,b]$? Are they well-defined, i.e. how can we know the limits always exist?

I can't visualize the deeper meaning behind $h$ and $H$, what they represent or whatever.

Jakobian

@psie If $x\notin [a, b]$ then you would have to take supremum/infimum of the empty set at some point

Xander Henderson

The limit superior and inferior always exist (though either could be infinite).

Jakobian

the limits are well-defined basically because monotone sequences, but they could be limits in the extended real line

oh sorry, $f$ is bounded

@psie they should represent lower and upper semicontinuity

Xander Henderson

And you can always assume that delta is small enough to live in your interval (as long as you aren't sitting at an end point).

psie

12:29 AM

ok

Xander Henderson

Oh! f is bounded. Easy peasy.

So the limits both exist. They are monotone as delta goes to zero, and bounded by the difference between the to and bottom of f(x).

Jakobian

$H = f$ should be equivalent to $f$ being upper continuous, and $h = f$ should be equivalent to $f$ being lower continuous

or, pointwise if we fix a point. And both imply that $f$ is continuous

Xander Henderson

Yup.

psie

@XanderHenderson I see how $\sup_{|y-x|\leq\delta}f(y)$ is a monotone decreasing function, and so the limit exists because...$f$ is bounded?

Xander Henderson

Yes.

psie

12:35 AM

Cool beans. Fava beans!

Xander Henderson

For any particular choice of $\delta$, then $\sup\{ f(y) : |x-y| < \delta\} < \sup\{ f(x) : x \in [a,b]\}$, and $f$ is bounded, so...

(My mixed notation is GOOD! GOOD, I TELL YOU!)

Also, someone else took a shot at Trump?!

VulpesInculta

2:06 AM

Is there anywhere on the site where I would be able to have someone check my work for answers?

*mistakes

Jakobian

2:17 AM

@VulpesInculta only in this chat

this is not allowed on the q&a system

unless you specify an exact step which needs to be explained, see wiki for the solution-verification tag

Soumik Mukherjee

8:10 AM

Hello

nickbros123

8:33 AM

how do I prove the following identity: if $a,b>0$ and $0<p<1$, then $(a+b)^p \leq a^p +b^p$. My initial attempt was to use concavity of $x \mapsto x^p$, which gives us $tf(x)+(1-t)f(y) \leq f(tx+(1-t)y) \implies ta^p+(1-t)b^p \leq (ta+(1-t)b)^p$ but that seems dead end. Is there a substitution I can do in Holder or Minkowski inequality?

Jakobian

@nickbros123 divide both sides by $b$, consider $f(x) = x^p-(x+1)^p+1$

show that its increasing

nickbros123

8:49 AM

right, got it

Sine of the Time

9:10 AM

or writing $p=1-q$ for $q\in ]0,1[$ you have $(a+b)^p=a(a+b)^{-q}+b(a+b)^{-q}<aa^{-q}+bb^{-q}$

nickbros123

@SineoftheTime neat!

Sine of the Time

:)

one potato two potato

10:17 AM

Interesting that some people study black holes in 5 (or higher) dimensional space

Ben Steffan

@onepotatotwopotato 4-space is hard :(

well at least 4-manifolds are f'ed up

one potato two potato

I'm not sure how high-dimensional study helps the understanding of black holes in dim 4

Ben Steffan

you could hope that the things you learn in higher dimensions somehow generalize down

if you can develop a solid theory of thing in high(er) dimensions you can then study the question of how the theory translates to lower dimensions (and how much of it)

something like that, maybe

one potato two potato

I don't know any single example that kind of "generalization" actually worked

Ben Steffan

10:34 AM

it's not my area of study, so my knowledge is rather limited, but something like $h$-cobordism comes to mind

the topological versions of it that hold in low dimensions, as opposed to the smooth one that only holds in dimensions $\geq 6$ (was it $\geq 5$?)

that's mathematics of course, not physics

the $h$-cobordism theorem was first proved in the smooth setting, and because it's so powerful people thought about how to prove analogues in lower dimensions, and actually succeeded (asterisks apply)

proof techniques for showing that there exist manifolds without smooth structures, and manifolds with more than one smooth structure were also pioneered in dimension 7 and above by Milnor and then studied and transferred downwards

one potato two potato

10:48 AM

Hmm. If that black hole paper I mentioned above actually is for the understanding of 4 dimensions, it sounds convincing but I don't know if that's the case.

That $h$-cobordism, as far as I know is for higher dimensional Poincare conjecture. But regardless of that, it feels to me that $h$-cobordism is for high dimensions and people try to generalize to lower dimensions after knowing the usefulness of that. The main target or purpose of that theory is not for the understanding of the lower dimensional manifold.

Ben Steffan

The $h$-cobordism is not just for the Poincare conjecture

It's very, very useful

people initially started out trying to prove it in all dimensions, but the proof that emerged only works in dimensions $\geq 6$

one potato two potato

The point is the last sentence.

Ben Steffan

yes, but I'm not done

people then thought about how to generalize downwards, because it is so useful

this lead to some results in lower dimensions, even though the conjecture in the smooth category is not true below that dimension

but the point really is that it provides precedent for studying things in higher dimensions

studying thing in large dimensions with the hope of being able to later study how eventual results transfer down is a valid strategy

whether this was the goal or not of the people studying $h$-cobordism initially is irrelevant

all that's important is that this pattern worked, and one can try to emulate it

whether you can justify this approach or not sufficiently or not is ultimately something for the people reading your grant proposal to decide :)

Bubaya

12:23 PM

Hi. I have a statement in mind that goes like "A matrix A (not necessarily square) has linearly independent columns iff A^T A is non-singular". This seems to be wrong, but does someone here recognize it and can tell me what the correct statement is?

Soumik Mukherjee

12:54 PM

@Bubaya Why do you think this is wrong?

Bubaya

@SoumikMukherjee Because with A = (1 \\ i), we get det (A^T A) = 0

Thorgott

is your $A$ supposed to be diagonal? that seems like a wrong computation

Bubaya

@Thorgott No, A is the 2 x 1-matrix with the single column (1 \\ i).

Thorgott

ah, of course

Bubaya

This is a non-zero matrix, so (because it has only one column), it has lin. indep. columns.

Thorgott

1:01 PM

I suspect you're forgetting to conjugate when transposing here

Mad

Is there a short course introduction for Differential geometry for physicists?

like 100 pages or less

Sine of the Time

Ted's book?

Bubaya

@Thorgott Oh, sure. That was stupid. Which raises the question: how to see that the statement (with conjugation) is true?

Or, say, how do I know there's no other field for which I have to do some other operation?

Mad

@SineoftheTime is it suited for physicsts tho?

Sine of the Time

dunno, try to see the topics in the index

PM 2Ring

1:13 PM

@onepotatotwopotato Even in Newtonian physics, it can be instructive to analyze orbits in a higher dimensional space, and with a different parameterization of time. From John Baez's blog, johncarlosbaez.wordpress.com/2015/03/17/…

> You probably [know] that planets go around the sun in elliptical orbits. But do you know why?

In fact, they’re moving in circles in 4 dimensions. But when these circles are projected down to 3-dimensional space, they become ellipses!

Soumik Mukherjee

@Bubaya The statement is not true for finite fields

Bubaya

@SoumikMukherjee Do you know where it holds? Only char = 0?

Soumik Mukherjee

2

A: Why is $A^TA$ invertible if $A$ has independent columns?

I think it need be mentioned that we deal with the real field, as the mentioned both facts may not be true for arbitrary field $F$. Example 1. Let $F=\mathbb Z_5$ and $A=\left[\begin{smallmatrix} 1 & 3\\ 2 & 1\\ 0 & 1 \end{smallmatrix}\right]$. Then $A$ has two independent columns, but $A^T A=\l...

@Bubaya it's true for reals, and for complex I think you have to take the adjoint

Bubaya

@SoumikMukherjee Thank you very much for the link; that helps. Do you know of any "similar" statement that works over arbitrary fields?

Of course, with additional hypotheses then and/or weaker statements.

Soumik Mukherjee

@Bubaya No, I don't know any

Bubaya

1:29 PM

OK, thank you.

Xander Henderson

@PM2Ring Oh, hey! That dude! He was on my PhD committee!

PM 2Ring

1:44 PM

@XanderHenderson Wow! He's currently in Edinburgh, living in Maxwell's house.

Thorgott

1:55 PM

There's two ingredients, best highlighted from a multi-linear perspective. If we do this in a coordinate-free manner, the question is when $m$ vectors $v_1,\dotsc,v_m$ in an $n$-dimensional vector space $V$ over a field $F$ are linearly independent. This is the case if and only if the element $v_1\land\dots\land v_m\neq0$ in the $m$-th exterior power of $V$.
Now, if $\langle-,-\rangle$ is a bilinear form on $V$, then $(v_1\land\dotsc\land v_k,w_1\land\dotsc\land w_k)\mapsto\det(\langle v_i,w_j\rangle)_{i,j}$ induces a bilinear form on $\Lambda^kV$. The result of pairing $v_1\land\dots\land …

If you replace "bilinear" with "sesquilinear" and "symmetric" with "conjugate-symmetric", you obtain the complex-flavored version.

Xander Henderson

@PM2Ring Good for him.

Thorgott

oh, I forgot to mention that the Gramian determinant is the same thing as the determinant of $A^TA$ if you take orthonormal coordinates because transposing the matrix corresponds to taking the adjoint

PM 2Ring

I like to read his Mastodon feed from time to time. mathstodon.xyz/@johncarlosbaez

Jakobian

@XanderHenderson good to have such prominent mathematicians as That Guy on your PhD committee

That D. Guy

nickbros123

@PM2Ring John baez was what I used to dream of becoming- a physicist and a mathematician:)

PM 2Ring

2:12 PM

He was also one of the inventors of blogging. Back in the day of Usenet, he had a regular post named This Week's Finds in Mathematical Physics, which combined technical info with a little bit of personal info. math.ucr.edu/home/baez/TWF.html

He's a Stack Exchange member, but doesn't post very often.

Xander Henderson

3:03 PM

@PM2Ring Honestly, I applied to the PhD program at UCR in large part because of his earlier work in math/physics. Then I got there and found out that he was mostly working in category theory, which I don't find to be all that interesting. I attended his seminar for a couple of semester, and got a fair amount out of it (I think), but ultimately ended up working with Michel (the other person I applied to UCR to potentially work with).

John is an extremely charismatic lecturer---I really enjoyed his seminars---but... category theory. Ew. X(

:P

Mad

how would one go around calculating the sigma algebra of $ $ $ E_n \subset \mathcal{P}(\mathbb{N}) $ with
$ E_n :=({2},{4}...,{2n}) $

Ben Steffan

what do you think the algebra should be? :)

Mad

I managed to find the ring
$ R(E_n):= \mathcal{P}(E_n)$

And the Sigma Ring is equal to it.

When i start taking compliments, i get confused.

my guess is $A(E_n ):= \mathcal{P}(\mathbb{N})$

But i can not seem to replicate the set containing {1} so it can not be P(N)

PM 2Ring

3:21 PM

@XanderHenderson Fair enough. I get why some people are excited by category theory, but it definitely doesn't inspire me.

Mad

Def tho $ N, N\{2k}, N\{2,4,...} $ and so on are in the Algebra for sure.

Jakobian

@BenSteffan hmm... anything that deals with multiplication or generalization of such, and talking about things relating to it. So for example topological vector spaces wouldn't talk about operations or things relating to it so its not algebra, but linear algebra would

Ben Steffan

haha

since I said "the" algebra clearly we're talking about universal algebra :^)

Mad

@BenSteffan i do not know that term, i know tho the Generated algebra, which is what i am searching for

Ben Steffan

universal algebra is a field of study

but yeah, as you essentially note, you can isolate even numbers, i.e. $\{2n\} \in E_n$ for all $n > 0$. Can you get a set that doesn't contain all odd numbers?

Mad

3:30 PM

n is finite. so it is not all even numbers.

Ben Steffan

wdym

oh, I might have misunderstood the problem, nvm

Mad

the set $ E_n$ extends to some n, lets say $n=2$ so $E_2:=\{\{2\}\{4\}\}$
so you have in the generated algebra $ N\setminus 2, N\setminus 4, N\setminus {2,4}$

Also you have $ N, \emptyset $. But for example you do not (seem to atleast)
get $\{1\}$

Ben Steffan

uhh

I don't think you get $N \setminus \{2\}$ or $N \setminus \{4\}$

Mad

But the ring as i mentioned (maybe i got it wrong) contains $\{2\}$ the compliment would be that?

Ben Steffan

about the ring thing I have no idea, but you start out with $\emptyset, \mathbb{N}, \{2, 4\}$

(also please be a bit more careful with notation)

what else can you get? $\mathbb{N} \setminus \{2, 4\}$

That's all.

Mad

3:36 PM

Sorry i might have wrote it badly
$E_n$ contains singletons...

Ben Steffan

Ah, okay

So you start out with $\emptyset, \mathbb{N}, \{2\}, \{4\}$, gotcha

Mad

No you start with only the set containing 2 and four.
when you build the generated ring of those, you eventually get the empty set. i am not sure where you get N from?

Ben Steffan

????

Mad

Your inital set is :
$E_n := \{\{2\}, \cdots \{2n\} \}$

Ben Steffan

You're trying to a calculate a sigma algebra on $\mathbb{N}$. This is given as a family of subsets of $\mathbb{N}$

$\emptyset$ and $\mathbb{N}$ you always get for free

2 mins ago, by Ben Steffan

So you start out with $\emptyset, \mathbb{N}, \{2\}, \{4\}$, gotcha

so this is correct

Mad

3:39 PM

I understand what you are saying, i was trying to be more methodical and calculate the ring, sigma ring, the algebra, then the sigma algebra.

You jumped to the sigma algebra. which is ok!

Ben Steffan

oh, is that the distinction?

Mad

No i am just trying to be more methodical. But let us continue! Yes you start with those in the sigma Algebra.

Ben Steffan

so you start out as above, and take all possible unions; this gives you $\{2, 4\}$ as an additional set

what about complements? well, that's easy

you get $\mathbb{N}$ sans $\{2\}$, $\{4\}$, and $\{2, 4\}$, respectively

Mad

Yes and the whole set minus these elements or sets respectively.

Ben Steffan

@Mad that doesn't give a set we don't have already

Mad

3:45 PM

The compliments of $\{2\} \cdots$ and so on...

Ben Steffan

that is in fact what I wrote

when I wrote $\mathbb{N}$ sans ...

at least what I meant :)

Mad

i am not sure what SANS means lol

Ben Steffan

"without"

Mad

Oh ok.

Ben Steffan

anyways, that's all

there's no more sets we can form

Mad

3:46 PM

Interesting. so you do not get the power set.

Thanks!

Ben Steffan

you're welcome :)

monoidaltransform

4:02 PM

If $(X,A)$ is space , $A$ closed in $X$, such that for some $U$ open in $X$, with $A\subseteq U$, $(U,A)$ is an NDR pair, does it follow that $(X,A)$ is an NDR pair?

Jakobian

@monoidaltransform If $X$ is normal, then I think so

monoidaltransform

How to show this?

Jakobian

4:17 PM

I didn't say that is true, but that it might be true if $X$ is normal

my intuition tells me that normality and tube lemma might be important

Ben Steffan

yeah, this should be false in general

monoidaltransform

Im wondering if $X$ is metric

if this holds

Ben Steffan

what, as a consequence?

Mad

@BenSteffan
in this post, they have a similiar problem
https://math.stackexchange.com/questions/1248973/of-which-sets-does-the-sigma-algebra-generated-by-the-first-n-singletons-of?rq=1

if we apply his argument to our case, we would have
$A_\sigma (E_2 )= P(\{2\}) \cup P^c(\{2\}$
But $P^c(\{2\}= P(N\setminus \{2\})$
And that would contain elements like $\{1\}$
Which we shown to not be part of it.

monoidaltransform

If $(X,A)$ pair, $X$ metric, $A$ closed, $A\subseteq U$, $U$ open, if $(U,A)$ NDR is $(X,A)$ NDR?

Ben Steffan

4:25 PM

@Mad no, read the definition of $P^c$ again.

it's not the powerset of the complement

Mad

@BenSteffan Ahh i see! i missed that!

Ben Steffan

the calculation in the post agrees with what I did above :)

VulpesInculta

Anyone willing to help me with some trigonometry I am working on? One question I am having some trouble with & a few others I am not overly confident in my answers.

Mad

@BenSteffan can you also help me understand this short proof, while we are still on the subject?
https://math.stackexchange.com/questions/1978907/about-the-sigma-algebra-generated-by-n-first-integers?rq=1

in this question, they look at the sigma algebra generated by the first n elements.

i understand the authors idea, but the Answer is confusing me. i am not sure how it follows from that, that the union is not a sigma algebra

Ben Steffan

It took me a while to parse what's going on, too :)

What the answer is showing is that the infinite sets $\mathcal{A}_k$ are of the form stated in the question

so complements of finite sets

Thorgott

4:39 PM

@monoidaltransform If $U$ is a 'halo' around $A$, i.e. there is a continuous function $\tau\colon X\rightarrow[0,1]$ s.t. $\tau\vert_A=0$ and $\tau\vert_{X-U}=0$, then $(U,A)$ is an NDR-pair iff $(X,A)$ is an NDR-pair. This neither requires you to assume that $A$ is closed or $U$ is open, but if you do assume those, then $U$ being a halo around $A$ is automatic as soon as $X$ is a normal space, as Jakobian intuited.

Ben Steffan

you can do this by observing that you know sigma algebras $\mathcal{B}_k$ containing the $\mathcal{A}_k$ for which this observation is true

monoidaltransform

@Thorgott doesn't $\tau \equiv 0$ work?

should one of those restrictions be $1$?

Thorgott

ah, typo, it should be $\tau\vert_A=1$

monoidaltransform

thanks @Thorgott !

ill try to prove this

Ben Steffan

and then the idea from the question goes through. $2\mathbb{N}$ is an infinite set that is not the complement of a finite subset of $\mathbb{N}$, and if $\bigcup_k \mathcal{A}_k$ was a sigma algebra then it would need to contain $2\mathbb{N}$ (in fact, the smallest sigma algebra containing all the $\mathcal{A}_k$ is $\mathcal{P}(\mathbb{N})$ since it contains all singletons), contradiction.

monoidaltransform

4:55 PM

@Thorgott how do you extend the homotopy to all of $X$?

VulpesInculta

I have a 2 part question where I need to find the number of solutions of an SSA triangle. Then in the next step determine the measure(s) of angle B. i am given side a, side b, & angle A. Would the measure(s) of angle B not have to be determined to find the number of solutions?

Mad

@BenSteffan and why would this be a contradiction?

monoidaltransform

@Thorgott $H'(x,t)=H(x,t\tau(x))$?

Ben Steffan

see what I wrote a few messages below that

Mad

5:13 PM

@BenSteffan i see!

Thorgott

@monoidaltransform Its not so clear to me if you can give an explicit argument

Mad

@BenSteffan do you have a hint on why it needs to contain the 2N?

@BenSteffan oh because it is an infinite union.

Thorgott

ah wait, you can

recall that a closed subspace $A$ of a space $X$ is cofibered in $X$ iff there is a continuous function $\varphi\colon X\rightarrow I$ s.t. $A=\varphi^{-1}(0)$ and $A$ is an ambient deformation retract of $\varphi^{-1}([0,1))$

so in the previous scenario, let $\tau\colon X\rightarrow I$ witness $U$ as a halo of $A$ and $\varphi\colon U\rightarrow I$ witness $A$ as cofibered in $U$, then $\psi\colon=\max(1-\tau,\varphi)\colon X\rightarrow I$ is continuous and witnesses $A$ as cofibered in $X$

monoidaltransform

5:37 PM

In terms of the homotopy $H:U\times I\rightarrow U$ rel $A$, where $A=\phi^{-1}(0)$, how is the extended $H$ related to $\psi$?

@Thorgott

Ben Steffan

@Mad yeah, if you have all singletons then you have all sets since $\mathbb{N}$ is countable, and sigma algebras are closed under countable unions

Mad

@BenSteffan Yea! i can make the same argument for my case.

My brain has a hard time accepting this proof, because the process of going to infinty messes me up.>
So in the countable union, you do have N or 2N in this case.
but then it needs to spesifically be part of one those sets, which is not.

somehow my brain rejects this proof. But whatever.

Ben Steffan

give it a good night's rest and come back to it tomorrow

that sometimes helps

Mad

Thats true, but honestly, the concept of going to infintity never computed in my brain.

But yet again, my brain has no obligation to be able to understand such concepts at all.

Thorgott

5:53 PM

@monoidaltransform there is no extension necessary, that is the point of my characterization

Thorgott

6:03 PM

when I say "ambient deformation retract" in my message, I mean a homotopy $\varphi^{-1}([0,1))\times I\rightarrow X$ rel $A$ from the inclusion to a retraction onto $A$

the way you construct from that in turn a homotopy as in the definition of an NDR-pair is by piecewise-linearly reparametrizing the second coordinate so that the homotopy is unchanged for points with small $\phi$-value and constant the identity for points with large $\phi$-value

Obliv

8:41 PM

I have a problem: "Prove rigorously that $\bigcap\limits_{n=1}^{\infty}(0,1/n)=\varnothing.$ Hint: Use the Archimedean property." But I don't see why I'd need the archimedean property for this proof

oh nvm, I think I'm interpreting this wrong. I thought this was the nested interval property which is the same thing but with closed intervals (and this intersection is nonempty)

we just used axiom of completeness to prove that

Thorgott

8:59 PM

this is essentially the definition of the Archimedean property

Jakobian

Yeah. But if you write it in an equivalent form

for fields that is

there exists rings which obey Archimedean property but this intersection is non-empty

Obliv

wait.. I accidentally proved it's nonempty

I'm sure it's logically flawed somehow though

Ben Steffan

it better be

:-)

Obliv

Recall: The Archimedean property states that for all $\epsilon\in K$, $\exists n \in \mathbb{N}$ such that $n>\epsilon$ ($\mathbb{N}$ is unbounded). It also states that $\forall \epsilon\in K,\exists n \in \mathbb{N}$, $\epsilon>0 \implies \epsilon>\frac{1}{n}$ (rational numbers are dense in $\mathbb{R}$).

Begin proof: Let $K = \bigcap\limits_{n=1}^{\infty}(0,\frac{1}{n})$. We will use induction to prove $K$ is nonempty for all $n \in \mathbb{N}$.

Ben Steffan

stop stop stop

Obliv

9:10 PM

ok

Ben Steffan

there's nothing to induct over!

$K$ does not depend on $n$

so "$K$ is nonempty for all $n \in \mathbb{N}$" does not parse

Jakobian

I suppose you could interpret it as $\mathbb{N}$ being unbounded, but you can also interpret it as $\mathbb{N}$ being cofinal

Obliv

oh

Jakobian

I don't know why $K$ suddenly became the intersection. I thought $K$ was the field?

Obliv

if this were $K = \bigcap\limits_{n=1}^N(0,\frac{1}{n})$ then I could use induction?

Ben Steffan

9:12 PM

yes, you could use induction on $N$

and this $K_N$ would in fact be non-empty for all $N$

Obliv

@Jakobian yes, I should use a different letter for that intersection.

@BenSteffan that's...strange to me but oh well.

Ben Steffan

is it?

in fact, $K_N = (0, 1/N)$

Jakobian

The negation of "there exists $n$ with $x > 1/n$" is that "for all $n$ we have $x\leq 1/n$"

Ben Steffan

think about it: an element of an intersection is an element that is in each of the sets you're taking the intersection over

Obliv

oh I see, I think you should write $K_n$ instead of $K_N$ because in my mind $N \subseteq \mathbb{N}$ which could be $\mathbb{N}$ itself

Jakobian

9:16 PM

An element in $K$ would have to be $\leq 1/n$ for every natural $n$, but we know there exists a natural with $x > 1/n$

Ben Steffan

I think this notation is good exercise for your mental flexibility :-)

also you were the one to suggest $N$, I just put it in the subscript

In any case, you can't have $n$ as it's written right now, because that's already the index variable in the intersection

you could have $K_k = \bigcap_{n = 1}^k (0, 1/n)$, say, or $K_n = \bigcap_{k = 1}^n (0, 1/k)$

but it's all tomato tomato

Obliv

@Jakobian This is not true for some $x \in \mathbb{R}$ though. For example, $x = 2$, then $2\leq 1/1$ is not true.

oh nvm

Ben Steffan

"for every"

Obliv

you have to swap the quantifiers on $x$ too so we get $\neg\forall x \to \exists x$

@jakobian is this right: $$\neg\left(\forall \epsilon\in\mathbb{R},\exists n\in\mathbb{N}\left(\epsilon>0\implies\frac{1}{n}<\epsilon\right)\right)=\exists\epsilon\in\mathbb{R},\forall n\in\mathbb{N}\left(\epsilon<0\implies\frac{1}{n}\geq \epsilon\right)$$

for negation of archimedean property

I should probably keep inequalities pointing in the same direction oh well

gotta run now, bbl

psie

10:45 PM

I'm quoting Folland:

> For example, if $f$ is Riemann integrable on $[0,b]$ for all $b>0$ and Lebesgue integrable on $[0,\infty)$, then $\int_{[0,\infty)}f\,dm=\lim_{b\to\infty}\int_0^b f(x)\,dx$ (by the dominated convergence theorem), but the limit on the right can exist even when $f$ is not integrable (Example: $f=\sum_1^\infty n^{-1}(-1)^n\chi_{(n,n+1]}$.)

Here $m$ is Lebesgue measure and $dx$ denotes Riemann integral. I don't see how $\int_{[0,\infty)}f\,dm=\lim_{b\to\infty}\int_0^b f(x)\,dx$ follows from DCT. Moreover, why is the infinite sum not integrable?

In DCT, we have a sequence, but I don't see any sequence in $\lim_{b\to\infty}\int_0^b f(x)\,dx$.

Ah, ok, I think I understand my first trouble.

$$\int_{[0, \infty)} f \,dm= \lim_{c \to \infty} \int_{[0,\infty)} f \chi_{[0,c]}\,dm = \lim_{c \to \infty}\int_{[0,c]} f \,dm= \lim_{c \to \infty}\int_0^cf(x) \, dx = \int_0^\infty f(x) \, dx.$$

As for $f=\sum_1^\infty n^{-1}(-1)^n\chi_{(n,n+1]}$, my brain's a bit slow today, but is the integral of this function simply $0$?

I don't understand what he means by "...the limit on the right can exist even when $f$ is not integrable."

Ben Steffan

$\int_{[0, \infty)} f^+ dm$ essentially computes the harmonic series I think?

same for $f^-$

because $\int_{(n, n + 1]} n^{-1} (-1)^n d m = (-1)^n/n$

so $\int_{[0, \infty)} f^+ dm = \infty = \int_{[0, \infty)} f^- dm$ and the lebesgue integral fails to materialize

but the limit on the r.h.s computes $\sum_{i = 0}^\infty (-1)/n = \ln(2)$

psie

11:04 PM

@BenSteffan ok, makes sense, thanks. I still do not quite see what $\int_0^b f(x)\,dx$ evaluates to under the $f$ specified. Is it simply $\sum_1^b (-1)^n/n$?

Ben Steffan

if $b$ is an integer, yeah

you can split the integral up into $\int_0^1 f(x) dx + \int_1^2 f(x) dx + \ldots + \int_{b - 1}^b f(x) dx$

which gives $\int_1^2 (-1) dx + \int_2^3 1/2 dx + \ldots + \int_{b - 1}^b (-1)^{b - 1} / (b - 1) dx$

and you're integrating constants over intervals of length 1 so you get exactly the sum you wrote, save for the fact it should end at $b - 1$

psie

ah ok 👍

psie

11:29 PM

@BenSteffan did you know that the harmonic series converges if you remove every number with a digit $9$ in it? :) Here we are not really getting the harmonic series, but a subseries. I still think it diverges, but I'm always cautious due to this fact.

Ben Steffan

oh that's fun, I didn't know that

yeah here you're essentially looking at $\sum_{n = 1}^\infty 1 / (2n)$ for $f^+$, and this diverges since you can just pull the $1 / 2$ out and get the harmonic series, and similar for the negative part

psie

yeah 👍

Jakobian

11:59 PM

@psie You want to check for $\int |f|$. Then its clear this is just $\sum n^{-1}$

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