00:00 - 23:0023:00 - 00:00

12:40 AM
@leslietownes are you claiming we can reverse engineer the real numbers
1โ 0

1:19 AM
My brother actually has an innocent client! And was all ready to go to trial today, but got all chargers dismissed at the last minute! Yay!

Raise your hand if you suspect cheating.
Good for the client!

Indeed. Dude was looking at serious jail time.
It has been a good day for him.
But sometimes, justice prevails.

1:40 AM
We should all toast his good fortune with a cocktail.

2:04 AM
๐ธ

๐ธ๐พ๐ฅ

2:21 AM
@TedShifrin quite bold to do it under a real name
my favorite cheating on MSE moment was when someone took an exam with Emily Riehl, tried cheating on it and she herself posted an answer on the MSE question saying "Meet me in my office."

1 hour later…
3:52 AM
I just wrote a totally unnecessarily complicated proof that $\mathbb{Z}$ and $\mathbb{Q}$ are not isomorphic by analyzing their lattices of subgroups.
two, actually

1 hour later…
5:13 AM
They say laymen care about only the mean, while statisticians care about also the standard deviation. But the strange thing is, no one cares about the skewness or the kurtosis.
And I'm in the situation where to care about the skewness.

5:53 AM
Hello everyone, is it appropriate if we use probability formula in order to understand whether a dog breed is likely to attack us? Can we use probability formula instead of crime per capita in order to understand if some place is more dangerous than others?

6:47 AM
No idea.
Btw I have cynophobia so I am afraid of every dog breed equally.

2 hours later…
8:23 AM
hello
must continuous functions map an open ball to an open ball?
they are proving that, if f(X)-->Y is a continuous function between metric spaces, then $f^{-1}(U)$ is open if $U$ is open
one of the steps is to take a point $f(x) \in U$, take an open ball $B_{a}(f(x))$, and argue that $f^{-1}(B_a (f(x))=B_b (x)$ for some $b$
why must the last relation be true
i doubt this is true because continuous functions need not map circles to circles

Most continuous functions don't preserve openness. Counterexample: Constant maps.
That said, what is presented in your exercise is actually the definition of a continuous function.
At least in topology.

8:47 AM
oh. but then why r they trying to prove it? and what about this result theyre using : "for $B_a(f(x))\subset U$, there exists some $b$ such that $f^{-1}(U)\supset f(B_b(x))=B_a(f(x))$
is this result even true
they say it's true for continuous functions
@DannyuNDos what did u mean by constant maps btw

The equality need not to hold.
@RyderRude Just another name for constant functions. They're called maps because they're always continuous.

so like f(X)=5 and we interpret the rhs as a point on a manifold, so that this qualifies as a function between metric spaces?

Yeah.

thanks
or something like f(X)=x such that $x\in Y$ for any other metric space will also do
@DannyuNDos is this equality true for the definition of continuous functions in topology?
If f is continuous and U $\subset$ Y is open and f(x)$\in$ U then there is an
$e$ > 0 such that B$_e$(f(x)) $\subset$ U. By continuity, there is a $d$ > 0 such that f maps
the $d$-ball about x into B$_e$(f(x)). This means that B$_d$(X) $\subset f ^{-1}$(U). This implies
that$f^{-1}(U)$ is open.
the above is the proof
it is from Bredon's topology and geometry
do u think it's correct? @DannyuNDos
especially the part where they say the d-ball in X gets mapped to the e-ball in Y

9:44 AM
i now think the book means $f(B_d(x))\subset B_e(f(x))$ is satisfied for some $d$ instead of the equality. is this correct?
these proofs are hard to read because of the vagueness

10:25 AM
What really holds is the subset relation, namely $f^{-1}(U) \supset f(B_b(x))$.

4 hours later…
2:15 PM
@DannyuNDos are you sure about that

2:38 PM
@SnoopyKid This makes no sense to me.
What is the difference between a "probability formula" and "crime per capita"?
Where does your "probability formula" come from?
You seem to have some confusion between the roles of "probability" and "statistics" in modeling the world.

Hi Jakobian.
Bye Jakobian.

@XanderHenderson it was given to us by chance
@Sahaj bye

2:56 PM
Bye
y
e

3:25 PM
Hi again.

@XanderHenderson someone, who is obviously not a statistician, in a comment section on social media said higher crime per capita didn't tell us about the likelihood if we get robbed or murdered. Something like that. Instead he said we should use probabilities formula like this youtube.com/…
@XanderHenderson does that makes sense to you?

@SnoopyKid No, and I have no interest in watching some rando's YouTube video to figure out what you are on about.

3:41 PM
That person with about 8M subscribers is likely not a random.

Per capita is a Latin phrase literally meaning "by heads" or "for each head", and idiomatically used to mean "per person". The term is used in a wide variety of social sciences and statistical research contexts, including government statistics, economic indicators, and built environment studies. It is commonly used in the field of statistics in place of saying "per person" (although per caput is the Latin for "per head"). It is also used in wills to indicate that each of the named beneficiaries should receive, by devise or bequest, equal shares of the estate. This is in contrast to a per stirpes...
@SnoopyKid

1 hour later…
4:49 PM
@Sahaj From my point of view, they certainly are.

Let $\alpha,\beta \in \mathbb{R}$ and assume that $a_n + a_{n+1} \to \alpha$ and $a_{n+1} + a_{n+2} \to \beta$ both when $n \to +\infty$. Prove that $\alpha=\beta$. My proof: let $\epsilon>0$. By hypothesis there exist $N_\alpha, N_{\beta} \in \mathbb{N}$ such that $n \ge N_\alpha$ implies $|a_n+a_{n+1}-\alpha|<\epsilon/2$ and $n \ge N_\beta$ implies $|a_{n+1}+a_{n+2}-\beta|<\epsilon/2$.
Put $\overline{N}=\max\{N_\alpha,N_\beta\}$, since $\overline{N}+1>\overline{N} \ge N_\alpha$ we have $|a_{\overline{N}+1}+a_{\overline{N}+2}-\alpha|<\epsilon/2$ and so $|\alpha-\beta|=|-a_{\overline{N}+1}-a_{\overline{N}+2}+\alpha+a_{\overline{N}+1}+a_{\overline{N}+2}-\beta| \le |a_{\overline{N}+1}+a_{\overline{N}+2}-\alpha|+|a_{\overline{N}+1}+a_{\overline{N}+2}-\beta|<\epsilon$. Since this holds for each $\epsilon>0$, this implies $\alpha=\beta$.
Is this proof correct?

@XanderHenderson ok its fine if you don't want to watch the video. But is it appropriate to use probability formulas instead of using crime per capita in order to find the likelihood whether we are more likely to experience crime in a town with a high crime rate per capita?

5:17 PM
@XanderHenderson you're not alone, I think so too. Popularity doesn't mean someone has good sources or opinions
It just means appeal, but it can appeal to kids or lie about certain things for all we know

i think you all missed the point
it's not even about what the video said, but about what somebody in the comment section said

@Thorgott which is a comment about what...?

dont know, dont care

could be, so what?

5:22 PM
Sure I've been talking about wrong stuff
But Xander's comment is very reasonable

sure

Comment should relate to video so to give good opinion about the comment you should watch the video

im just saying whether the video is credible or not is besides the question

6:23 PM
in Helpful Commentary, 22 secs ago, by Shaun
Can I get some feedback on this question, please?

6:41 PM
@TedShifrin they're doing it again...

7:05 PM
@SnoopyKid I have no idea. You aren't formulating a question which makes sense to me. Again, you seem to be having some problem with the distinction between probability and statistics.

@Jakobian Who?

Hi @TedShifrin

Howdy, a Balarka.

I'm having framing issues, Ted.

Well, pass from SO to O and your issues will disappear.

7:11 PM
Ah, but I am in SO(2)

Well, whose fault is that? :D

I blame dimension 3

@TedShifrin you know... the one above

I won't be able to help, but what's the question, a Balarka? :)

Given a Legendrian knot $\gamma$ in $(\Bbb R^3, \xi = \ker(dz - y dx))$, the tubular neighborhood theorem a la Gray-Moser states a neighborhood is contactomorphic to $S^1 \times D^2$ with contact form $\sin(2\pi \theta) dx + \cos(2\pi \theta) dy$. However, this distribution twists once (1 time) as one goes around $S^1 \times \{0\}$, whereas the same need not be true for the knot $\gamma$.
What does twisting mean for $\gamma$? Twisting with respect to what? Well, the framing given by choice of a Seifert surface of $\gamma$.

7:25 PM
Your $S^1$ is $\Bbb R/\Bbb Z$, I take it.

Yes, indeed. I guess I am going to hell for denoting $\theta \in [0, 1)$.

So you obviously have to calculate something using $\xi$.
I don't see how you can "see" the Seifert surface and $\xi$ and just declare that there's no twisting.
see $\xi$ ... tee hee.

Typically $\xi$ does twist with respect to the Seifert framing. The number is called the Thurston-Bennequin number.
@TedShifrin Lol

Well, I'm not used to picturing what a Legendrian knot looks like.
But you do seem to have a contradiction if your blatant assertions are right.

I guess not, because the contactomorphism of the tubular neighborhoods may be like $S^1 \times D^2 \to S^1 \times D^2$, $(\theta, z) \mapsto (\theta, e^{i k \theta} z)$.
A nontrivial mapping class, but a diffeomorphism nevertheless

7:29 PM
But that doesn't pull back the contact form to itself.

But it may pullback $\sin(\theta) dx + \cos(\theta) dy$ to $dz - ydx$, where $(x, y, z)$ in the domain are the coordinates coming from restricting from $\Bbb R^3$ and $(x, y, \theta)$ on the range are radial coordinates on the solid torus
This is the point I was getting confused about

@BalarkaSen Should we send in a search party to rescue you?

@XanderHenderson Send them in the clockwise orientation

Oh, now $\theta\in [0,2\pi)$. :P

I overcorrected for my sins
Or rather, for my sines...

7:32 PM
I'm still confused by your $k$-fold twist in the contactomorphism.
Of course, using $x,y$ for two completely different coordinate systems confuses me, too.

Lol
Yeah, that was the source of the problem

Oh, so you have now resolved the problem?

I think so.

So you didn't need me at all.

Not true, it helped to say the problem it out loud.
@Ted: A nice way to draw a Legendrian knot $\gamma \subset (\Bbb R^3, dz - ydx)$ is to project to the $xz$-plane. The $y$-coordinate is determined: $y = dz/dx$ is slope.
You like the $xy$-projection better, I think, because then $z$ is signed area.
But much harder to compute signed area under a complicated curve.

7:41 PM
Yeah, I've even taught the project and use Stokes' stuff numerous times, but that doesn't help me relate to a putative Seifert surface.

A good way to think about the Seifert surface near the knot is: If I push the knot by inward normal to the Seifert surface near the boundary, then the pushed off copy has 0 linking number with the knot.
Whereas pushing it along an oriented normal to the contact planes might give a pushoff with different linking number (this is the Thurston-Bennequin number).
To visualize this in the $xz$-projection: The Seifert pushoff comes from the Seifert algorithm and can be written down by Seifert algo.
The contact plane pushoff is the same as the Reeb pushoff, where Reeb is $\partial/\partial z$ for $dz - ydx$. So just push the knot diagram in the $xz$-plane horizontally up in the $z$-direction, and count crossings.

Hey @Balarka

Hi @Alessandro

Hi, demonic Alessandro.
OK, @Balarka. That sounds good and concrete, if you have the actual knot. So I still don't see an answer to my complaint about the $e^{ik\theta}$ and pullback.

I have a surface with boundary $X$ which is a disk $D$ from which $n$ open disks have been removed (in a nice way, they had disjoint closures before being removed). I have a homeomorphism $f\colon\partial X\to\partial X$. I think I should be able to extend it to $X$, unless it is messing up the orientation of some boundary circle but not all of them. Does this sound reasonable? What are the precise conditions I need on $f$?
Hi @Ted

7:53 PM
Wow, @Alessandro. It's been something like 6 years since you were learning to drive and threatening to run over me.

I got my licence five years ago I think

Can you do this smoothly and work with vector fields?

(why did you remember now?)

Because I don't call you demonic that often.

Maybe? I don't understand smoothness! But $f$ isn't necessarily smooth

7:54 PM
Can you isotop your homeomorphism to the identity in a collar?
Well, I wanted to do the case where $f$ is smooth. I am not good with the strange homeomorphisms you use.

Every homeomorphism of the circle is isotopic to either the identity or the reflection (orientation reversing for the latter).
So it suffices to do the case where $f$ is identity on some of them and refl on the others.

If we reverse orientation, then we'll have to make sure the boundaries of the holes are appropriately symmetric.
Oh, sorry, $X$ has all those boundary components.
So when you say reflection you're already assuming the symmetry.
So, no, total degree should be an obstruction to extending to all $X$, I suspect. The boundary theorem applies in the smooth case.
You need total degree $0$ on the boundary for the map to extend.

Sorry what do you mean with degree here?

I mean smooth degree, but look at induced map on $H_1$ of the boundary.
Something I'm saying is wrong. The identity extends.
Ah, I'm thinking about extending $\partial X\to S^1$ to $X\to S^1$. So what I'm saying here is nonsense.

But that sounds like a good approach. Run the relative homology sequence on $H_2(X, \partial X)$ assuming $f$ extends to $X$. Assume WLOG $f$ doesn't permute the components of $\partial X$, by taking a power of $f$.

8:00 PM
But, still, if $\partial X$ has two components and $f$ is orientation-preserving on one and -reversing on the other, I don't see how it extends.
Oh, I wasn't even thinking about permuting.

Right I think $f$ should be either orientation preserving or reversing on every boundary component. But I don't know if this is sufficient to guarantee that $f$ extends

That suffices, no problem.
You can easily write down an extension for when $f$ = identity or $f$ = reflection on all boundary components.

Is there a global homeomorphism that reduces to that case?

For the latter case, think of a disk with n holes punched out from equally spaced points along the diameter. Then reflect along the diameter.

I know it's true component by component on $\partial X$.
Your result on homeomorphisms of the circle I believe, but how do you apply that in this global setting?

8:04 PM
@BalarkaSen Even if $f$ is permuting the components?

So Balarka took a power of $f$ to get rid of that, but I don't see how it solves the problem.

Right but I want to extend $f$ not a power of it

A permutation of the boundary components (ignoring orientation) can be realized by a braid homeomorphism. There is an orientation-preserving homeomorphism of $S^2$ with $n$ punctured that realizes any permutation of the $n$ punctures you want.
After you have done that, use the flip map along the diameter.
@TedShifrin Right, this reduction was a suggestion for proving the part "Is this condition necessary?".

A braid homeomorphism?

Ah ... some actual topology :)

8:08 PM
@AlessandroCodenotti Say you have $\Bbb R^2$ with $3$ punctures, $(-1, 0)$, $(0, 0)$ and $(0, 1)$. Can you give me a homeomorphism which switches the first two?

One cannot ask a simple concrete question about a simple concrete space without the geometers invoking some five different definitions and two deep theorems
@BalarkaSen Well I'd enclose the first two in a disk, swap them inside it with a homeo that becomes the identity on the boundary

Right. So using that, we can get any two punctures to switch. And your guy also preserves orientations.

@TedShifrin: we've gotten some rain here the last few days. How are things going there?

The symmetric group is generated by transpositions, so all permutations of any number of punctures is realized.

We're getting light rain, @robjohn. I don't think you'll be going up to Mammoth for a while, though.

8:09 PM
Right
Gotta go for dinner, sorry, I'll be back soon

But this preserves orientation. To reverse all of them simultaneously, compose with the flip $\Bbb R^2 \to \Bbb R^2$, $z \mapsto \overline{z}$.

@Alessandro That isn't geometry. It's geometric topology.

@TedShifrin Nah, we only go in the summer time. We're not big skiers.

Topology: a geometric approach
@BalarkaSen sure

So this proves all your suggested homeomorphisms are realized.
As in, it's certainly sufficient.

8:10 PM
@robjohn I saw this morning that they'd closed 395 because of snow blindness.

@TedShifrin On some pass highway north of here, there were winds of 190 mph.

Crazy.
I don't think my car would survive that.

It seems to me it is also clearly necessary, because if $h : X \to X$ is an orientation-preserving homeomorphism, $h$ preserves the orientation of $\partial X$ induced from the boundary orientation on $X$. EDIT: Eh, kind of suspect.

395 north of Bishop
We travel that road on our way to Mammoth in the summer

That's what I was thinking, @robjohn.
@Balarka I have only thought about boundary orientation in the smooth case, but I guess the homology exact sequence of a pair gives you something meaningful in the topological case?

8:16 PM
Yes, that works. Or you can isotope the homeomorphism a little bit to make it smooth, since 2D TOP=DIFF.
I think it's actually OK. I was confused for a bit about the homeomorphism $h$ of $S^1 \times [-1, 1]$ given by $h(x, t) = (x, -t)$.
But that reverses the boundary orientation, too :)
Outward-pointing becomes inward-pointing

The attached is a very nice and useful recursion theorem
to Ted's usual point, I don't want to waste time proving the following to myself but want to be sure it's true: there is nothing special about 0 here right?
That is, If I "started" at some $n_0 \in \mathbb{N}$, the sort of construction above would still pick out a unique function from $\Bbb N_{\ge n_0} \to X$, right?

Well, for me $0\notin\Bbb N$.

Oh, hold on, no, $h(x, t) = (x, -t)$ preserves boundary-orientation.
Outward pointing remains outward pointing after the flip.

Modulo those considerations let's say Ted :)

So what you said is wrong, EE.
You have to use $\Bbb N_{\ge n_0}$, not $\Bbb N$.

8:25 PM
Ah good catch, I've edited that
would you now agree with the claim as stated?

But, yes, then it's correct.
Maybe.

the maybe is ominous

Do you have a particular application in mind?

Application is probably stretching this, but the book uses the above to define $\sum_{k = 0}^n x_k$ rigorously as the value $f(n)$ of the function defined by Prop 5.11 (when $a = x_0$ and $V_n$ is defined by $V_n(y_0,...,y_{n-1}) = y_{n-1} + x_n$)
But they then go on to use notation like $\sum_{k = 1}^n x_k$
As is standard, once they teach you about the notation it's expected that you're able to "interpret" slight variations thereon, so I wanted to be sure I was doing that correctly

No, the index represents how many terms you're adding, not where you start.
You'd better write out your generalization of the result before you misapply it.

8:30 PM
BBL

I would say the appropriate thing is to define $\sum_{k=1}^n x_k$ to be $\sum_{j=0}^{n-1} x_j$ with $j=k-1$.
Take care, @robjohn.

@Ted: The resolution for your $e^{ik\theta}$ question is: the contact tubular neighborhood theorem gives a diffeomorphism $f : \nu(\gamma) \to S^1 \times D^2$ sending $\gamma$ to $S^1 \times \{0\}$, such that $f^*\xi_{\mathrm{ran}} = \xi_{\mathrm{dom}}$, where ram and dom indicate domain and range, respectively.
There is a further identification $g : \nu(\gamma) \to S^1 \times D^2$ sending $\gamma$ to $S^1 \times \{0\}$ and $g(\sigma) = S^1 \times \{1\}$, where $\sigma$ is the Seifert pushoff of $\gamma \subset \Bbb R^3$. Whether or not $f \circ g^{-1}$ is a $k$-fold Dehn twist of $S^1 \times D^2$ depends on how many times the contact planes $\xi_{\mathrm{dom.}}$ twist with respect to the Seifert surface.

Oh, so you're using different contact forms.

Right.

That is confuzling.
When the domain and range are the same space.

8:33 PM
Yep.

But this seems like trying to write down a symplectomorphism that does not preserve area.
I'm suspicious.
How does area smoothly change to $k$ times area?
We're changing integral cohomology classes continuously.

The $k$-fold Dehn twist of $S^1 \times D^2$ is given by cutting it along $\{p\} \times D^2$, twisting $k$ times, and gluing back -- that seems like it's volume-preserving to me.
Not that the contact form records area... $\ker(\alpha)$ is the same as $\ker( c\alpha)$ for any constant $c$. $d\alpha$ is only a conformally well-defined symplectic form on $\xi = \ker \alpha$.

Right, so a contact form is not a symplectic form.
Of course, there is no integral cohomology for a contact form.

Right.

Only for its derivative.
So my intuitive analogy is bunk, of course, but still ...

8:45 PM
@Ted @AlessandroCodenotti Here is the nail on the coffin. Suppose $f : \partial X \to \partial X$ is a homeomorphism, where $\partial X = \sqcup^k S^1$ is a disjoint union of oriented circles. Suppose $f$ extends to $h : X \to X$, where $X = S^2 \setminus \{k \, \mathrm{disks}\}$. Then $h$ also extends to a homeomorphism $H : S^2 \to S^2$. This is because every homeomorphism of the circle extends to a homeomorphism of the disk -- cap off by disks using these extensions.
Fix an orientation on $S^2$. $H$ must either preserve or reverse this orientation. If it preserves orientation, then for any pair of points $x, y \in S^2$ such that $H(x) = y$ and tiny oriented circles drawn around $x$ and $y$, $H$ must preserve the orientation of these circles. Same for if it reverses.
More formally, $H$ is multiplication by $\pm 1$ on the local homology groups $\Bbb Z \cong H_2(S^2, S^2 - x) \to H_2(S^2, S^2 - y) \cong \Bbb Z$, and whether it's $+1$ or $-1$ is independent of $x, y$. Let $x$ and $y$ be the centers of the capped disks. This implies $h$ must preserve all the orientation on the circular components of $\partial X$, or reverse all the orientation on the circular components of $\partial X$.

Ah, the continuous world is easier than the smooth world, in some respects. Easy to glue.

Right. It's also true smoothly but significantly (but not awfully) harder.

@BalarkaSen Ah nice, thanks!

Yeah, you need collars and bumps.

In essence, it is the local degree (at $y$ and its preimage $x$) that we computed. And we argued by invariance of local degree.

8:51 PM
Just to avoid being mistaken for someone who thinks about nice spaces let me say that the context for this question is really a question about the Sierpinski carpet, which reduces to the disk situation above

We would never mistake you for a human, demonic Alessandro.
Was dinner good?

Yeah I never doubted you

It was indeed. My flatmate made some great pizza

I recently had pinsa for the first time. I shall have it again.

Is there a tractable way to find the set of all (as opposed to any single) s-t paths with maximum flow in a flow network?

8:56 PM
I actually never ate it! But it does look good

Just a lighter, crisper dough.
@user10478 Isn't this ultimately a computer search question?

I think it would be hard as a brute force search.
Not sure if a linear programming formulation would return this set, analogous to f'(x) = 0 simultaneously finding all extrema?

@user10478 What if you just run Ford-Fulkerson, and when it terminates, delete the path found, then run it again on the new graph, so on?

No, linear programming, as I recall, if there are ambiguities at any step, you just choose one option and proceed. You could keep track of all those choices and go back and follow up on them.

@BalarkaSen What do you mean by delete the path found? If you delete all edges in the path found then you are surely losing more solutions than just the one you found.

9:02 PM
Fair enough!

Right. I don't know the algorithm Balarka is referring to, but you'd have to remove only the part that followed from a place with options.

Maybe you can take the path found, then set -infty capacity to each edge of the path, run it again
Not each edge, but one edge. Then another edge. etc

@AlessandroCodenotti lol

It's an exponential time algorithm

Anyhow, this sounds like something a graph theory or OR expert would know. I sure am not one.

9:04 PM
@Thorgott I have a reputation

Maybe finding all paths with max flow just is exponential. I don't know that it isn't.

@TedShifrin Just so I understand Ted: when we define $\sum_{k=1}^n x_k$ we have in mind some "underlying sequence" $(x_1,...,x_k,...) \in X^{\mathbb{N}}$ from which we will take elements of the the sum. Then you are saying that I use Prop 5.11 to define $f$ using $a := x_1$ and $V_n(y_0,...,y_{n-1}) = y_{n-1} + x_{n+1}$? And then I define $\sum_{k=1}^n x_k$ as $f(n-1)$ under this definition?

Does the linear program return a path with max flow or just the numeric max flow? I'm a bit confused because an LP should have 0, 1, or infinite solutions, but flow networks can have multiple but finite s-t paths with max flow.

Ford-Fulkerson actually returns a path, if I remember correctly

Yes

9:08 PM
The linear program just returns the value

Ahh okie, that makes sense since there's obviously 1 max flow value.

@EE18 That doesn't look right to me.

Isn't EE18 supposed to be doing analysis? Why are they doing set theory again?

Good question.

I have two books on the go, both analysis! Zorich and Amann Escher. Zorich is into the analysis already, Amann Escher is in a foundations part but gets to analysis later.

9:10 PM
If we can define $x_0+\dots+x_k$ recursively, then certainly we can by reindexing define $x_1+\dots+x_{k+1}$ recursively.

I agree Ted, I am just trying to make the notion of reindexing precise in my head I guess
Certainly when working through physics books I know how to do it mechanically, but want to relate it to the underlying recursive definitions

Just define a different sequence. $z_k = x_{k-1}$ with $k=1,2,\dots$ for $x$ and $k=0,1,\dots$ for $z$.
Luckily, it's time for Ted's lunch.

OK, I will think that through. Thanks as always Ted, and enjoy the lunch:)

Imagine if we had to write a sequence as a function $f : I \to S$ from an indexing set $I$ to a set $S$ every time
Subsequence being a composition $J \to I \to S$ where $J \to I$ is cofinal... eek

monotonically increasing and cofinal*

9:19 PM
@BalarkaSen I didn't know you are such a big fan of nets

@BalarkaSen I ducked out of Halmos before seeing that penultimate word

"Ass" - Xander, probably

cofinality isn't scary

I used to like nets

BTW @TedShifrin, I am going to cease worrying about this too much but this is the "generalization" I wrote down in my own notes after our discussion:

9:22 PM
why used to? They're useful in analysis

I don't do analysis

oof, the above should say $n \in \Bbb{N}_{n \geq n_0}$

Or the kind of analysis I engage is PDE theory and has no need of point-set oddities

what I don't like, though, is people arguing with sequences when they should be using nets

$n\in\mathbb{N}_{n>n_0}$ is, uh, subpar notation, id say
just use $\mathbb{N}_{>n_0}$ for the set of natural numbers $>n_0$

9:24 PM
this happened once on this site with an argument about weak topology on a Banach space

@Thorgott I am just using Ted's notation ;) but how come? Isn't this pretty standard?

You are not using Ted's notation

@BalarkaSen well, maybe you don't need to use concept of weak topology on Banach spaces in your arguments

Ted said $\Bbb N_{\geq n_0}$.

not sure why you mentioned point-set oddities

9:25 PM
Ah I guess my extra $n$ is there

I don't think you fully realize why we use nets

I see
OK, will amend

@Jakobian I have never needed it in too much generality. I use distributions sometimes.
@Jakobian It's possible, but again: I have never needed to realize that.

yeah. Perhaps you used a result that is proved using nets though

I love the nets proof of Tychonoff theorem.

9:27 PM
and you don't even know

But it's been such a long time that I barely remember it

I like the proof using Alexander subbase theorem

I honesty don't even remember the last time I used Tychonoff theorem

@TedShifrin Did...did you misspell pizza or...

like pizza but different

9:31 PM
@Jakobian Well, they are oddities to me. Far out of the realm of the kind of mathematics that I tend to think about.

@BalarkaSen nets aren't exclusive to realm of set-theoretic topology
even something simple like $\lim_{x\to a} f(x)$ is a specific instance of a net

Regardless, it has never really been part of the realm of mathematics I think about.

or Riemann sums
or Rayman sums, if you will
@BalarkaSen you don't think its useful to have a concept that you can build specific types of convergence around?

It could be, I am agnostic. I have never thought of convergence in terms of nets.
Never needed to
Generalities tend to not impress me either.
I'm a cut and paste topologist, you're looking for the wrong place to find approval for nets.

This is just my intuition but, does $f * \delta^{(n)} = f^{(n)}$ actually hold?

9:43 PM
@DannyuNDos Sounds right to me. Why not check it using the definition of distributional derivatives?
For $n = 1$, say.

I've never had a chance to learn about distributions in math classes. Rather, I was introduced to them in an electronic engineering class, namely signals & systems.

The formal definition is not terribly different from what you already know about them
Also, you probably mean $f * \delta^{(n)} = f^{(n)}(0)$, where $\delta$ is the Dirac mass at $0$?

nets are nice to have, but i like open sets
open sets are awesome

@BalarkaSen I mean the convolution by $*$.

I actually prefer arguments using nets than open sets

9:50 PM
I know, @DannyuNDos

Isn't the Dirac delta function the identity element of convolution?

I think so
10

The delta "function" is the multiplicative identity of the convolution algebra. That is, $$\int f(\tau)\delta(t-\tau)d\tau=\int f(t-\tau)\delta(\tau)d\tau=f(t)$$ This is essentially the definition of $\delta$: the distribution with integral $1$ supported only at $0$.

Yes, @DannyuNDos

$(\delta*f)(t) = ... = f(t)$

I think you meant $\int (f \cdot \delta^{(n)}) = f^{(n)}(0)$.

9:54 PM
in some contexts one would hesitate to call it an "identity element" of convolution without first specifying a set on which convolution was a binary operation
maybe not here, but for future reference [the more you know]

@leslietownes The $L^2$ space, I think.

@DannyuNDos no

okay, maybe here :)

@DannyuNDos Ah sure, I agree $f * \delta^{(n)} = f^{(n)}$ is correct. Integrating $f$ against $\delta^{(n)}$ is $f^{(n)}(0)$, upto a sign.

Hmm?

9:56 PM
$\delta$ isn't a function in $L^2$

You can use that to prove $f * \delta^{(n)} = f^{(n)}$ (try writing the formula for convolution)

also, how is convolution a binary operation on L^2
psie isn't here but i'm going to blame wikipedia for all of this

I think I should dislike the $L^2$ space then.

@leslietownes we should blame Ted while we have something we can blame him about

oh definitely, let's drag ted into this

9:58 PM
$L^2$ is a nice space. Not all nice "functions" belong there, but that doesn't mean it's not a nice space.
Product of two $L^2$ functions typically do not belong in $L^2$, but that is OK

to elaborate on my previous point, the phrase "A is an identity element for B" sometimes carries with it both a connotation that A belongs an unstated but implicit set, and that B is a binary operation on that set. which maybe isn't how people always use the phrase, but almost any modern book that discuss binary operations in the abstract will do this and define things like this, and people internalize it, and if you are speaking more loosely you should know
e.g. because "the dirac delta function" is not in L^2(R) (or any L^p space) and "convolution" does not define a binary operation on L^2(R)
i'm willing to believe that a lot of people use "identity element" more loosely, but in any situation where domains of functions matter, you might choose not to do that
see also the woe that wikipedia has brought to psie because of its discussion of "the fourier transform" (also sometimes a term for a unary operation on an unstated but implicit set)
as an exercise, use "the fourier transform" to show that there is no "identity element" for "convolution" on something like L^1(R) :)

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