Mathematics - 2024-11-01

leslie townes

00:01

oh yeah. up to equivalence, exactly 14 of them. of course

you can move between most of them using some pretty natural operations/constructions, but i think its still open whether, if you don't "up to equivalence," you might still only be able to get finitely many norms out of a given norm and those operations.

just very normal stuff that people who aren't serial killers study as something other than an outlet to keep them from serial killing

Ben Steffan

maybe that's why Grothendieck pivoted

leslie townes

somewhere in france there's a house with 14 runaways buried under the driveway

Thorgott

14 is such a funny number for such a result

reminder it's the same as the number of sets you can obtain from a subset of a top. space by closure and complementation, but the two results have no formal analogy :)

Ben Steffan

I thought of that as well haha

I'm still amazed that this maximal number is achieve by a subset of $\mathbb{R}$

Thorgott

00:21

I'd be more surprised if it wasn't achievable, $\mathbb{R}$ is big

but it's not like I ever verified the example or have any intuition for it

Ben Steffan

$\mathbb{R}$ is big, but it's also nice

Thorgott

hmm, I don't feel like maximizing the number is a pathological sort of behavior, but at the same time I lack intuition

psie

I'm studying the famous change of variable formula. The proof is, to say the least, long and convoluted. At one point Folland has proven an inequality for a function $G$ which is a $C^1$ diffeomorphism and then goes that we can simply replace $G$ with $T^{-1}\circ G$, where $T\in GL(n,\mathbb R)$ (the general linear group). I don't know why this is possible, but I suspect that $T^{-1}\circ G$ is also a $C^1$ diffeomorphism. Is that correct?

(A $C^1$ diffeomorphism in this case is a function from $\Omega\subset\mathbb R^n\to\mathbb R^n$ which is $C^1$, injective and the Jacobian is invertible for all $x\in\Omega$.)

Ben Steffan

@Thorgott me too

Thorgott

I wonder if there's any result quantifying 'how many' subsets of $\mathbb{R}$ achieve the maximal amount

Ben Steffan

00:24

(the intuition part)

exciting new research venues

Thorgott

@Jakobian do you happen to know if there's any result on how generic Kuratowski 14-sets (in $\mathbb{R}$) are?

leslie townes

00:52

psie: yes. you should either (a) be able to prove that post-composition with an invertible linear map preserves being "a C^1 diffeomorphism" [if you care about what that is, using whatever definitions you have of those things], or (b) take it as part of an assumed true background that is not the subject of discussion

Ben Steffan

more generally, you should prove that (c) invertible linear maps are $C^1$-diffeos and that (d) the composition of two $C^1$-diffeos is a $C^1$-diffeo :)

leslie townes

the degree to which you care about stuff like this should be a function of what you plan on doing with it. e.g. more, if you plan on using multidimensional changes of variable with maps that might have all sorts of irregularities, or less, if you plan on just understanding the vibe of the theorem in cases that do not put pressure on what exact set of hypotheses you have

ben's point (d) would be a very good exercise if you care

Ben Steffan

neither (c) nor (d) are very difficult; perhaps this is worth pointing out

they are exercises in the definition and basic facts about derivatives

psie

Ok, I guess the chain rule is key here.

And the definition of the inverse of a composition.

Jakobian

02:30

@Thorgott I don't

Jakobian

11:06

Kuratowski 14-set problem is algebraic in nature. There is not much papers in general topology about it

Not only that but the notion of being a generic set is not always a precisely defined concept. This means that such problems don't really have much research going into them

Thorgott

understandable, thanks anyway

psie

12:17

What is the justification behind $y\mapsto D_yG$ being continuous if $G$ is $C^1$? Here $D_yG$ is the Jacobian at $y$. I know a vector-valued function is continuous iff all its entries are. Do we have to identify $\mathbb R^{m\times n}$ with $\mathbb R^{mn}$ (the former being the space of all $m\times n$ matrices and the latter simply $mn$ copies of $\mathbb R$) in order to justify continuity of $y\mapsto D_yG$?

Jakobian

@psie what we "have to" and what we can are totally different matters. But yes, entries of $D_y G$ will be partial derivatives of $G$

psie

@Jakobian Ok. That $\mathbb R^{m\times n}$ and $\mathbb R^{mn}$ are isomorphic seems to be an "advanced" statement (of course, that's subjective), but I feel it is hard to understand, hence why I was asking.

Ben Steffan

Advanced? It's essentially trivial!

It's obvious that these are isomorphic as vector spaces, and you already proved that all norms on finite dimensional spaces are equivalent.

If you even define the topologies here via norms

psie

Ok, I have to simply accept it then. Maybe not advanced, but I'm looking at FIS's Linear Algebra (Friedberg, Insel and Spence) and it requires quite some definitions and theorems to finally get at. In their book, I think it follows as a corollary to the theorem that states if $V,W$ are finite dimensional, then $V$ is isomorphic to $W$ iff $\mathrm{dim}(V)=\mathrm{dim}(W)$.

Ben Steffan

Psie, sit down and write down an isomorphism.

psie

12:29

Ok :)

Vladimir Lysikov

12:40

Even if we use the statement you quoted, it's not "advanced", it's usually the main point of the first chapter on vector spaces: vector spaces have bases, dimension is the size of the basis, vector spaces of the same dimension are isomorphic.
I think you should have basic understanding of this when you study multivariate analysis.

Or rather not "vector spaces of the same dimension are isomorphic" but "a vector space of dimension $n$ is isomorphic to $\mathbb{R}^n$", maybe without saying these words

Xander Henderson

13:22

@VladimirLysikov Vector spaces over the same field of the same dimension are isomorphic.

Vladimir Lysikov

I know :)
This is in responce to above

Thorgott

14:23

even that is kind of overselling the difficulty, you can explicitly write down the isomorphism with ease

Thorgott

15:02

@Ben you really should post your HA questions on MO instead of MSE

leslie townes

15:17

@psie in the case m = n = 2 could you convince yourself that the map sending [[a,b],[c,d]] to [a,b,c,d] is a vector space isomorphism? this is the kind of statement you are talking about. it reflects the fact that the addition and scalar multiplication in both spaces are defined 'entrywise.' i.e. it is right out of definitions, no theory.

theory maybe provides the background thing, that "all that is going on" is dimension counting, which removes the apparent arbitrariness of my choice of how to rewrite a matrix as one row of numbers. the map sending [[a,b],[c,d]] to [a,c,b,d] would also be an isomorphism for example.

Frieren

How can I express in symbols the proposition "each neighborhood of $x$ contains infinite points of the set $E$"? I am working in $\mathbb{R}^n$, so if we denote $B_r(x)$ a neighborhood of $x$ with radius $r>0$, I was thinking about "for each $r>0$ there exists $A_r \subseteq E$ such that $|A_r|=+\infty$ and $A_r \subseteq B_r (x)$". Is this correct?

leslie townes

15:32

frieren: looks OK to me. depending on your purpose you could simplify that a little, e.g. for all r > 0 one has |E intersect B_r(x)|= infty. there is no need to introduce a separate A_r because such a set A_r will exist if and only if the set E intersect B_r(x) has the same property.

but your formulation is also fine, it captures the necessary information.

Ryder Rude

16:19

hi

think_meaning_builds

@leslietownes how's munchkin doing with inequalities?

Jakobian

16:35

@Frieren "neighborhood of x with radius r>0", I don't think such concept exists

Ryder Rude

@think_meaning_builds in 12 angry men, do u think the kid is innocent or guilty

i think the uncertainty is 1 or 2%, which is high

Frieren

16:53

@Jakobian Thanks for answer. Aren't in $\mathbb{R}$^n$ neighborhoods defined as open balls with the Euclidean metric?

Jakobian

@Frieren No, those are only some of the neighborhoods

The definition is that $U$ is a neighborhood of $x$ if $x\in \text{int}(U)$

alternatively we can speak of open neighborhoods at which point the definition reduces to $x\in U$

thats why the proper concept is that of a closed or open ball in a metric space

you want to speak one of those when you discuss $B_r(x)$

nickbros123

@Jakobian i was thinking along the lines of $B_{2^{-1}}(q_n)$ actually but I did not think of the lebesgue measure thing, but that seems way clean approach :thumbs up:

Jakobian

@nickbros123 Do note that if $D\subseteq X$ is dense and $r > 0$ is fixed then $B(x, r)$ for $x\in D$ still cover $X$. The point of this example was to see what happens when $r$ is not fixed

The important thing to notice here is $x\in B(y, r) \iff y\in B(x, r)$

Density says that every neighborhood of $x\in X$ contains an element of $D$. This is the same as, for every $r > 0$ and $x\in X$ there is $y\in D$ with $y\in B(x, r)$, or, $x\in B(y, r)$.

In other words, for every $r > 0$, $\bigcup_{y\in D} B(y, r) = X$

So you can see this is equivalent to density of $D$

This actually makes me curious. Call $D$ strongly dense in $X$ if for every function $r:X\to (0, \infty)$, $\bigcup_{y\in D} B(y, r(y)) = X$. Characterize this in some way

@nickbros123 exercise for you

I already know

think_meaning_builds

17:20

@RyderRude I really couldn't say, since they keep bringing in new facts throughout the movie. What I can be sure of is how biased people can be when they bring in their own personal prejudices into a situation like this.

Kavin Ishwaran

Can anyone suggest a good book which covers singular points (cusp, nodes, conjugate points) in detail?

leslie townes

17:45

@think_meaning_builds not a lot of math these days but she is very sensitive to the idea that someone might be getting better treatment than she is. she came third place in a costume contest yesterday, and boy did i hear about it.

Ben Steffan

18:15

@Thorgott I should, yeah

I guess I feel some anxiety about posting on MO

Thorgott

I did too, but then my stupid question about a Prop. in HTT got 12 upvotes

Ben Steffan

my approach so far has been to post on MSE, wait a day or two, then take it to MO

except that I haven't had to do the latter so far

unless Daniël swings around to answer it tonight I'll bring it to MO

leslie townes

with something like your most recent question, i think the only people on MSE who see it and are capable of responding would also be on MO. i do generally understand wanting to maintain a kind of separation between 'thinking out loud'/MSE and 'serious stuff'/MO, but i would think of that as a self imposed constraint and maybe one at odds with getting a quick answer to questions like that

Thorgott

yeah, all the active MSE users in this tag are also on MO, but you get a lot of additional traffic on the latter

Ben Steffan

fair points, fair points

I'll take it over there in a bit

leslie townes

18:23

also, what's with all the fancy questions. do you think you're better than me?

Ben Steffan

let me put it like this: not understanding difficult mathematics is easy :)

leslie townes

you "higher" people aren't the boss of me, you can say "higher" all you want. you're not. i'm higher.

Thorgott

at our uni, there's a lecture called "higher probability theory", which I've always found really funny

leslie townes

is it a seminar where they debate which, of two events, is the more probable? all of you stoners out there may want to "light up" for this one.

wagering on the outcome of a sporting event is an exercise in higher probability theory

Sine of the Time

19:04

site is in only-read mode but a lot of people are editing/commenting

how is this possible?

Ben Steffan

site is in read-only mode?

leslie townes

that's just for you, sine. the mods got together and decided that enough was enough. :)

more seriously the site seems to be operating normally for me.

Ben Steffan

yeah, for me too

leslie townes

there's this big banner about how "Sine of the Time is gone forever" and "a new era is upon us." i assume that's part of the usual experience

Ben Steffan

maybe it's a (drumroll) sine of the times

Sine of the Time

19:07

Ben Steffan

something strange is going on

remember my profile picture issue from yesterday

Sine of the Time

@BenSteffan yes

Soumik Mukherjee

Site under parasite attack

Sine of the Time

can it be a virus in my computer?

Ben Steffan

a virus that specifically blocks your access to MSE by pretending as if the site was read-only?

I mean technically...

maybe someone is playing a very elaborate prank on you

Sine of the Time

19:12

only Math.SE

Ben Steffan

@SoumikMukherjee this is the day we realize that MSE was just a collective hallucination

Soumik Mukherjee

@SineoftheTime what is the 6th site in the HNQ?

Sine of the Time

other communities work fine

Ben Steffan

you could try asking on meta

leslie townes

maybe there's some kind of ddos protection or something on the other end, where someone within your family of IP addresses is antagonizing MSE and you're caught up in the response.

Sine of the Time

19:14

@SoumikMukherjee this ?

In incognito mode it does not seem read only mode

Soumik Mukherjee

Yes that

@BenSteffan :0

Thorgott

@Ben proposition (I think): if $K\times\Delta^1\rightarrow\mathbf{Cat}_{\infty}$ is a natural transformation of diagrams that is object-wise fully faithful, then the induced functor on limits is fully faithful

Ben Steffan

19:29

sounds reasonable

Thorgott

I think I need this to see a claim that Lurie describes as "follows immediately from the definitions"

Ben Steffan

:^)

Sine of the Time

I've asked on meta

psie

19:51

I'm reading a proof of the change of variables formula for a $C^1$ diffeomorphism $G$ in Folland, that is $$\int_{G(\Omega)}f(x)\,dx=\int_\Omega f\circ G(x)|\det D_xG|\,dx.$$There's a statement I struggle with. Let $Q=\{x:\|x-a\|\leq h\}$, i.e. a cube of side length $2h$ in $\mathbb R^n$ since $\|\cdot\|$ is the $\max$-norm. Divide this cube into subcubes $Q_1,\ldots,Q_N$ of side length at most $\delta>0$ and with centers $x_1,\ldots,x_N$.

Folland derives the following inequality $$m(G(Q))\leq (1+\epsilon)\sum_1^N |\det D_{x_j}G|m(Q_j)$$and then he claims that $\sum_1^N |\det D_{x_j}G|\chi_{Q_j}(x)$, the function whose integral is the sum above, tends uniformly on $Q$ to $|\det D_xG|$ as $\delta\to0$ because of continuity of $D_xG$. This is Theorem 2.47 on page 75. If someone could explain how $\sum_1^N |\det D_{x_j}G|\chi_{Q_j}(x)\to |\det D_xG|$ uniformly as $\delta\to0$, I'd be very very grateful.

psie

20:15

I think it has to do with uniform continuity of $D_xG$ on $Q$, a compact set. This allows us to get a uniform bound on $Q$ of $||\det D_yG|-|\det D_xG||<\epsilon$ for all $x,y\in Q$. And then something something... (I'm thinking)

Thorgott

yeah, that is correct

leslie townes

well, you won't get a bound like that for all x and y in Q. given e, you'll get d so that no matter how x and y are chosen within Q, if |x - y| < d [which is far more restrictive than x and y being in Q], you'll have |det(D_x G) - det(D_y G)| < e. so if the only term mattering to that sum is within delta of x_j, okay yes, please finish that thought.

Thorgott

then use that $x$ is contained in exactly one $Q_i$ and then $|x-x_i|<\delta$

leslie townes

what thorgott is saying.

if you have some development of the riemann integral in R^n that is also very close to a rephrasing of what is driving all of this. i'm not sure if folland bothers with that.

i don't mean to be rude, but many of these questions about folland have something in common, which is folland very much assuming a reader to be OK with relying on this kind of reasoning to not derail some broader point of what he is doing, and it seems like maybe you are still more concerned with fundamentals, in which case i'd recommend R^1 before R^n and something other than folland before folland.

i forget if rudin's PMA takes the riemann integral in R^n seriously, you could very much derive what folland is using from that. i just don't remember if he does that.

folland's introductory section refers to PMA almost as if this stuff should be in there, although i'm not convinced that it actually is.

folland is super slippery about how much linear algebra he is assuming, haha. "a bit of linear algebra - actually, not much beyond the definitions of vector spaces, linear mappings, and determinants." ooookay.

psie

20:34

@leslietownes Folland says the uniform convergence of the function should follow from the continuity of $D_xG$, but we are also using the continuity of the determinant here, right?

PM 2Ring

Jul 9, 2023 at 4:21, by leslie townes

the showstopping number "There's No Trick To This, It's Just This One Little Trick"

Sine of the Time

20:59

@PM2Ring I've solved the issue

Claudio

Let $K \subset R^n$ be a closed set and $f : K \to R$ a continuous function. I wanna show that given a closed set $C \subset R$, its preimage under f ($f^{-1}(C)$) is a closed set as well. Im trying to prove that $R^n-f^{-1}(C)$ is open

I've arrived at the following but can't seem to go on: $$ M:=R^n-f^{-1}(C) \Rightarrow M-K$$ is open since K is closed and $M-K \subseteq M$

that means I can find an open ball of radius $r>0$ around $x^{\ast} \in M-K \subseteq M$

PM 2Ring

@SineoftheTime Oh, good. But I thought you (& future readera) might like a bit of background info on that glitch.

Claudio

but then I realized that maybe I only need to take a point $x \in M$ and not $M-K$ since its image will be $\in C^{c}$ which is open since $C$ is closed

Sine of the Time

@PM2Ring yes, Thank you for the links

the question turned out to be a duplicate, so I deleted it

Claudio

well I'm stuck here

PM 2Ring

21:14

@SineoftheTime Fair enough. Maybe post a comment on one of those Meta Stack Exchange posts, so the devs know it's still happening.

Claudio

maybe I can use the fact that the preimage of $\mathbb{R}-C$ is an open set

Sine of the Time

15

Q: Continuity $\iff$ Preimage is closed whenever set is closed

I'm trying to do the following: Prove that a function $f : S\rightarrow T$ between two topological spaces is continuous iff $f^{-1}(C)$ is closed whenever $C\subset T$ is closed. To show that the closed-set relationship implies continuity, notice that $T-C$ is open. The preimage of this ope...

Claudio

wait: is this correct since $f^{-1}(R-C)+f^{-1}(C) = K$ then $f^{-1}(C)$ must be closed?

Sine of the Time

@PM2Ring ok :)

Claudio

@SineoftheTime I don't think I'm allowed to look at the solution online ahahha

I mean, I could since this is just from the exercise sheet but it feels wrong

@SineoftheTime last reasoning is wrong right?

Ben Steffan

21:23

@Claudio let's not think about "asking people in chatrooms" :)

the last reasoning is indeed wrong

Claudio

oh my bad, completely forgot I won't anymore :P

@BenSteffan My mind is telling me it is correct hahaha

But I know it is not

Ben Steffan

I mean, you haven't really presented an argument as to why what you wrote should imply $f^{-1}(C)$ is closed so

Claudio

I was thinking that its boundary in $K$ is contained in it

Ben Steffan

maybe you were, but you weren't arguing that this is true

Claudio

But the sum of the two must be $K$?

Ben Steffan

21:29

does it? you tell me :)

why form the sum to begin with

Claudio

No, I only know about continuity, the function should be biijective probably

Ben Steffan

why would bijectivity matter

Claudio

the preimage of the codomain is the domain?

no wait that's wrong too

Sine of the Time

$f^{-1}(\Bbb R\setminus C)=K\setminus f^{-1}(C)$

try to see if this is correct in general

Claudio

wait maybe I'm already done: $B:=f^{-1}(\mathbb{R} - C) \subset A:=(\mathbb{R}^n - f^{-1}(C))$ so I can find a ball of radius $\epsilon$ contained in $B \subset A$ and I'm done

@SineoftheTime Oh I didn't see the message, sorry

maybe I should start using $\setminus $ instead of $-$

Sine of the Time

21:41

I've seen authors use both \ and -

Ben Steffan

yeah, they're both fairly common

but using $\setminus$ will lead to fewer temptations to write a $+$ somewhere it shouldn't belong ;)

Claudio

ah ok I thought I was comitting a sacrilegious act for a moment

@BenSteffan hahah, I'll keep using setminus then

@SineoftheTime I feel like your last equality is trivial, but then you would'nt be suggesting me to prove this to be true :)

the answers to this question do in fact say that the equality is true directly from the how a function from $A \to B$ is defined

Ben Steffan

at your level one might still ask for an argument

after all, if this identity is really that trivial, then you shouldn't have struggled to come up with a proof here ;)

Sine of the Time

@Claudio why does it feel trivial? Is it always true that $f^{-1}(A\setminus B)=f^{-1}(A)\setminus f^{-1}(B)$ ?

Claudio

21:58

Oh I see, that's where I need to use the continuity hyp

Ben Steffan

uhh

[doubt]

Sine of the Time

It's not always true for continuous maps

Claudio

I'm loosing track of what I'm trying to show :p

Sine of the Time

@Claudio this question

it solves your problems

Claudio

Oh not continuity, I guess injectivity since the arrows can't mix things from the two sets

Sine of the Time

22:07

what?

Claudio

f must be one-to-one for that equality to be true

if that equality holds, then I can just use this $\text{ open set }=f^{-1}( \mathbb{R} \setminus C)=K \setminus f^{-1}(C) \subset R^{n} \setminus f^{-1}(C) $

Ryder Rude

@think_meaning_builds yes. i think the movie wants to establish the latter

sku

22:51

for set C = {x in R, x < 1/x}, I then assumed x^2 < 1 and came to conclusion -1 < x < 1 and inf(C) = -1. But if I had just kept x < 1/x, then inf(C) does not exist. why this weirdness?

Ben Steffan

$x^2 < 1$ is not equivalent to $x < 1 / x$, so there's no reason to be surprised

also your set $C$ is not actually well-defined, for a start

Sine of the Time

@Claudio If $f:X\to Y$ and $A,B \subseteq Y$, then it's not true that $f^{-1}(A\setminus B)=f^{-1}(A)\setminus f^{-1}(B)$. But if $A=Y$, the equality holds hence $f^{-1}(A\setminus B)=f^{-1}(A)\setminus f^{-1}(B)=X\setminus f^{-1}(B)$

So you're not using continuity or injectivity, this is a general result about maps

sku

darn. True. I need to look at sign of x before multiplying by x on both sides. thanks. may I ask why C is not well defined?

Ben Steffan

what happens if $x = 0$? :)

Sine of the Time

Flashbacks of a conversation of a couple of weeks ago

Ben Steffan

23:11

there was a man who tried to invert 0 once and his ring died. sad story :(

Claudio

23:32

@SineoftheTime I see, in the end I did look at the solution written by my professor: it was much easier: $f^{-1}(C)$ is closed by assumption therefore $f^{-1}(\bar{C})$ is open but $f^{-1}(\bar{C}) = \overline{f^{-1}(C)}$, thus $f^{-1}(C)$ is closed by def

bruh moment

Ben Steffan

...but this uses what sine told you above

Sine of the Time

How is $f^{-1}(C)$ closed by assumption ?

Claudio

you suppose it I guess

Ben Steffan

...that's what you're supposed to show though?

Sure, if I assume what I want to show is true then it will be true :)

Sine of the Time

yeah

Claudio

23:35

lol wait

Sine of the Time

Maybe you're confusing the other direction

Ben Steffan

what you want to say is $f^{-1}(\bar{C})$ is open since $\bar{C}$ is open and take it from there

Claudio

oh yeah

my direction is proved via sequences

yeah the solution uses the other definition of closedness

But I didn't even consider it to begin with

Sine of the Time

it's commonly used in analysis

but yes one usually does not think about it the first times

Claudio

sigh

my brain is like mashed potatoes rn

Sine of the Time

23:43

I'm hungry

Ben Steffan

brain eating day was yesterday I'm afraid

Claudio

lol

Sine of the Time

skull emoji

Claudio

@BenSteffan I had the terrible idea of looking at your profile, I took a quick look at your questions, then I looked at my question, and now I feel even more defeated hahaha

Ben Steffan

haha :P

Dannyu NDos

23:46

There is no uniformly continuous homeomorphism between $\mathbb{R}$ and the open interval $(0,1)$, right?

Ben Steffan

don't, I've been at this whole mathematics thing for quite a while :P

@DannyuNDos what's $I$?

also I've ended up in an area of mathematics that has many big words

Dannyu NDos

@BenSteffan Sorry about that, I didn't mean that.

$I = [0,1]$, so it's not even homeomorphic to the open interval.

Sine of the Time

@BenSteffan you're still in time to leave the dark side :)

Ben Steffan

@SineoftheTime why would I

Sine of the Time

yes

Ben Steffan

23:48

we have unlimited power >:)

Claudio

we as algebraic topologists?

Ben Steffan

homotopy theorists, more specifically

Sine of the Time

7

Q: Uniformly continuous homeomorphism from open set to $\mathbb{R}^n$.

Let $U \subset \mathbb{R}^n$ be an open set. Suppose that the map $h:U \to \mathbb{R}^n$ is a homeomorphism from $U$ onto $\mathbb{R}^n$, which is uniformly continuous. Prove $U = \mathbb{R}^n$. My first attempt guided by intuition was to look at covering $\mathbb{R}^{n}$ by balls of radius $=\f...

multivariable-calculus uniform-continuity

Dannyu NDos

Thanks!

Sine of the Time

:)

Thorgott

23:51

for some reason I read that as "unironically continuous" and chuckled

Dannyu NDos

That said, I just noticed that a homeomorphism being uniformly continuous doesn't suffice. Its inverse must also be uniformly continous.

$\tanh$ is uniform continuous, but $\mathrm{arctanh}$ isn't.

(Suck it, LaTeX.)

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