Mathematics - 2022-06-17 (page 1 of 3)

Ted Shifrin

1:36 AM

Morgue 2.0

Joe Shmo

huh?

I take it youre not a fan

Koro

2:14 AM

X is a complete metric space.

I want to show that the set of all isolated points of X can't be countably infinite.

Suppose on the contrary that {x_i: i in N, i>=1} is the set of all isolated points.

leslie townes

baire category theorem comes to mind.

Koro

I claim that $X\setminus \{x_i: i\in N, i\ge 1\}$ is closed.

@leslietownes By BCT, the set $\cap_{i=1}^\infty X\setminus \{x_i\}$ is open.

Since $X\setminus \{x_i: i\in N, i\ge 1\}=\cap X\setminus \{x_i\}$

It follows that RHS is X (because by BCT, RHS can't be empty) or in general equal a set V that is both closed and open.

This gives contradiction in case X is connected.

I think that there is no contradiction if X is not connected.

leslie townes

seems right. consider Z as a subset of R with the subspace topology.

Koro

because in connected X, there are sets apart from emptyset and the full X which are both open and closed.

nitsua60

Maybe someone can help me find the term (which I suppose exists) that I'm looking for?
It's the idea of a function defined by a small set of algebraic properties of the function. E.g. let f be a function with the properties that (i) f(a+b)=(f(a)+f(b))/(f(-a)+f(-b)) and (ii) f(x)>0 for all real x.
Or e.g. f has the properties that (i) f(ab)=f(a)+f(b) and (ii) there exists b>1: f(b)=1. (Logarithm, IIRC.)

Calvin Khor

2:24 AM

Functional equation

In mathematics, a functional equation is, in the broadest meaning, an equation in which one or several functions appear as unknowns. So, differential equations and integral equations are functional equations. However, a more restricted meaning is often used, where a functional equation is an equation that relates several values of the same function. For example, the logarithm functions are essentially characterized by the logarithmic functional equation log ⁡ ( x y ) = log ⁡ ( x )...

leslie townes

nitsua sometimes these things are called 'functional equations'. the cauchy functional equation is a classic example. things get fairly subtle fairly quickly. no known general theory that i am aware of.

nitsua60

Functional is totally the word I was thinking, but I kept ending up at en.wikipedia.org/wiki/Functional_(mathematics)

leslie townes

calvin beat me to it.

Koro

what do I do with my question? :(

Calvin Khor

i recall seeing books specifically on functional equations but idk how deep the hole is yeah

nitsua60

2:25 AM

Thank you! I'm not crazy!!!

Calvin Khor

@leslietownes by a microsecond

@nitsua60 yw!

nitsua60

(All day I've been going "functional, right?" And then thinking I was wrong.)

Calvin Khor

that's another word that has a million meanings now that i think of it...

leslie townes

yeah, 'functional' is something of an overloaded term. although many functionals do satisfy the cauchy functional equation.

:D

Calvin Khor

does this mean im going to start minting nfts

leslie townes

2:26 AM

nitsua you might still be crazy, just not for the reason that you think there's a math term for this.

you could be crazy for a separate, independent reason.

nitsua60

@leslietownes You mean, like, for thinking that writing a six-page letter to my boss and their boss about understaffing and its knock-on detrimental effects to students might be a good idea?

Yeah, pretty crazy.

leslie townes

calvin a lot of people are minting NFTs on the lesliecoin blockchain and living their dreams. akiva weinberger just bought his third yacht. ted shifrin drives a solid gold car that gets 10 gallons per mile.

nitsua: four pages would have been one thing, but six pages is crazy.

Ted Shifrin

@Koro Huh?

Koro

^in not connected

nitsua60

@leslietownes Sooth. Wisely, wisely.

Koro

2:33 AM

the question was: If X is complete, can the set of all isolated points be countably infinite?

I was trying to prove that it can't be countably infinite.

Joe Shmo

X is what?

an arbitrary metric space?

leslie townes

koro i think you were able to show it under more hypotheses but in general it can happen.

Koro

X is a complete metric space.

Joe Shmo

does an interval with all the integers on the real line work?

leslie townes

15 mins ago, by leslie townes

seems right. consider Z as a subset of R with the subspace topology.

Joe Shmo

2:36 AM

it would have to be a closed interval

leslie townes

we're still assessing that.

Joe Shmo

leslie is on fireeee

Koro

Z is complete.

Joe Shmo

so why not? the interval would have to be closed

Koro

any cauchy sequence in Z is eventually a constant sequence.

Joe Shmo

2:38 AM

so like $[1/2, 3/4] \cup \mathbb Z$

yes it is

Koro

yeah, I think that works :-)

@JoeShmo yes, but it was not relevant here?

Joe Shmo

no, it wasn't :-)

Koro

all points of Z are isolated.

Joe Shmo

yes

Koro

:-)

leslie townes

2:39 AM

sometimes i feel like a point of Z.

Joe Shmo

in fact you only need one integer

well false

Koro

me, a point of Cantor set

Joe Shmo

2 integers

Koro

:D

Joe Shmo

false yet again

all integers

Calvin Khor

2:40 AM

countably many at least.

Joe Shmo

you dont need that interval I gave you. adds nothing

the answer that leslie gave you 15 minutes ago... :-)

Koro

BCT is amazing :-)

Joe Shmo

BC who?

Koro

Baire Category Theorem

leslie townes

you can use it to prove a lot of things.

Joe Shmo

2:46 AM

topology fell out of favor for me a long time ago

I used to love it. then one day I wasn't getting the fix anymore

I started falling for analysis

Calvin Khor

hm im just now realising my cat has attempted to become my co-author probably a couple days ago

leslie townes

my cat has a superpower, she can tell when my wife wakes up. she can't seem to tell when i wake up.

i was up around 5:30 this morning, just sitting in bed. my wife wakes up around 6 and the cat immediately runs into the room and is meowing at her and trilling and winding between her legs.

Joe Shmo

I think the cat likes your wife a little better les

leslie townes

they spend a lot of time together. during work from home she usually sits behind my wife's laptop. you just see paws sticking out from each side.

Joe Shmo

cute

I wake up with my dog sitting on my head every morning. He is not a cat.

leslie townes

2:58 AM

she blocks the fan exit from the laptop so the thing often revs into high gear and sometimes overheats.

Ted Shifrin

3:10 AM

@leslietownes by design

Joe Shmo

must feel cosy

Akiva Weinberger

3:21 AM

@JoeShmo Can't relate

So here's a question

Joe Shmo

to which part?

Akiva Weinberger

Leaving topology for analysis

Maybe one day I'll know enough analysis to like it, but so far I'm pretty meh towards it

leslie townes

you should hear what analysis says about you.

Joe Shmo

harmonic analysis is the way to go

algebra is fun too, the little of it I know

Akiva Weinberger

@JoeShmo Where can I learn it?

Joe Shmo

3:24 AM

harmonic? first Stein, then Katzanelson

leslie townes

yitzhak katznelson has a good book on the classical theory. dover reprint i think there is also a free pdf online.

Joe Shmo

jinx

leslie townes

anyway i somehow have a pdf of the book.

Joe Shmo

its available online

rep theory looks like a looot of fun

but I just dont know when I'll get around to it

leslie townes

fell and doran have a good book on the not necessarily commutative case.

Joe Shmo

3:28 AM

rep theory or harmonic?

leslie townes

representation theory. although some of it is definitely what you might call noncommutative harmonic analysis.

Joe Shmo

sure

leslie townes

mackey had a good book about this stuff too. or set of notes.

Joe Shmo

nice

I actually recently came across a set of notes online I really liked

it was an introduction, rather concise, well written, and well organized

leslie townes

serre's linear representations of finite groups is good. maybe a little too concise, but good.

Joe Shmo

3:31 AM

@AkivaWeinberger basically if I were to do mathematics all over again, the advice I'd be looking for would be: drop everything else, do harmonic analysis. But that's me... if you happen to be a particularly talented topologist I dont want to take that away from the world... :-)

SHASHAANK B.H.

Hello Guys

Joe Shmo

hey!

Calvin Khor

@JoeShmo i want to cite something for the L^p boundedness of CZ operators on say (R/Z)^d. Assuming I didnt just explain how dumb i am, got a suggestion?

leslie townes

cite calderon or zygmund.

Calvin Khor

lol

robjohn

3:38 AM

@JoeShmo I went the other way, though I took it from Stein in person.

@CalvinKhor I think Harmonic Analysis on Euclidean Spaces by Stein might cover that.

Joe Shmo

@CalvinKhor no clue, sorry

leslie townes

i have a set of notes somewhere from the guy who taught me most of what i learned about harmonic analysis that may have that. but i don't think he ever published them.

he was paranoid about putting stuff on the web, i don't know why. it's not like people are combing academic websites for this stuff. or that any of it was new.

Calvin Khor

@robjohn it does in fact. but its more a consequence of the R^n theory + some result that lets you pass to periodic stuff, so not stated cleanly

or maybe i didnt see it. i'll check again!

Joe Shmo

@robjohn very interesting.. any particular problem that took you to the dark side?

robjohn

@CalvinKhor It has been a long time since I went through that stuff, so I might be mixing things up.

Calvin Khor

3:42 AM

@robjohn well i can tell you its in one of the books for sure :) its where i learned it from

robjohn

@JoeShmo the dark side? you mean analysis?

Joe Shmo

:-)

you went from analysis to topology as I understood it. Any particular reason why?

Calvin Khor

@leslietownes funny you say that, lots of now-offline professor websites are backed up on the wayback machine

sometimes but not always with their notes

leslie townes

yes. i think there is a complete archive of quizzes and midterms i wrote 20 years ago.

Calvin Khor

ive got some shit i dont really want to put out but its cuz i think its not good lol

leslie townes

3:45 AM

it's stopped somewhat in recent years. in the mid 2000s a lot of universities began using 'course management software' which keeps that stuff one step away from the public web. before that it was all online.

Koro

leslicoin :-)

Koro

4:40 AM

If a metric space X has a countable base, then why is it true that every open cover of X has countable subcover?

if X is not separable (i.e., X does not have a countable dense subset), then can X have a countable basis?

leslie townes

math.stackexchange.com/questions/967454/… introduces relevant terminology.

Koro

Thanks a lot leslie.

copper.hat

5:36 AM

i guess everyone knows convex now. dearth of questions.

robjohn

6:11 AM

@copper.hat they're just below average functions.

Koro

Let P be the set of all condensation points of an uncountable set E, then P is perfect set.

I can show that P is closed.

robjohn

well, it definitely contains all its limit points.

Koro

For if x is any limit point of P, then for every r>0, B(x,r) contains a $p\in P\setminus \{x\}$ and a sufficiently small ball around p contains uncountable many points of E.

that is, B(x,r) contains uncountable many points of E for every r>0

Hence x is a condensation point of E. So x is in P.

Hence P is closed.

But how do I show that every element of P is a limit point of P?

2

Q: Baby Rudin condensation points

$\newcommand{\seq}[1]{\left\{#1_n\right\}}$ Question 27 says Suppose $E \subset \mathbb{R}^k$, $E$ is uncountable, and let $P$ be the set of all condensation points of $E$. Prove that $P$ is perfect and that at most countably many points of $E$ are not in $P$. I seemed to be able to prove t...

I found some more similar posts, but nowhere was a clear answer.

Also, if B(x,r) contains uncountable points of E then there is no reason to believe why any sub-ball will contain countable points of E.

Koro

6:36 AM

I checked my notes. I got it now.

$E\subset \mathbb R^k$.

pejel1967

8:20 AM

if you follow a post, does the OP get notified?

pejel1967

8:30 AM

how much reputation is earned if I get my answer as "accepted" answer?

SHASHAANK B.H.

8:44 AM

10

pejel1967

8:54 AM

@SHASHAANKB.H. has anyone tried to tell you about redemption?

SHASHAANK B.H.

?

robjohn

9:23 AM

@pejel1967 15

SHASHAANK B.H.

@robjohn We get only 10 right?

robjohn

no. 15

SHASHAANK B.H.

Oh

robjohn

upvote gives 10

acceptance gives 15

SHASHAANK B.H.

Oh Ok

I got confused

pejel1967

9:33 AM

@SHASHAANKB.H. it's okay at least you got redemption

SHASHAANK B.H.

redemption?

Wdym?

pejel1967

i won't tell you, it'd be better if you know it yourself

maybe some time in the future

SHASHAANK B.H.

oo

Oh

robjohn

math.stackexchange.com/help/whats-reputation

automorp15m

10:34 AM

we're tasked to prove that $x \in \text{int} A$ iff there exists a neighborhood $U$ of $x$ such that $U \subseteq A$.

Attempt:

$(\Rightarrow)$ Let $x \in A$, then by definition of $\text{int} A$, there exists a neighborhood $U$ of $x$ such that $U \subseteq A$.

$(\Leftarrow)$ Let there be a neighborhood $U$ of $x$ such that $U \subseteq A$, then by the definition of an interior point, $x$ is an interior point of $A$. Thus, $x \in A$.

Is that correct or is something wrong or are there any lapses?

we're tasked to prove that $x \in \text{int} A$ iff there exists a neighborhood $U$ of $x$ such that $U \subseteq A$.

Attempt:

$(\Rightarrow)$ Let $x \in A$, then by definition of $\text{int} A$, there exists a neighborhood $U$ of $x$ such that $U \subseteq A$.

$(\Leftarrow)$ Let there be a neighborhood $U$ of $x$ such that $U \subseteq A$, then by the definition of an interior point, $x$ is an interior point of $A$. Thus, $x \in \text{int} A$.

Is that correct or is something wrong or are there any lapses?

edit: $\text{int} A$

automorp15m

11:37 AM

seems correct to me but

anyone?

Calvin Khor

@automorp15m what is your definition of int A?

sorry typo on what i wanted to ask

automorp15m

the union of all open sets
contained in A.

Mad Spaces

I am having brain Freeze, please endure this Stupid question

Calvin Khor

sure, kinda silly if thats the definition to ask. @automorp15m

Mad Spaces

Is it possible for a holomorphic function, to have a derivative to be equal to zero at some points and not at others? i keep thinking that if a complex function has derivative zero at ome points then it is zero everywhere

I am not sure why, can someone confirm or deny this

Calvin Khor

11:43 AM

@MadSpaces yes, but they must be isolated points

e.g. z^2 is holomorphic and has derivative zero at 0

Mad Spaces

So what would this mean?

In analysis language.

Calvin Khor

what would what mean?

Mad Spaces

You can only have zero derivatives for none constant function if they are isolated?

Calvin Khor

that's the identity theorem

either f=0 identically or the set for which f=0 consists of isolated points

in particular then sin(1/z) whose zeros accumulate at 0 is not analytic

Mad Spaces

How many isolated points can you have, without breaking holomorphicity

automorp15m

11:47 AM

@CalvinKhor $$\text{int} A = \bigcup_{V_i \subseteq A, V_i \text{is open}} {V_i}$$

Mad Spaces

I got to go, but i will check back in thirty minutes or so

Calvin Khor

infinitely many. @MadSpaces take sin(z).

automorp15m

@CalvinKhor sorry I don't get what you mean

Calvin Khor

@automorp15m yes, but maybe you should explain precisely what the set U is in your "=>", and perhaps explain why rather than just saying 'by definition' in "<="

@automorp15m i mean the question seems like it is just trying to check you understand the definition. and maybe if you know how to write a proof

automorp15m

will elaborate it thanks

Koro

11:52 AM

Let X be a complete metric space and Y be a completion of X. What can be concluded from this information?

Calvin Khor

how do you complete a complete metric space and get something different?

Koro

I think: Y is also a complete metric space.

Calvin Khor

if words mean anything, then the completion of a metric space better be a complete metric space. In particular also if you started with a complete metric space

Xander Henderson

@CalvinKhor Since when do words mean anything?

Calvin Khor

@XanderHenderson uh oh.

Koro

12:00 PM

Definition: Y is said to be a completion of X if 1) there exists a isometry F from X to Y, 2) F(X) is dense in Y.

@CalvinKhor I think you mean using equivalence relation on Cauchy sequences.

Calvin Khor

sorry, gotta bounce! but yes

Koro

I don't yet know that completely :(

Alessandro Codenotti

12:44 PM

@Koro Y=X

Akiva Weinberger

$\Bbb R^\infty\setminus\{p\}$ is homeomorphic to $\Bbb R^\infty$

The infinite product of lines is homeomorphic to itself minus a point

copper.hat

1:00 PM

@robjohn :-)

Alessandro Codenotti

1:11 PM

@AkivaWeinberger from the statement alone I will bet that this is due to Jan van Mill

automorp15m

1:41 PM

@CalvinKhor By definition, $\displaystyle{ \text{int} A = \bigcup_{V_i \subseteq A, V_i \text{is open}} {V_i}}$. Now, suppose $x \in \text{int} A$. Then $x \in V, \exists V \subseteq \text{int} A$. Thus, there exists an open neighborhood $U$ of $x$ such that $x \in U \subseteq V \subseteq A$. Conversely, suppose that there exists a neighborhood $U$ of $x$ such that $U \subseteq A$, then $x \in \text{int} U \subseteq \text{int} A \Rightarrow x \in \text{int} A$.

kung naa cm nko dri pls laina lang ayaw kopya tanan lamats

is it correct now?

Problem:
Prove that $x \in \text{int} A$ iff there exists a neighborhood $U$ of $x$ such that $U \subseteq A$.

Attempt:

By definition, $\displaystyle{ \text{int} A = \bigcup_{V_i \subseteq A, V_i \text{is open}} {V_i}}$. Now, suppose $x \in \text{int} A$. Then $x \in V, \exists V \subseteq \text{int} A$. Thus, there exists an open neighborhood $U$ of $x$ such that $x \in U \subseteq V \subseteq A$. Conversely, suppose that there exists a neighborhood $U$ of $x$ such that $U \subseteq A$, then $x \in \text{int} U \subseteq \text{int} A \Rightarrow x \in \text{int} A$.

or this part: Then $x \in V, \exists V \subseteq \text{int} A$.

should be Then $x \in V, \exists V \in \text{int} A$. right?

oh ignore that i think the above is already right

anyone?

karnop

2:17 PM

Hello everyone. I am self learning combinatorics by the book ' a walk through combinatorics', so, i frequently come across questions i cannot understand and there are no answers for them on stackexchange. Moreover, there are also questions that i am able to solve but when i try to check on stack exchange for that question, it isnt on the site.

Is it okay if i write those questions and then self answer them? or is it not allowed on the site ?

Akiva Weinberger

2:37 PM

Idea

Instead of writing "Lemma: the product of blahs are blah. Proof: Obvious" (which as we all know is bad pedagogy for using the word 'obvious'), simply write "Remark: the product of blahs are blah" without proof

@karnop Self-answering is allowed

Just check that box

It’s OK to Ask and Answer Your Own Questions

Jeff Atwood on July 01, 2011

The FAQ has contained one key bit of advice from the very beginning: It’s also perfectly fine to ask and answer your own question, as long as you pretend you’re on Jeopardy! — phrase it in the form of a question. So … if you have a question that you already know the answer to…

karnop

Thanks dude

Monty

3:02 PM

anyone help me find the inverse of $(1-t)\exp(-tA)-tI$? for an invertible matrix $A$ and constant $t$, $I$ is the identity

sorry sign error- I meant to write $(1-t)\exp(-tA)+tI$

I can write it like this by factoring out a constant $(t-1)\big(\frac{t}{t-1}I-\exp(-tA)\big)$

and then calculating the inverse is calculating the resolvent $\big(\frac{t}{t-1}I-\exp(-tA)\big)^{-1}$ but im stuck

Mad Spaces

$\forall x \exists y \\
\exists y \forall x $

The first one means, there is idividual y to every x and the second one, one y for all x, right? or am i mixing up

Monty

@MadSpaces that is correct

Mad Spaces

thanks

Monty

The first could say to be more clear for each $x$ there exists a $y$ denoted $y_x$ or something liek that. but yeh you interpreted them correct.

Akiva Weinberger

3:17 PM

Is it just me or is this prose really dense

($Z$ is $\Bbb Z$)

A subpartition is a collection of disjoint sets. A simple subpartition is a collection of disjoint finite sets

Basically, if for each $i$, $A_i$ is an infinite family of disjoint subsets of $\Bbb Z$, then I can create an infinite family of disjoint sets $B$ made up of infinitely many elements of each $A_i$

*$A_i$ is an infinite family of disjoint finite subsets of $\Bbb Z$

Akiva Weinberger

4:02 PM

(The proof is basically just "there's nothing stopping you")

Ted Shifrin

4:18 PM

@leslie Perhaps I missed the mention of lesliecoin in this.

@AkivaWeinberger I'd settle for horribly written, on various counts.

leslie townes

ted: all crypto is a scam except lesliecoin.

Mad Spaces

ted

i am having a little fight with my tutor

Would you like to tell me what you think before i go ape nuts on him

its about writing a complex $ z = x+iy$ i have always, being the autistic rigorious dude i am, wrote the isomorphy on top of the equality sign, since the way i understood it, its a notation agreed upon using the isomorphy between the complex and 2d Reals.
He keeps telling me not to do it, since and i quote " Z is actually x +iy"

Whos right.

leslie townes

4:34 PM

mm, i wouldn't get too caught up in the details of formalism. e.g. whether R "actually is" a subset of C or simply isomorphic to one. but in most constructions of C from R^2, you would have literal equality in z = x + iy due to the way that + and complex multiplication (i*y) is defined

e.g. if C is elements of R^2 with componentwise addition and a funny multiplication, it's expressing (x,y) = (x,0) + (0,1)(y,0) as a literal equality of ordered pairs of reals.

Mad Spaces

yes, but the way you wrote them, are vectors.

when you write z = x + i y you are implying real scalars

thats literally the isomorphy is it not

And people would really go on and write x ,y elements of the reals....

leslie townes

i don't see the vector/scalar distinction as very informative here. in this view, "x + iy" is just notation for (x,0)+(0,1)(y,0). you're identifying the real number x with (x,0), but this is not a distinction worth preserving in notation.

yes, implicitly x and y are real in the above. otherwise x might not be "(x,0)"

Mad Spaces

What happens when you start diving with i x and y?

No vector division is defined, arent you using the isomorphy

leslie townes

again, i don't think it's helpful to think of ordered pairs of reals as 'vectors' in this setting, for example, because you do have a notion of dividing by them when they are nonzero.

Mad Spaces

I am talking from a point of view of Rigor. not "helpful", in sense of Linear algebra

leslie townes

4:41 PM

from the point of view of rigor, then, if you want to think of complex numbers as vectors, you do have a defined notion of 'vector division.'

i don't think it's helpful to think of it that way, but it exists as a concept.

Joe Shmo

@AkivaWeinberger oof

Mad Spaces

Leslie.
even with that, then your + and * are not the same on the reals.

Joe Shmo

believe it or not, I think bitcoin is here to stay

and once it weathers the storm, it gains an immortal status

Mad Spaces

this is obvious notational abuse, which is why stating the isomorphy is essential

an when you literally say "identify (x,o) with x" thats LITERALLY the isomorphy

"autistic happiness screeching in my heart"

leslie townes

if you want to notationally preserve all of those distinctions you are welcome to do so. i don't think it helps and i don't think it really aids in 'rigor,' whatever that means.

Mad Spaces

4:51 PM

Leslie, say it.

I am.. Right

leslie townes

indeed, outside of the cauchy-riemann equations, i can't think of a beginning topic in complex analysis that is made easier by thinking of complex numbers as pairs of real numbers. it's better if they are first class citizens and numbers in their own right.

Mad Spaces

Yes but no one is talking about the aspect of helpfullness, i am talking pure riogor. I know they are helpful.

leslie townes

would you write integers as equivalence classes of ordered pairs of natural numbers? if not, why not?

Mad Spaces

No thats not the point i am making.

you are right its not helpful

But i will not accept the statement " writing z = x+iy " does not use the isomorphy and are the same, thats bullocks

leslie townes

it's just my original point. if you think of x and y in that expression as the complex numbers x and y (that just happen to have zero imaginary part), it makes literal sense. like you wouldn't nationally distinguish between real 0 and complex 0.

unless you would. i don't know.

Mad Spaces

4:55 PM

You would. a complex zero is literally a vector

leslie townes

and again, i wouldn't distinguish between real 0 and rational 0 or integer 0, but you're certainly welcome to do so.

Mad Spaces

You are completly missing the point of my argument and reverting to your argument about helpfullness

Joe Shmo

wait @MadSpaces why would you do that? how does that help you?

yes its the 0 vector. so what?

Akiva Weinberger

Whether to distinguish between the integer 0 and real 0 and rational 0 etc is one of those things that computer scientists and mathematicians tend to disagree on

leslie townes

you're saying that 'rigor' demands that you preserve this distinction and i guess i'm wondering why it starts at complex numbers and not sooner.

Akiva Weinberger

4:58 PM

I think computer scientists are more used to typed systems than mathematicians are

(unless those mathematicians happen to be type theorists, I suppose)

leslie townes

it's a choice about where to start. i think if you're gonna do complex analysis you should think of complex numbers like any other kind of numbers, and not something constructed out of lower-level data. but that's just my opinion.

Akiva Weinberger

In any case, it never really matters all that much

Joe Shmo

yes, @AkivaWeinberger is on the money

Ted Shifrin

@MadSpaces He is right.

Joe Shmo

it's only technically true that theyre not the same

Transcript for

$Mathematics$ Mathematics