for example, my linear algebra book uses $\{a_1,a_2,\ldots,a_n\} = \{a_i\}_{i=1}^n$

which makes sense

so i'm complaining to you guys because the book's publisher isnt here

@Daminark "Your gminushimon is evolving"

@GFauxPas I mean the ordering is just furnished by the index, so you don't need to worry about that too much. I've seen both and neither causes a problem

but the ordering of terms changes the sum of a series in general

aannyway

@orbit-stabilizer Well, let's be particular. What does g - h vanish on, from definition?

How's the Biology going? @Daminark

Ah I mean series is another story, but like, that's where you determine the order of the sum just by the index, so the notation for the sequence still seems to me as being somewhat immaterial

the proportion of ambiguous notation just increases the more hard math you do

Over $E$ @BalarkaSen

@skullpatrol I am now done with that class. We had 3 tests during the quarter and no final in final's week

So, basically the rationals in [0,1]

@orbit-stabilizer Right. So do you know a characteristic property of $E$, as a subset of $[0, 1]$?

Right, rationals in [0, 1]

@EricSilva Which brings to mind the joke we had a day or so ago

But yeah I dunno, some mix of my having other things to do and not being particularly interested in bio made this probably the class that I was least engaged with, at least in college and possibly ever

Ok

It's not connected for one. It's also not complete. It's also not closed. It's not open.

Let $X$ be a subset of $\Bbb R$ so that every continuous function $X \to \Bbb R$ can be extended to a continuous function $\Bbb R \to \Bbb R$. What can we say about $X$?

@orbit-stabilizer Keep listing :D

@Semiclassical which joke?

"What domain is the study of abuse of notation?" "Physics."

lol

Not compact.

lol

You're getting ahead of yourself, though. Let's stop. Why is it not closed?

@Vrouvrou $\frac{a-b}2$ need not be in $K$. What's important is that $\frac{a+b}2$ is in $K$.

It doesn't contain its limit points.

What are it's limit points?

All points in [0,1]

Ah.

That has a name

Do you know the name?

Martha?

LMAO

That was beautiful

I lol'd IRL

that hit me at a time i was not expecting it

nice

me too hahaha

I am become dead

At least BvS gave us some memes

kg/m^3

Martha is kind of dense though.

What did Avengers ever do

@BalarkaSen ohhh righttt

I hope I'm not being annoying or something..

Semi I was feeling rushed in my homework for analysis so for one of the problems i extended the function to $\mathbb C \to \mathbb C$ so I could use the identity theorem on it

@FuzzyPixelz I hope that too

Okay, so E is dense in [0,1]

hope the teacher will be okay with that

But if you have time for this,

What domain is the study of the abuse of terminology? @Semiclassical "Philosophy" :P

@BalarkaSen i was actually thinking of that scene haha. So pumped for Infinity War

I really love MCU

Oh you should see that meme

was deriving the power series for arctan by integrating a geometric series, but the geometric series was only valid for $r < 1$, but I made a complex argument that it should hold for $r = 1$ as well

not sure how to do it otherwise

Aced my geometry final woot woot

@Faust which type of geometry?

garts

Congrats

Euclidean and Axiom geometry

@orbit-stabilizer But yeah E is dense in [0, 1]. Or, as you say, that means every element of [0, 1] is a limit point of E. What does that say about g - h?

ancient greek geometry ofc, what other kind of geometry is there?

using NUMBERS? pfft

@BalarkaSen hah just watched it.

straightedge and compass or gtfo

Sick @Faust, congrats!

:-)

@orbit-stabilizer I think this is my favorite one so far

thanks meant that

joke ruined

Still got 3 more though =(

@BalarkaSen hmm. I don't know, I mean g - h is still defined over [0,1]

hows everyone elses exams going?

starting this week, algebra tuesday followed by analysis wednesday

so bleh

I've got two exams Friday morning, which is gonna be less than fun

and school just ended this friday - so shitty

got the very first two days of exams

@orbit-stabilizer Here's an idea. Suppose $x \to x_n$, and $f$ is continuous. Then $f(x) \to f(x_n)$.

But what is Friday? [Vsauce music]

@BalarkaSen okay ill think about that

It's Friday, friday... ugh I can't even do that

thanks

Ouch! @Daminark

i got algebra real analysis and number theory left, but at least geometry and graph theory are done ^^

Let $f\left(x\right)=\arctan\left(x\right)-\frac{n}{x}+1$ and let $a_n$ be the solution of the equation $f(x)=\frac{\pi}{2}$ on $(0,\infty)$ Prove that $\arctan\left(\frac{1}{a_n}\right)=\frac{n}{a_n}-1$

having exams the first day sucks

Number theory woot

@FuzzyPixelz $\arctan x + \arctan x^{-1} = \frac \pi 2$?

Yes, I tried using that, but I end up getting $\arctan\left(\frac{1}{a_n}\right)=1-\frac{n}{a_n}$ @LeakyNun

you're big enough to deal with sign errors on your own

$\arctan x+\arctan\frac1x=\frac\pi2$ if $x$ is positive and $-\frac\pi2$ if $x$ is negative

FIVE math finals? @Faust

though I don't know if that's what happened

@AkivaWeinberger $(0,\infty)$

@skullpatrol one was an on the last day of class open book final

So sort of only 4

Still...

Oh we know that $a_n \gt n$ for all $n$ so there couldn't be a problem with that, I just don't see what's the issue.

what is this gfoty album

my next two are the 12th and 13th the last is the 16th so at leas ti got some time between my geometry one was the first day of finals :'(

this is like music but syrup of ipecac

No doubt.

@FuzzyPixelz I think you're right, I think the problem is wrong

Numerically, $a_2\approx2.964$

No way, this was on my exam this morning...

:O

$\arctan(\frac1{2.964})\approx0.3254$

How funny, I din't think about this carefully back then.

and $1-\frac2{2.964}\approx0.3252$

are all compact connected subsets of $\Bbb R$ homeomorphic?

And 2.964 is a solution of the equation ?

@LeakyNun Yes

@FuzzyPixelz It's $a_2$

@AkivaWeinberger and if we change "compact" to "open"?

Oh wait I'm wrong

Duh, $\{0\}$ and $[0,1]$

lol

Point and interval

@BalarkaSen: So, $h,g$ cts implies that $\lim_{x \to a} g(x) = g(a)$ and $\lim_{x \to a} h(x) = h(a)$. This happens iff $\lim_{n \to \inf} g(p_n) = g(a)$ and $\lim_{n \to \inf} h(q_n) = h(a)$ for every sequence $\{p_n\}$ such that $\lim_{n \to \inf} p_n = g(a)$ and $\lim_{n \to \inf} q_n = h(a)$.

The only connected subsets of $\Bbb R$ are intervals (where we count $\{c\}$ and $[c,c]$ I guess)

@LeakyNun

(Strange how points are both connected and totally disconnected)

@AkivaWeinberger so it's true when we change compact to open?

Should be yeah

you forgot $\Bbb R$

but alright it's an interval

$(-\infty,\infty)$

Are point sets connected?

@orbit-stabilizer yes

Hm? What do you mean

(Also I don't know if the empty set counts as connected)

(so I'mma assume "neither")

because you can't divide it into two non-empty sets, period (you don't even need open)

@orbit-stabilizer That's true

So if it's connected then, all nonempty connected open subsets of $\Bbb R$ are homeomorphic.

Okay, then I'm stuck. I don't know what to do from here

What are you trying to show?

@orbit-stabilizer What does $E$ being dense in $[0, 1]$ mean?

If something's zero on the rationals it's zero everywhere?

In sequence lingo?

Every point of [0,1] is a limit point of E or contained in E.

@orbit-stabilizer "sequence"

But what does being a limit point of E mean?

Say $f(x)\ne0$. Then choose an open set around $f(x)$ that doesn't hit zero

and look at its preimage

Uh, for every point in [0,1] there exists a rational sequence that converges to it.

If all the rationals map to zero, the preimage will contain no rational

@orbit-stabilizer Bingo

and thus not be open

and so that's an open set whose preimage is not open so it's not continuous

So let your $a$ be any thing in $[0, 1]$

Pick a sequence $\{p_n\}$ in rationals converging to $a$

Now use what you said

@AkivaWeinberger I want to show there's only 1 continuous extension of $f$ on [0,1]

Namely, $g(p_n) \to g(a)$ and $h(p_n) \to h(a)$

@orbit-stabilizer If $g$ and $h$ are different thingies than look at $g-h$

@orbit-stabilizer should one do real analysis before topology?

That's what we're doing right now haha

It's continuous so preimages of open sets should be open

Say $g(x)-h(x)\ne0$

@AkivaWeinberger We're trying to get him to do a much easier argument

@orbit-stabilizer so it isn't topology?

@LeakyNun uh idk... ive never done topology, but im in a real analysis class rn

Using sequences

Oh

@orbit-stabilizer then why are you using words like "dense" and "limit point" and "compact"?

Ah, I thought this was a topology thingy

Fine, sequences then

@LeakyNun Because that's chapter 2 in Rudin!

^can confirm

real analysis use those words? :o

He has a weird definition of connectedness though

I mean, compact is acceptable, but "limit point"?

Chapter 2 of Rudin is honest topology, I believe

Yeah it's metric topology

@LeakyNun I mean, you'll also find limits in real analysis

$\lim$

but not limit point

Like it's not just the analyst's take on topology even, it's topology topology

Analysts need a good bit of topology

in the sense that a set has many limit points but a sequence has only one limit

@BalarkaSen okay looking at the rational sequences, the functions must agree on each point in the sequence, and since they're cts, they must agree on their limits as well?

@orbit-stabilizer exactly

@orbit-stabilizer Bingo

@Daminark I don't remember, how does it define compactness?

Rudin?

Every open cover has a finite subcover

Yeah

Yep

@AkivaWeinberger how does he define connectedness?

Oh, OK, it doesn't butcher it then

:P

There's nothing wrong with sequential compactness

That's how normal people think about compactness

@Leaky$A\cap $ closure($B$) is empty set and $B \cap $ closure($A$) is also empty set

@Balarka the thing is, in analysis you often think about the weak topology on a space

Idk how to do the bar

sequential compactness is compactness in analysis, until you get to the dual of Banach spaces

\overline{whatever}

Fair, fair, fair

Functional analysts aren't normal people anyway

@BalarkaSen thanks! I'll have to think a bit more to make it rigorous, but I get the idea

But yeah the weak topology on a space is basically the smallest one that makes all bounded linear functionals continuous

No problem, you have the idea

And I think those topologies are not metrizable

(For an infinite dimensional space ofc)

Sounds true

The funny part is, in $\ell^1(\mathbb{N})$, it turns out that weak and strong convergence are the same

@orbit-stabilizer so a set $S$ is connected if there is no $A$ and $B$ such that $A \cup B \supseteq S$, $A \cap \overline B = \varnothing$, and $B \cap \overline A = \varnothing$?

It takes blood to prove it, you have to use this thing called the gliding hump (or at least that's how I've seen it)

Actually wait @Alessandro do you know if Banach-Alaoglu gives you just sequential compactness of the unit ball? Or is it legit compactness

@Daminark There are also other spaces with this property (which is called Schur's property) apparently even though I don't know any

Oh. There's a Christmas remix of Everyday Bro

I wanna die

@Daminark Doesn't it give weak* compactness?

Well yeah it's def weak*

@BalarkaSen No, I mean

There's another theorem, I think by Banach, that gives sequential compactness, let me look it up

I was worried he'd define it as "closed and bounded"

@LeakyNun: Two subsets $A$ and $B$ of a metric space $X$ are said to be seperated if both $A\cap \overline{B} = \emptyset$ and $\overline{A} \cap B = \emptyset$.

I'm just wondering like, because weak* is not metrizable, sequential may not be the same as compactness anymore

So I'm wondering which one it gives

@Akiva oh

A set $E \subset X$ is said to be connected if $E$ is not the union of two nonempty seperated sets.

@orbit-stabilizer aha

In particular they don't need to be open

and $A\cup B$ equals the set

Wiki says that the if $X$ is separable the weak* topology on the closed unit ball of its dual is metrizable, that can't be!

Wait what? Hold on the weak* topology on a Hilbert space is gonna be the weak topology

Oh wait maybe it's only on the unit ball and not on the whole space

It is on the ball

Oh wait weak* is not weak?

still very weird

Like that I could buy somehow

This is a horrible offense on the English language

en.wikipedia.org/wiki/Banach%E2%80%93Alaoglu_theorem look at the "sequential Banach-Alaoglu" paragraph

@Akiva so if you have a Banach space $X$, you can embed it naturally into its double dual

By evaluation maps

@Daminark After doing so you hope real strong that the embedding is actually an isomorphism

So, if you look at $X^*$, you have the normal weak convergence, which is a condition about every sequence in the double dual

But weak* convergence just requires that all the evaluation maps in the double dual (the embedded copy of $X$) satisfy the condition

@Alessandro true

which is why reflexive spaces, that is spaces canonically isomorphic to their double dual, are super nice

I mean the only Banach spaces I've ever interacted with are the $L^p$ spaces and they're all good so woo

how do i link and show something i said earlier?

$1<p<+\infty$ @Dami

Fair

Anyway $L^2$ and $\ell^2$ are the only Banach spaces

So we're good :P

I said this earlier:

So, $h,g$ cts implies that $\lim_{x \to a} g(x) = g(a)$ and $\lim_{x \to a} h(x) = h(a)$. This happens iff $\lim_{n \to \inf} g(p_n) = g(a)$ and $\lim_{n \to \inf} h(q_n) = h(a)$ for every sequence $\{p_n\}$ such that $\lim_{n \to \inf} p_n = g(a)$ and $\lim_{n \to \inf} q_n = h(a)$.

@Alessandro kek

That's wrong right. The last two equalities should be $a$, right>

I mean what about $L^2(L^2([0,1]))$?

@Daminark There's also $c$ and $c_0$ (and $c_{00}$ but that one isn't Banach) which have messy duals I think

Both their duals are $\ell^1$

@Daminark You have to define a measure on $L^2([0,1])$ for that to make sense, if you really want to...

@Daminark wtf

Oh yeah, and that's near hopeless

Like it'll be a garbage measure, not translation invariant (and possibly you'll lose a lot more)

Yeah I can pull up the proof, I had that as a pset problem

Yeah, I don't think there's really a point in doing so

Thanks, I can't read it right now, but I saved it for later

Aight

What's the source by the way? I'm looking for a good book or notes on functional analysis

Well that was just my solution to a pset problem

The book we technically used last year was a kind of baby book

The more introductive the better

Kolmogorov & Fomin: Elements of the Theory of Functions and Functional Analysis

But it does all the Banach space theory immediately after sets/metric spaces so it doesn't work as much with L^p, and generally feels less comprehensive

The book we're using next quarter in functional analysis is this: people.math.ethz.ch/~salamon/PREPRINTS/funcana.pdf

Traditionally the class used Brezis but our prof felt like it didn't do stuff like spectral theory enough, and treated functional analysis as a bunch of tools that you had to slog through in order to do PDEs, especially those related to fluid dynamics

Ah, yeah, I looked at Kolmogorov and Fomin already. The first $3$ chapter of the second one look exactly like the stuff we covered so far, thanks!

No problem! But yeah he also said that Lax Functional Analysis isn't too bad, though I'm liking this one a lot so I'll just roll with it

Brezis seemed a bit too terse and steep especially in the first part

Lol yeah

Dumb question: Suppose $x \in [k,k+1)$ for some $k \in \Bbb{N}$. I need to find $N \in \Bbb{N}$ such that $(x - \frac{1}{n}, x + \frac{1}{n}) \subseteq [k,k+1)$ for every $n \ge N$.

I mean he does Hahn-Banach on like page 3

For some reason I can't find such an $N$.

@user193319 Try drawing it

@AlessandroCodenotti champion

@user193319 You want $k<x-\frac1n$ and $x+\frac1n\le k+1$

You can solve for $n$ in each case

Ok, gotta go, I'll be back later, bye everyone!

Those are both $\le$ actually

See you @Alessandro!

later

@Daminark

I''m looking at this:

Since $G/P$ has order 26, why does that tell us that our 13-sylow subgroup has index 2?

@orbit-stabilizer Lagrange theorem?

How?

$P$ is a 7-sylow subgroup

no, the 13-sylow subgroup inside G/P

That subgroup can't be contained within the quotient.

But I guess, you're saying look at aP, where a is in the 13-sylow subgroup?

G/P is a group of order 26

its 13-subgroup has index 2

Oh. Yes. Then how can we transfer that information back into our original group G?

by taking the preimage?

Hm okay.

Let's try Sylowing the 13-subgroup

The only options are 1 or 14

AH, correspondence theorem apparently applies

@Daminark yeah

If there were 14, then you'd have 168 elements of order 13. This leaves 12 elements, 6 of which have order 7

Wait no

14 elements left. 6 have order 7

So there are 8 left

8 left. One has order 0.

7 left.

I call the identity order 1

8 little monkeys, jumping on the bed.

But yeah so the remaining ones all have to be order 2

Sorry, yeah

So then n_2 = 7

But, that's a contradiction, no?

Wait, that's fine.

@orbit-stabilizer kek

Okay, say I have this group. How do I classify it?

Hey! I need help. I'm trying to find a stackexchange site that will help me figure out a clever way to teach solving system of linear equations by substitution because I'm somehow failing with my explanation. I have done so many examples with them. I'm beginning to think that their cognitive ability is not up to the task or I'm also thinking I might suck at reaching them in this topic. :(

Where $n_2 = 7, n_7 = 1, n_{13} = 14$.

@randomgirl matheducation?

thank you @orbit-stabilizer

Askaway right here

Oh wait here's an idea

Take a Sylow 2-subgroup, its normalizer has size 26

That'll intersect trivially with the normal subgroup of order 7

How are you approaching the topic? @randomgirl

How do we know its normalizer has size 26?

Meaning the whole group is a semidirect product of $C_7$ and some group of order 26, so if we can figure out all of those...

It has index 7 by the third Sylow theorem

So it is an 8th grade class just so you know. In a way I do think the topic is advanced but you know state standards. I teach them we need to rearrange one equation of the two for one variable and insert it into the other and boom it is just like solving a regular linear equation from there.

Or wait maybe there's something easier we can do

I think they are not used to math being so long

You take a group of order 13 and then take its semidirect product with the group of order 7, that'll be normal because Index 2

So then take a semidirect product of that and you're good

And the best part is

Groups of order 91 are cyclic

Okay, I need to go over semidirect products bcz I can't follow this. I still don't understand the part where you said the normalizer has size 26

So I'm abandoning that approach

Oh, okay

But, how did you know it had size 26?

But the idea was that n_2=7

So by the third Sylow theorem, the index of the normalizer is 7

Oh, I didn't learn that part of the theorems...

Rip

But yeah here we've got something

Our group necessarily has a subgroup of order 91

But groups of order 91 are cyclic by Sylow counting

So a group of order 182 is a semidirect product of $C_2$ and $C_{91}$

@Daminark thanks, I'll read this over

So we want to look at the automorphism group of $C_{91}$, which is $C_6\times C_{12}$

So either the homomorphism from $C_2$ into it is trivial, so we're cyclic, or we have to look at a subgroup of order 2

And I think there are just 3 of those? Thinking about it in additive notation, the elements of order 2 should just be (3,0), (3,6), and (0,6)

So we've narrowed ourselves to at most 4 possibilities

@randomgirl sorry I was doing something else but I can try to help out here

So this system is just 2 linear equations, right?

yep.

I also posted the question on math educators

Okay so, this may be a bit strange, but have you tried explaining it in a sort of, "what can we guarantee from each" way?

Keep in mind that I haven't done any pedagogical research so anything backed up by cognitive studies should immediately take precedence over what I'm saying

I'm worried it might be too advanced and that is why there is so much torture but it could also mean I just suck at getting the explanation across. I feel like I have explained it myself real well but to 8th graders and how they perceive things they most likely don't feel the same way.

And are you talking about the three differ ways?

But when I first learned it, I found it helpful to think about it by looking at the equation and saying well, if we know this equation should be true, then we can rearrange it to guarantee something else is true

we haven't even made it to elimination

Which textbook are you using? @randomgirl

So you have some $ax+by=c$, and then if that has to be satisfied, so much $y = \frac{c-ax}{b}$

prentice hall course 3 mathematics... i probably made a mistake and am teaching them too much at one time. The book jumps from slope-intercept form without doing any rearranging to solving systems of equations that do require rearranging.

Now this and another equation both have to be true, this allows us to plug one thing in another. If students have a conceptual problem, that might be a way to approach it

I wouldn't expect this to be too advanced, like I think this is basically standard fare for 8th grade, but maybe

I'm sorry I mean the book doesn't have them rearrange equations and then figure out slope and y-intercept.

Oh hmm

like we jump from things like y=mx+b to solve the system 3x+y=3 and 2x-y=4

And I probably shouldn't have made that jump with the book

Originally I was thinking it was conceptually unclear, but to be fair, the algorithm doesn't require much thought

And if they haven't had practice with generically rearranging equations, I'd guess that's more likely the problem