Mathematics - 2017-02-27 (page 2 of 5)

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2:00 AM

Anyone here comfortable with convex optimization. I have been struggling with a problem for hours without much progress.

Show that if the method of steepest descent is started from $\mathbf{x_0}=(0,0)^T$ it cannot converge to $\mathbf{x^*}$ in a finite number of steps. The function is $f(\mathbf{x})=x_1^2+4x_2^2-4x_1-8x_2$.

Given that the level sets of that are ellipses, I wonder if the following simpler problem would be informative: If $x_0=(1,0)^T$ and $f(x)=x_1^2+x_2^2$, can the method of steepest descent converge to the minimum at $x=(0,0)^T$?

uhh can someone help me with this little linear algebra thing with quadratic forms

I have to find the symmetric matrix $A$ such that for $\vec{x} = (x_0,x_1)$, $A\vec{x} \cdot \vec{x} = 3x_0^2 - 2x_0x_1 + 3x_1^2$

I got it, but $A$ has complex values so it doesn't work

You mean, you can't find $A$ without it being complex?

How did you come to that conclusion?

Say what?!

assume $A = \begin{pmatrix} a & b \\ b & c\end{pmatrix}$

2:09 AM

(Guess Zach wishes I hadn't entered just now.)

@TedShifrin oh snapo

:P

I meant "Wow that problem was really easy! Ted will be so surprised!"

That's better.

anyways, $a = 3$ and $c = 3$

i know that

2:09 AM

And $b=-1$.

OH

LOL I JUST REALIZED

Nothing like Ted embarrassment :P

nevermind, i thought it was square root of the middle term, not half

that makes this a whole lot easier

Hence why I asked you to explain your work. I figured it was some simple slip :)

Write it out one time and you'll get it.

2:10 AM

yeah i get $2bx_0x_1$ for the middle term when dotting

You can also watch my lectures on this stuff (and tying it into $LDL^\top$ form rather than $P\Delta P^{-1}$ form).

Righto, mr @Zach

I'm forget terminology, is one of them a similarity and the other a congruence?

The latter is similarity

the former, I don't know...

@Semiclassic: The latter is similarity — we're looking at linear transformations. In the former case, we're looking at bilinear forms.

Ah.

Yeah.

2:12 AM

It's just called diagonalization of a symmetric bilinear form. ... @Zach, there's some cool stuff with Sylvester's law. ... BTW, the $LDL^\top$ decomposition corresponds to completing the square. I did all this in one of my linear algebra lectures on line.

yay inertia

The amazing thing is that the entries of $D$ and $\Delta$ disagree, in general, but the numbers of positive, zero, and negative line up.

(I have literally never actually used Sylvester's law myself, but I know what it's about)

Show that if the method of steepest descent is started from $\mathbf{x_0}=(0,0)^T$ it cannot converge to $\mathbf{x^*}$ in a finite number of steps. The function is $f(\mathbf{x})=x_1^2+4x_2^2-4x_1-8x_2$.

mumbles moment of inertia tensor @Semiclassic

2:13 AM

I know that the minimizer is (2,1) with $f^*=-8$.

Sounds right.

For number 3, we have that $\vec{v}, \vec{w}$ are both eigenvectors, and thus are members of an orthonormal basis for $\Bbb R^n$ by the Spectral Theorem... does that show that $\vec{v} \cdot \vec{w} = 0$?

I have willfully forgotten most things involving the moment of inertia tensor.

@ozarka: I'm not looking at specifics, but it's probably best to draw the level curves of your function and actually see the (negative) gradient flow.

@Zach: As best I remember, it said to give a self-contained argument. Ahem.

Evenin'

2:15 AM

@Semiclassic: I wish I had gotten to teach the instability about the middle axis for rotating a book :)

oh, so without that wonky theorem?

0

Q: Prove that $G$ embeds as proper subgroup into group of units of the group ring $\mathbb{Z}(G)$

I need to prove that a group $G$ embeds as a proper subgroup into the group of units of the group ring $\mathbb{Z}(G)$. The group ring $\mathbb{Z}(G)$ consists of the set of formal sums $z_{1}g_{1} + z_{2}g_{2} + \cdots + z_{k}g_{k}$ where the $z_{i}$ are integers, and the $g_{i}$ are elements o...

abstract-algebra ring-theory group-rings

nods @Zach

I think it may help to do a simpler problem first. (Hence my suggestion above)

user image

@TedShifrin This is the contour lines for the respective function.

2:15 AM

so let's see what we can work with... Lemma 3.1 seems relevant here

Hello, nerds! :D

Now start at $(0,0)$, draw the negative gradient and keep following it, @ozarka.

@Fargle hey nerd

We can do this with differential equations, of course.

@Fargle: I respond to many names, but not that one.

@Ted, I mean it with only the utmost respect and gravitas.

2:16 AM

The negative gradient would be $-(2x_1-4,8x_2-8)$.

So we get $T\vec{v} \cdot \vec{w} = \vec{v} \cdot T\vec{w}$

Bull**** @Fargle

...sometimes I wonder if I'd be better suited as a diplomat than as a mathematician.

chokes

Trump-style, yes, @Fargle.

Judging by this conversation, I think you chose correctly.

2:17 AM

@TedShifrin I understand that I can just iterate over and over but is there a succinct proof to show that it doesn't converge in $n$ steps?

LOL, @Semiclassic.

Come now, one denigration at a time, please.

and so, we have $\lambda \vec{v} \cdot \vec{w} = \vec{v} \cdot \mu \vec{w}$

@Fargle watch out you're going full trump here

Akiva Weinberger

@ozarka That picture looks uncomfortably like we're about to fall into a star

2:18 AM

@ozarka: Maybe I'm missing some subtlety.

Heya, DogAteMy.

Akiva Weinberger

Or at least orbiting one way too closely

Hey

At least I'm not disparaging the differently abled, racial minorities, gender/sexual minorities, or those who hold different opinions than me, but that's beside the point. ;)

@AkivaWeinberger I can e-mail the math supervisor

but that's all i can do

I'd call that a low bar, but it's one which seems to get missed all too often these days...

I've just been watching too much of The West Wing and, by gum, it really makes me want to be politically involved.

2:19 AM

Indeed.

Well, good, @Fargle. We need that.

Especially in states like yours.

@TedShifrin you are not missing a subtlety. I just thought there was a better way to approach the solution than doing it numerically.

Especially in states like mine, indeed.

I've always thought I would make a good lawyer, but I'd never want to be in politics.

i saw an anti-trump demonstration outside of the New York Times building, in NYC today

@Fargle Build up a political career and then suicide it to get nuclear power moving again

2:20 AM

Another convex optimization problem.

Show that $\tau_k$ defined by
$$\tau_k =
\begin{cases}
1 &\quad\mathbf{g_k^TB_kg_k}\le0 \\ \min(\frac{\vert\vert\mathbf{g_k}\vert\vert^3}{\mathbf{\Delta_k g_k^T B_k g_k}},1) &\quad \text{otherwise}
\end{cases}$$
gives the minimizer of
$$m_k=f_k+\mathbf{g_k^Tp}+\frac{1}{2}\mathbf{p^TB_kp}\quad \text{s.t.}\quad \vert\vert p \vert\vert\le\Delta_k$$
along the direction $-\mathbf{g_k}$.

Tennessee is a public policy disaster with the cherry of "beneficial privatization" on top.

Hey everyone!

@ozarka: I guess it you solve the differential equation, it will in fact lead you to that point, but (maybe discretizing) you'll never get there in finite time.

Hey @arctictern

hi @Daminark

2:20 AM

Hi @Daminark

@MickLH: you have no idea how close that is to what I'd like to do.

@Daminark, hello.

I think what you'd want to do is find some measure of distance from the minimum which falls off like $a^{-n}$ where $n$ is the number of steps.

@TedShifrin I got as far as "eigenvalues are equal"

@Fargle It would be useful, shits safe by now people are just tripping

2:21 AM

hey

In what, @Zach?

problem 3

It's Jessy.

(I probably shouldn't say it's exponential but something which goes to zero at infinity but not otherwise)

No more Bourne?

How's it going for you guys?

2:22 AM

Hi tern. Thanks for you response earlier. It's Ali who needs to talk with you.

@Semiclassical are you referring to my first or second problem?

Nah, I'm getting ready for Game of Thrones season now.

The first.

@Zach: But they're not!

Anyway, I was tired of looking at Cat Damon

2:23 AM

nevermind

i meant

Oh, if i can prove they're not

@MickLH: there are environmental concerns with true merit with regard to the handling of fission power, but fusion is the future.

then it shows that $\vec{v} \cdot \vec{w} = 0$

Jessy changes names more often that Skull/Skill/etc.

But you're given that they're not, @Zach.

oh shit that's in the problem statement. Q.E.D.!

tells Zach politely that he needs to learn to read and retain

2:24 AM

well I only have one tab open at a time!

Don't complain to me.

ok I'll complain to the Chrome developers

(or just use Windows' windowing feature...)

@ALannister (1) You don't need to know what the group of units of Z[G] is. (In fact, the answer to that is extremely complicated.) (2) G is already a subset of Z[G].

Would that we could power our cars that way, but I don't trust the average consumer; beyond that, I'd gladly commit political suicide to remove fossil fuel dependency, create high-tech jobs, and in general revitalize our power grid infrastructure.

@Fargle I mean even in the fission case there's failsafe techniques

2:25 AM

@arctictern I'm trying to understand the answer that Aristide fellow posted.

Absolutely, @MickLH--I don't mean the environmental effects of a nuclear meltdown, but the problems with waste management.

But is it a subset of its units?

oops

@ALannister Every group element has an inverse. That's part of being a group.

@Fargle Even this is reduced by several orders of magnitude with modern designs

2:26 AM

Of course.

If $g\cdot g^{-1}=e$ inside $G$, then that holds true inside $\Bbb Z[G]$ as well.

And reduced yet further by a transition to fusion.

Silly Jessy

Now, but what about the embedding?

@ALannister $G$ is already a subset of $\Bbb Z[G]$

@TedShifrin I'm going to save problem 5 for when I'm sane... right now I can't risk any more damage

2:27 AM

Eh, I think it's a bit dubious to talk about further reduction based on technology which we haven't gotten to work yet.

I'll work on 6 instead :P

So...identity map?

yes, $g\mapsto g$.

ah

The old canard about fusion being "Fusion is the energy of the future, and it always will be."

2:27 AM

@Fargle I think it would be unwise to miss out on some current fission work, as we can convert existing waste with half lives in the tens of thousands of years, into much less radioactive waste with half lives in the hundreds of years. A sort of industrial scale "clean up"

@Zach: You don't need to do rote work ... maybe one or two. There are more interesting exercises (including ones you "missed" in section 2).

Well, $1\cdot g\in\Bbb Z[G]$, mr tern :P

Akiva Weinberger

@ZachHauk I feel like they're most likely to take you seriously if you're like "I already took a bunch of timed practice tests and got high marks", to be honest. 'Cause most middle schoolers don't have a very large chance on a test like that, I'm guessing

Thanks @arctictern

DogAteMy: You posted a day ago about my question "is DogAteMy lost?" ... Someone had just before said you were missing or something.

@Semiclassical You have a valid point. Perhaps I just see the future in rose-colored glasses. Or maybe, since it's moving at us really fast, it's redshifted.

Akiva Weinberger

2:29 AM

@ZachHauk I heard somewhere that that was developed by Apple, actually

what was?

@Fargle: I'm purple-shifted.

Note the link there, @Zach.

Akiva Weinberger

They thought Xerox or something figured out how to do it, and spent a lot of time trying to figure it out, and then it turns out that they were mistaken about Xerox having done it in the first place

@AkivaWeinberger I thought they both stole it from Xerox

Akiva Weinberger

Windows

2:29 AM

oh, I'm dumb

@Semi: pretend I didn't send that. I'm tired and forgot which one is red.

Akiva Weinberger

Specifically, overlapping windows

as in, the GUI

afaik the windowing system was originally invented by Xerox or someone like that

@Fargle: You're just too busy being a nerd. ... Any good analysis stuff to talk about?

@Ted: We've just covered Heine-Borel and Bolzano-Weierstrass, and proved the latter with the former.

Akiva Weinberger

2:30 AM

So they were able to do it because they didn't know it hadn't been done before. Of course, @MickLH might be right, which would make the whole story less exciting

So, basically, you're snoring a lot, @Fargle.

And discussed lim sup/lim inf. I got to make a cannibalism joke.

@MickLH oh, you said it before me

@TedShifrin God, always with the analysis. What of the number theory? :'(

didn't see sorry :P

2:31 AM

@Fargle Interesting, I usually see it proved the other way around

The Star workstation, officially named Xerox 8010 Information System, was the first commercial system to incorporate various technologies that have since become standard in personal computers, including a bitmapped display, a window-based graphical user interface, icons, folders, mouse (two-button), Ethernet networking, file servers, print servers, and e-mail. Introduced by Xerox Corporation in 1981, the name Star technically refers only to the software sold with the system for the office automation market. The 8010 workstations were also sold with software based on the programming languages Lisp...

Akiva Weinberger

Or maybe Xerox had windows, and they stole that, but they invented the overlapping thingy

@MickLH: I have $<\epsilon$ interest in number theory.

@Daminark My professor is odd.

The dog ate him @TedShifrin

2:32 AM

@TedShifrin $\forall \epsilon > 0$? That's surreal.

(heh heh heh heh heh)

Probably so, Jessy :)

oh and @AkivaWeinberger said it before me too!

Bad dog.

damn I'm slow

I imagine you proved Bolzano-Weierstrass by the whole, cutting up boxes?

2:33 AM

@TedShifrin :'( I can't avoid it in the work I'm doing

@AkivaWeinberger i'll be honest i haven't touched calculus in a long while

And yet he's 14...

@MickLH: That's fine. It's not my work :)

@Ted But number theory looks so cool!

Akiva Weinberger

I'll ask my parents for details on how I got my math class set up, since I don't remember much of it

2:34 AM

do your parents support your interest in math?

To be fair your stuff doesn't use number theory

Akiva Weinberger

Very

@Daminark: It's never excited me, although I developed a healthy respect for it for pedagogical reasons in teaching algebra (namely, I started with number theory in my algebra book and returned to it a few times).

Are we having fun?

BRB gotta fold some laundry

2:34 AM

aww

Akiva Weinberger

My dad buys me so many math books. I have an overflowing bookshelf. It's my favorite possession, and I am eternally grateful to him

Heya, @MikeM

@Daminark That doesn't sound familiar, but the way my professor teaches really puts me in a fugue state, so I don't remember.

@AkivaWeinberger well that's good for you

:P

DogAteMy: Yes, I think you are exceedingly fortunate. And talented doesn't hurt.

2:35 AM

And lol, my parents are probably a bit concerned about the whole, will he be broke after graduation?

@Fargle: you're channeling Bach in class?

Akiva Weinberger

For some reason I've never wanted to believe in talent, though

@Daminark: No, you'll get your grad school paid for. It's after that that you'll be broke.

Tough diddles, DogAteMy.

@Ted: I wish I meant that kind of fugue. (I've had to play one or two!)

Akiva Weinberger

I feel like I just got interested in math earlier than other people so I got a head start, is all

2:36 AM

i usually conceal all of my study stuff because you know

But they have no knowledge of anything academic related so they just give a general reminder to make sure that I do something that has a future

family will think

less of me

Akiva Weinberger

@TedShifrin But thank you

I mean when I was young my mom had all the typical mom dreams of her son growing up to be a surgeon

@Zach: There was a guy who chatted here a lot but whom I haven't seen in a while. His mother literally burned and destroyed all his math stuff. She was super religious and a nut. Soooo sad.

2:36 AM

lol omg

@Zach: I wish I could help.

@Daminark become a manifold surgeon

Next you're gonna tell me they're part of the whole Trump anti-intellectualism band.

Not to mention other problems with it.

^^

But yeah

@TedShifrin wait what

2:37 AM

I was into that sorta thing for a while and eventually was like, look it's no happening

Akiva Weinberger

@TedShifrin Oh my god, wow

@Daminark Surgeons are often psychopaths

@Daminark: You should be fine.

It basically takes one to cut a human open with a steady calm hand

Akiva Weinberger

"Body carpenters," I think Scrubs called them

2:38 AM

@Daminark We used contradiction--let $E$ be a bounded infinite subset of $\Bbb R$ with no limit point--which must be closed--and consider a collection of neighborhoods $\{N(x)\}$ where each such neighborhood intersects $E$ only at $x$. This contradicts Heine-Borel.

Yup, tern. I can't remember his name. He was from Brooklyn, was living in a church because his mom basically kicked him out.

jesus

and not as in, religiously

Right, @Zach.

i mean, like, jesus christ (profanity fix!)

Lolol, yeah, I can't handle injury well, and my hands aren't steady at all, much less when I'm doing something of the sort

Wait wait wait @Fargle

You said you proved Bolzano-Weierstrass first

2:39 AM

And I won't mention the number of my former students who were disowned and kicked out of their houses when their parents discovered they were gay. There are some horrid parents out there.

yeah, that's sad

@Daminark The statement we proved is stronger and implies BW.

Akiva Weinberger

It was "corpse carpenter," actually

@Daminark: No surgery for you.

Akiva Weinberger

tvfanatic.com/quotes/…

2:39 AM

@Daminark I meant to say that we'd stated HB and proved BW with HB.

Yeah. Did you see the article lately about the drop in teen suicide rates?

I don't think we proved HB itself.

Yeah, when my mom realized this fact she let go of the surgery idea

That won't last long because of you-know-who and his hateful henchmen, @Semiclassic.

ugh.

2:40 AM

i heard some statistic that trans people have like a really really really high suicide rate

LOL, @Daminark.

Yup, @Zach. True.

My dad is kind of hoping that I'm going to pursue compsci

Consider the quadratic function $\mathbf{\frac{1}{2}x^TGx+b^Tx}$ in four variables where $$\mathbf{G}=
\begin{bmatrix} 2 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 2 \end{bmatrix}$$
and $\mathbf{b}=(-1,0,2,\sqrt{5})^T$. Apply the conjugate gradient method to this problem from $\mathbf{x_0=0}$ and show that it converges in two iterations. Verify that there are just two independent vectors in the sequence $\mathbf{g_0,Gg_0,G^2g_0,\dots}$.

would not surprise me @ZachHauk

He's not really pushing it

2:41 AM

@Daminark: There's a lot of fancy topology in big data/CS stuff.

Oh hey, a Toeplitz matrix.

@ZachHauk Yes. I don't recall the exact statistics, but I do recall that they tell me that every year I have a good chance to lose a close friend.

But he's all like, AI is the future, that's where the money lies.

41 percent... that's the statistic

Anyhow, @Daminark, I'm confident you'll do just fine ... the world, I'm not so sure about.

2:41 AM

I wonder if anyone's thought to send the White House a fiddle.

For reference, the article I have in mind is this: washingtonpost.com/news/morning-mix/wp/2017/02/21/…

Somehow I think that would fly over their heads anyway.

I'm going to cook dinner and see where the Oscars are. Talk to y'all soon.

@TedShifrin would you like to hear my rambling? (solution to p. 3)

Fiddletysticks @Fargle

2:42 AM

oh, no rambling for me today

You already told me, @Zach.

bye, have a good dinner :P

i mean problem 5

@Ted: enjoy your dinner!

Can't help but quote this bit of Yeats again (i'm sure i have before)

I'm pretty confident that it's not gonna be as bad as most people are worried the next few years, pretty sure most of the bullcrap was just to secure the ignoramus vote. On many of his worst statements he also contradicted himself later (or earlier)

2:42 AM

"The best lack all conviction, while the worst/ Are full of passionate intensity."

I think it's that my dad has seen more people who do what they have some passion for and find themselves unemployable than he did people who hated themselves for choosing better paid fields, so he tends to lean that direction somehow

not 3 lol

Bye, @Fargle.

No, @MickLH. You be totally wrong.

He said you shouldn't live hating yourself and having stress, but make sure you'll get paid

And yeah the whole topological data stuff is neat

@MickLH Eh, it's not just what he chooses to do. It's who he appoints to do the stuff.

2:43 AM

Talk soon, @Daminark.

On top of that, we've got this one guy named Razborov who does complexity theory + algebraic geometry, which is cool

See you @Ted

But yeah @Fargle

@MickLH: that would be more accurate if it weren't for the actions that have already been taken, the people who have already been appointed, and the people that've already been alienated.

I want to be the coolest algebraic geometer

So the way we proved Heine-Borel was basically this

We're over a month into this presidency. I think the time to hold out for "it might not be this bad" passed a month and five days ago.

2:44 AM

You prove that a closed box (or $k-$cell as Rudin calls it) is compact

$k$ minus cell?

@Daminark And then that closed subsets of compact sets are compact, no?

No, just a k-cell lol

I think I remember that proof.

Can anyone ELI5 conjugate gradient method?

2:44 AM

It's not too bad

Basically you take some open cover of a box that you assume has no finite subcover

Cut the box in half

One of the two sides also can't be finitely covered by assumption

Cut that up in half

i kid you Dam

Kek @Zach

Oh also at each stage of cutting, you choose one point in the side you will proceed to cut

Whence cometh the contradiction?

user image

Then you get a Cauchy sequence, which converges

That point is the intersection of all your boxes, but that's covered by a single open set

2:47 AM

@Daminark Aha.

(@Zach Where did you find that?)

I think he didn't teach us that because there are seven of us, and about five of us wouldn't have gotten that proof.

imgur

40keks

But yeah, so then that single open set basically handles all but finitely many stages of cutting

actually I might have some kek stuff in 4chan folder

2:48 AM

And then all you have to do is, like I said, prove that closed subsets of compact sets are compact, and that proves HB.

Yup

There's another way to do it actually

HB?

Yeah

You can use suprema to directly prove that $[a,b]$ is compact

Then prove that product of compact sets is compact

I saw a much longer and more arduous path, through "[a,b] is compact", "products of compact spaces are compact", etc

Yep, exactly.

Akiva Weinberger

@Daminark I like this proof

2:50 AM

Spivak does it in calc on manifolds, which I kinda liked

Akiva Weinberger

(Though I used infima)

Lol, well the way I saw it was

Take an open cover of $[a,b]$, call it $O$

what is HB?

@Zach: the Heine-Borel theorem, that a set in $\Bbb R^n$ is compact iff it is closed and bounded.

Akiva Weinberger

@ZachHauk Do you know what an open cover or what compactness are?

2:51 AM

i've heard of cover

Now let A = {x $\in [a,b]$ : [a,x] is covered by finitely many sets in $O$}

Hi there.

Akiva Weinberger

An open cover of $X$ is a collection of open sets such that $X$ is contained in their union.

compact contains all it's limit points

right?

@Zach: not quite. That's closed.

Akiva Weinberger

2:51 AM

A set $X$ is called compact if every open cover has a finite subcover.

subcover meaning

another cover whose union is a subset of the other open cover?

Nope

This is awesome, @AkivaWeinberger.

A subcollection of the open cover which is still sufficient to cover the whole set.

So a cover is a collection of open sets

Akiva Weinberger

2:52 AM

@ozarka What is?

A subcover is some subset of that collection

Which still covers what you want

Being able to learn analysis.

@AkivaWeinberger

what?

Akiva Weinberger

(1/2) So, like, if I give you $A_1,A_2,\dots$, where they're all open and where $A_1\cup A_2\cup\dotsb\supseteq[0,1]$,

@ZachHauk So, like, if we consider ALL intervals $(a,b)$ in $\Bbb R$, they cover $\Bbb R$, but the subcollection of intervals $(a,b)$ where $a,b$ are rational STILL covers $\Bbb R$. This subcollection is a subcover.

2:53 AM

then there exists some $B_1,B_2,\dots$ such that $A_1 \cup A_2 \cup \dots \supset B_1 \cup B_2 \cup \dots \supseteq [0,1]$?

(Actually $\mathbb{R}^n$ is Lindelof so the fact that you used countably many is completely general @Akiva :P)

Uh, more like

Akiva Weinberger

(2/2) then there's a finite collection of them that's a cover also. So, maybe, for example, $A_1\cup A_5\cup A_{12}\supseteq[0,1]$.

what if there's only one set in the cover?

@Daminark Is $\Bbb R^n$ Lindelof? Somehow that surprises me but I guess it shouldn't because it has countable base.

Akiva Weinberger

@Daminark Remind me what that means?

2:54 AM

Every open cover has a countable subcover

@ZachHauk Then it has a finite subcover automatically. But to be compact, EVERY cover had better have a finite subcover.

Akiva Weinberger

@Daminark Ah.

Yeah, so let's say $(a,b)$

cover being a set of countable open sets, right?

Or, let's just say $(0,1)$

Akiva Weinberger

2:55 AM

Not necessarily countable

@ZachHauk Could be uncountable. Could be countably infinite. Could be finite.

Akiva Weinberger

We need their union to contain the thing though

i'm confused.. what makes $\{(-0.25,1.25)\}$ a subcover?

You can cover it with $(\frac{1}{3},\frac{2}{3}) \cup (\frac{1}{4},\frac{3}{4})\cup \ldots$

rather than just a normal cover?

2:56 AM

But finitely many sets will never be enough

@ZachHauk Subcovers only exist relative to there being an original cover.

Akiva Weinberger

A subcover is a subset of the cover. Throw out some of the open sets.

@ZachHauk Example: Say a function on $[0,1]$ is "locally increasing" everywhere: for every $x$, there is an open interval containing $x$ such that the function is increasing on that interval.

yeah but isn't it a cover too?

@ZachHauk Every subcover is also a cover.

Akiva Weinberger

@Fargle ?

Might not contain the thing we're trying to cover

2:58 AM

ok, but then it doesnt have a finite subcover... right?

if it only has one element

@Akiva: which is why I said subcover and not subcollection.

Akiva Weinberger

The subcover doesn't need to be smaller

@ZachHauk Every cover is a subcover of itself

oh, so it's like a $\subseteq$ thing?

Akiva Weinberger

Yeah

and so the stickler is the "finite" fact

Akiva Weinberger

2:59 AM

@ZachHauk Actually, a different example: Say we want to show that continuous functions on $[0,1]$ are bounded.

because with infinite subcollections, you can't just take itself

Akiva Weinberger

Part of the extreme value theorem.

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