Mathematics - 2014-04-02 (page 1 of 4)

Jasper Loy

00:00

Anyway I have already deleted my 20k account, so now nobody can complain that I am cheating, lol.

Thomas Andrews

Of course more people know the old books. Isn't that the nature of "old" in text books? Not true in fiction, but one can only read so many introductory to algebra books and really get much new out of them. It's not like reading the latest mystery novel.

Jasper Loy

I spend a lot of time surfing amazon, so I know many new books, lol.

First I look at the contents page.

After that I try to find a copy somewhere online to browse.

There are some books where no matter how hard I try there is no copy online, lol.

Thomas Andrews

But knowing them well enough to recommend them over the current "standards"?

Jasper Loy

Well, I have not read too many books, but I have browsed through many, I think it is OK to recommend if you have spent about half an hour browsing.

Mike

00:37

@PedroTamaroff Are you with Fernando?

Pedro Tamaroff

No, why?

Jasper Loy

@Mike No, Pedro is straight, lol.

Mike

Not what I meant.

@PedroTamaroff I wanted you to tell him that homological algebra sucks.

Jasper Loy

@Mike Why?

Mike

Because it does.

And he should know.

Jasper Loy

00:41

OK. I was wondering whether there is cohomotopy...

Pedro Tamaroff

@Mike You suck.

Sucker.

Mike

That's a lie.

Studentmath

I guess he doesn't think it sucks.

Mike

He may not think it, but that doesn't mean it's true.

Pedro is fallible.

Pedro Tamaroff

@Mike Jacobson has forced me to read his chapter on Categories.

Dang it.

Jasper Loy

00:43

@PedroTamaroff I like Jacobson.

Pedro Tamaroff

@JasperLoy I like him too.

Jasper Loy

@PedroTamaroff But, as you know, I think the three volumes by Cohn is the best, lol.

Mike

I think you're the only one.

Cohn does too, I suppose.

Jasper Loy

His books really deserve more attention, sigh.

Studentmath

Maybe Cohn is incapable of appreciating his own work too..

Jasper Loy

00:49

Anyway, Cohn and Rudin are both Jewish, lol.

Anthony

@PedroTamaroff If I want a subspace of R containing Z where {-4} is open but {1} is not, why is [-5,-3] U Z wrong?

Studentmath

@JasperLoy Perhaps I have a future after all!

Jasper Loy

@Studentmath Hehe

Anthony

Nvm

Pedro Tamaroff

@Anthony $\{-4\}$ is not open in $[-5,-3]$ with the subspace topology.

It's the other way around. =)

Anthony

00:56

Waaaaat

Pedro Tamaroff

$\{1\}$ is open in your subspace.

Studentmath

Can I state that if a Tree has at least three vertices, its diameter is at least 2?

Pedro Tamaroff

Because it is $B(1,1/2)\cap \rm your\; set$, say.

Anthony

Oh.

Crap.

Hmph. Thanks.

Pedro Tamaroff

You see it, yes?

Jasper Loy

00:58

Just open your eyes and you will see.

Anthony

Yeah I see it.

Agh.

I still get confused about things like open subspaces, in terms of connectedness

Pedro Tamaroff

Time to time.

Anthony

Because they say it's disconnected if there are two disjoint proper clopen subsets.

Pedro Tamaroff

@Anthony Well, is your subspace connected?

Anthony

The one I constructed?

Pedro Tamaroff

01:00

Aha.

Anthony

What?

Pedro Tamaroff

Is $[-5,-3] \cup \bf Z$ connected?

Anthony

I mean, I would hope not, in R2, but I'm not sure how to prove that it's disconnected.

Pedro Tamaroff

Well, can you exhibit open sets or closed sets in your set, $A,B$ with $A\cup B=\rm that$ and $A\cap B=\varnothing$?

Anthony

open or closed?

I thought they had to be clopen!

Mike

01:02

@Anthony Open. Prove that they're clopen.

Pedro Tamaroff

In fact, it suffices you find open sets in your whole space with $A\cup B\supseteq \rm that $ and $A\cap B=\varnothing$.

@Anthony They will be clopen in your subspace, of course.

Mike

@PedroTamaroff Only give Anthony vague hints, he's good enough to do most of it on its own.

Anthony

@Mike <3

Mike

Probably all of it but hey what's MSE for if not vague hints

Alec Teal

My favourite hints are the useless ones

4

Anthony

01:11

Agh

Wait sorry

@PedroTamaroff I'm confused by the clopeness though.

An interval [0,1] is a closed interval, not open.

But as a subspace, it's clopen?

subset*

Mike

@Anthony What's the definition of a topology?

Anthony

@MAll open subsets of a set?

Pedro Tamaroff

@Anthony Yes, as a subspace on it's own it is closed and open.

Anthony

Isn't every subspace clopen, then?

Pedro Tamaroff

Yes.

But "subspace" is different from "subset".

Anthony

01:17

Er.

Crap.

Pedro Tamaroff

When you say "subspace" you're seeing the subset as a topological space on its own, with the subspace topology.

Anthony

So then only some subsets are clopen?

Pedro Tamaroff

Aha.

In fact a space is connected iff the only clopen subsets are...?

robjohn

@Anthony If the space is connected, there are only two

@PedroTamaroff oops

Anthony

Itself and the empty set?

Pedro Tamaroff

01:21

@Anthony YAS.

Why?

Proof?

Anthony

Well we define connected as not disconnected, so then there are not two disjoint clopen subsets.... I don't really know. And I'm still confused on what kind of subset wouldn't be clopen.

Mike

You do know.

Assume a subset of $X$ is clopen. Then...

Anthony

If a subset is clopen then it has it's limit points, and you can also make open balls around all points!

Mike

I mean I guess that's true but I don't care.

Relate this fact to connectedness.

Or, rather, disconnectedness.

Greg Ros

In the definition of an interior point in a metric space, it says that there exists an open ball that contains that point. Am I allowed to choose a specific, arbitrary open ball such as in order to construct a function between the set of points and the set of balls?

Anthony

01:26

Well it is connected...

Mike

No, there's no reason a clopen set should be connected.

Anthony

That's true, agh.

Mike

Don't just say the first thing that comes to mind. Write the definitions down and play with them.

Would you like to give me a theorem for you to prove, to focus your attention?

Anthony

I suppose, I just don't know what definitions to work with.

Mike

Topological ones.

Anthony

01:30

Well yes, just the only thing with connectedness I know is that a disconnected set is one with two proper disjoint clopen subsets

Mike

Clopen means both open and closed. Closed means "complement of an open set". Disconnected means "Can be written as a union of two proper disjoint open subsets". Connected means not disconnected.

Anthony

Why is open?

Mike

That's the definition I'm choosing.

Anthony

I thought my professor said clopen... Is there a reason?

Mike

Yes, and you're going to find it out on your own.

Anthony

01:32

Alright.

Mike

Now let's say you have a proper clopen subset $A \subset X$.

Anthony

Yes.

Mike

Edited.

@Anthony Prove that $X$ is disconnected.

Anthony

With just one subset?

Mike

You have everything you need.

You don't need to appeal to anything except that set of definitions I gave you above.

Pedro Tamaroff

01:35

Also chicken.

Anthony

You can take A and A complement to form X.

And they're both clopen.

Oh but they aren't disjoint.

Mike

Why aren't A and A complement disjoint?

Anthony

Oh

Oy vey.

Nevermind, they are.

Mike

:P

Anthony

So then one proper clopen subset implies disconnectedness...

Mike

01:38

Yep. Now let's do the other direction.

Assume $X = A \sqcup B$, where $A$ and $B$ are both open.

Prove that $A$ and $B$ are both clopen.

Pedro Tamaroff

@Mike Disjoint union.

Anthony

Well A complement is B

and B complement is A

And since they're both open

Mike

@Anthony Yep.

Anthony

They both gotta be closed too.

Mike

You've got most of the story about clopen sets now.

Do you know the definition of a connected component?

Nah, I'll just give you one.

Pedro Tamaroff

01:42

@Anthony Connected components are fun!

Mike

"Let $x \in X$. Then the connected component of $x$ is the union of all connected open sets that contain $x$."

I dunno of a notation for it but let's call it C(x).

Alexander Gruber

hey look, vi is using kalle numbers

Mike

Okay, @Anthony?

Pedro Tamaroff

@AlexanderGruber I luv her voice. Heh.

=)

Mike

@AlexanderGruber how does she feel about the following proof

Alexander Gruber

01:43

@PedroTamaroff yeah man!

Pedro Tamaroff

@AlexanderGruber I think the voice of a woman is so ridiculously important. Gotta love radio hosts.

Mike

"$(1, 1, \dots )$ is a Cauchy sequence and $(0, 0.9, 0.99, \dots)$ is too. Their difference is a Cauchy sequence converging to $0$. Thus $0.9999... = 1$"

Alexander Gruber

@PedroTamaroff do you watch house of cards?

Pedro Tamaroff

@AlexanderGruber No, but I should.

Alexander Gruber

@Mike it's an april fool's joke, she isn't actually a crank.

Mike

01:45

i know.

i didn't know it was an april fool's joke cuz i didn't watch the vid

Anthony

@Mike I see.

Mike

@Anthony You cool with the definition?

Pedro Tamaroff

I thought the Hendricks girl from Mad Men would have a nicer voice, but she sounds like she's 18. =O

Anthony

Maybe.

Mike

@Anthony Clearly $C(x)$ is open.

Anthony

01:46

Sure yeah.

Mike

Really? Or are you not sure?

Anthony

Union of open sets?

Mike

Okay, yeah.

Just wanted to make sure :P

@Anthony Prove that it's closed.

Alexander Gruber

@PedroTamaroff this chick has a crazy good voice, from house of cards.

Mike

This is trickier than the other stuff I asked you to prove but it's not too bad. Tell me when you've got a proof.

Pedro Tamaroff

01:50

@AlexanderGruber OK, my #1 choice is this.

Anthony

@Mike Can you just say that whatever remains has to be open because it is disconnected? So what we have, is closed?

Mike

Nope.

@Anthony Hint: Consider the set of connected components.

@AlexanderGruber is the kalle numbers question still visible to mods?

Alexander Gruber

@Mike eeeverything is visible to the mods

Anthony

The set of connected components is open... Is it's complement disconnected?

Mike

@Anthony Well, the union over ALL connected components is just $X$.

(In particular, $x \in C(x)$ for any $x \in X$)

Anthony

01:59

Oh true.

Mike

@AlexanderGruber are you willing to hook a buddy up

Alexander Gruber

@Mike what do you mean?

Mike

i wanna see it

Anthony

I don't see how that helps, though.

Mike

Well, I don't think you've considered it properly.

That's a good hint and I'm gonna stick with it.

Alexander Gruber

02:02

@Mike can't you?

Mike

This question was removed from Mathematics Stack Exchange for reasons of moderation. Please refer to the help center for possible explanations why a question might be removed.

Alexander Gruber

ahhh i see

Mike

viewing deleted posts is part of 10k-tools

Alexander Gruber

@Mike ohhh i see, yes it's part of those

Mike

oh well.

Anthony

02:08

@Mike Agh Mike I gotta do my actual homework, I guess I'll write you again some other day. Thanks for your time and help, though :/

Mike

@Anthony Anyway, the punchline is that a set is clopen iff it's a connected component.

Anthony

Clopen if it's a connected component... I see.

Mike

iff means if and only if.

Err.

Sorry, that's wrong.

A subset of a space $X$ is clopen iff it's the union of connected components.

Victor

Anyone could answer my question? math.stackexchange.com/questions/735980/…

Mike

I have no idea what you're asking

Victor

02:12

which part you misunderstood? @Mike

Anthony

I still don't understand, though, why my professor used clopen subsets for disconnected, and I also still don't get why [0,1] is clopen-given that they always just say it's closed.

Mike

@Anthony We covered that earlier!

If a set is disconnected, and we can write it as a disjoint union $A \cup B$ of open sets, then those sets are both clopen.

But if they're clopen clearly they're both open.

Anthony

Oh I see.

Mike

So your prof and I secretly agree on what it means to be disconnected.

Anthony

:P

But also, why is [0,1] open?

Mike

02:14

Remember that clopen is nonsense unless you're given an ambient space to be talking about it in.

[0,1] is clopen as a subset of what space?

Anthony

$\mathbb{R}$?

Mike

That's not true.

Especially since $\Bbb R$ is connected!

Anthony

Oh....

I see.

So then if I have [0,1]U[2,3]

Then [0,1] is clopen as a subset of that set

Mike

Well, remember that an open set is the complement of a closed set.

We agree that [0,1] and [2,3] are closed.

Anthony

Yep.

Mike

02:16

(You guys are talking about topology, right? You've defined a 'subspace topology'?

Anthony

Yeah.

Mike

What's an open set in the subspace topology?

Anthony

At least I think. It's just a subset with the topology of the parent set, right?

Mike

Nooooo.

Anthony

Hnnnnngh.

Mike

02:18

I'd keep talking but I gotta go.

Look in your notes and all that.

Anthony

Thank you.

Mike

02:34

@Victor I know what you're looking for - I'll write up an answer when I get home.

Victor

Appreciate @mike

Alexander Gruber

03:33

question is for norms in general, though the one i'm looking at is $\|\cdot\|_2$

Mike

what's the triangle

Alexander Gruber

@Mike when $a$ and $b$ are on the same ray, in other words

Mike

oh, the discriminant of the quadratic?

cubic? whatever.

math is hard.

Alexander Gruber

i guess. i don't know, i'm shit at analysis.

Mike

well i mean what does the $\Delta$ mean

i don't even know what that is.

Alexander Gruber

03:36

$\Delta abc$ is the triangle formed by $a$,$b$, and $c$

Mike

lame.

Anthony

Quick question!

How can I directly prove the 2-sphere is closed?

Alexander Gruber

it's giving me this as part of a numerical analysis proof and i have no idea where they got it.

@Anthony can you prove its complement is open?

Mike

you can do that explicitly @AlexanderGruber

Anthony

Would that be easier?

Mike

03:37

pick $(x,y)$ and $(x',y')$ in the plane...

Alexander Gruber

@Anthony i find it usually is

Anthony

I don't really know how to prove things are closed directly, in general...

Alexander Gruber

@Mike well i'm wondering about its generalization

to arbitrary norms

Mike

arbitrary norms on what

Alexander Gruber

normed spaces

Mike

03:38

normed spaces over what

cuz i'm pretty sure this is gonna be hopeless for a nonarchimedean norm

Alexander Gruber

@Mike like, normed vector spaces. over fields.

Mike

give it an archimedean norm and i bet it's true.

Alexander Gruber

gosh what an awful discipline. they should just seal all the analysis books in cement and banish them to the bottom of the ocean.

Mike

analysis does cool things sometimes.

but norms are algebra too, my friend.

Anthony

I'm still confused though-is there a good way of proving things are closed from the contains it's limits definition?

Mike

03:41

yeah

Alexander Gruber

@Anthony okay you're looking for closure of the $2$-sphere

Mike

you can prove that it contains its limits.

Anthony

But for an arbitrary set that's kind of hard, isn't it?

Alexander Gruber

@Anthony well you can't prove arbitrary sets are closed

Anthony

@AlexanderGruber I'm actually aiming to show that the 2sphere isn't homeomorphic to R2

well I mean some arbitrary closed set.

Mike

03:42

erm

@Anthony the point is there's no universal technique, other than "prove it's closed"

Anthony

Yeah... Alright.

Mike

and ihave no idea how proving that the sphere is a closed subset of $\Bbb R^3$ helps you prove it's not homeomorphic to $\Bbb R^2$

Anthony

Er...

I thought closed &bounded => compact by heine borel and R^2 is not compact

Mike

Okay, good answer.

I thought you were trying to use that fact alone :P

Anthony

Naw

I dunno how to go about closedness though, still. Show the complement is open sounds gross

Can I set up a continuous function to a closed set?

Mike

03:45

that's like

wronger than wrong

Anthony

Really?

Mike

showing that the complement is open is like snoozezone.

Anthony

Really? I'm so shaky on this stuff. :/

Mike

you'll learn

it's a matter of building intuition, and that takes time.

the best thing you can do as an aspiring mathematician is lots and lots of examples.

also, as a professional mathematician.

Anthony

I see.

I mean, I should just take a ball of radius half the distance to the nearest point on the sphere, and that ball is always contained

Right?

Mike

03:50

yes

prove it

I'm so glad I bought colorful comments.

Anthony

Yay

Now I need to see if R^2 without Q^2 is connected...

Mike

lol

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