MSE should allow LaTeX in usernames. that would be wicked.

So you could write your username in colors?

:D.

Hah.

Can anyone suggest a reasoned argument why the various things called "topology" ought to share that name?

On one end, people who study the large-scale behavior of manifolds under homeomorphism.

On the other, spaces with weird structure and odd cardinalities which are perhaps Hausdorff if you're lucky and the phase of the moon is right.

That's a good question. I always liked the doughnut=coffeemug kind of topology. But all the technical stuff on open sets and so forth was never appealing to me.

Topology began with the classic spaces, and whatnot.

Then people noted that you can define other kind of topologies.

Then algebraic topology rose from the ground and devoured most of the topology.

The logic-inclined topologist got bored and began working on set theoretic topology.

All those cardinal invariants, etc etc.

Alas you must also consider the analysts which saw further than their own foot and extended their research to spaces which behave like R^n, i.e. manifolds.

Ugh. I have to prepare for tomorrow's class. I have to talk about permutations. Why? I have no idea. It's gonna set me back a week further away from the professor too. :|

Thanks for the history lecture :-)

Still, somebody ought to have thought up a new name for one of the branches along the way.

This is why we have:

1. Geometric topology

2. Algebraic topology

3. Set theoretic topology

4. Classical topology

5. Yo momma so fat topology jokes.

E.g.: Yo momma so fat she ain't even first countable!

let # count the number of bits in the binary expansion of n. I noticed yesterday that #(a+b) <= #a+#b

@QED Not surprising.

I was excited because I thought I could solve the buckets problem with that... but it uses subtraction instead

Hmm.. Is the norm closure of a subset of a normed vector space really contained in the weak closure? Or is that a typo on the assignment?

sorry

I said it wrong

I meant: let # count the number of 1's in the binary expansion

@anon thanks for helping me realize that mistake

1111 + 1 = 10000

yes so #10000 <= #1111 + #1 (1 <= 4 + 1)

for that example

maybe it can be proved with induction

induction on the trace of the binary + algorithm

anyway I can't solve my puzzle with it

You should consider the binary expansion as encoding finite subsets of the natural numbers.

so I was wondering if there is anything else you can solve with it

@Asaf: How would that help on the # problem? You get things like 01+01=01, I would presume...

@anon: There is some encoding in which it works fine, I can't recall it right now though.

F2[x]?

No, Z2 maybe..

but regardless the # function doesn't seem to respect the algebraic structure very cleanly

I think it is the encoding of hereditarily finite sets which is useful.

But does addition have any nice interpretation under that encoding?

@Matt Not a typo.

@HenningMakholm Should correspond to union or symmetric difference or something like that.

@AsafKaragila As long as you don't have any carries, that's fine. But a carry will turn two copies of one member set into a completely different one.

I couldn't write enough "?"s when someone wrote (for all 0 < s < 1 and n in N) that sin(n pi s)^2 = 1/2!

@HenningMakholm I noticed that, yes.

I don't understand the Lang joke. Wasn't Lang part of Bourbaki?

No, but he wrote with them a book or two.

The Lang joke is because Lang has written so much.

OEIS has the finite von Neumann ordinals as A034797, but doesn't identify them as such.

Hi! You guys are awesome, have a lovely day ! Cheers =)

Neat!

Yes, I do know that Lang has written so much, but I don't understand the joke.

The joke is that even a team of people couldn't compete with Lang

You're not just killing the joke, Jonas, you're burying it in a shallow grave to exhume it quickly thereafter and kill it again.

Can one type LaTeX in chat?

$5x + 3 = 8$

One can type, the chat won't compile it though.

Darn shame

A Buddist walks over to a hot-dog stand and asks, can you make me one with everything?

@N3buchadnezzar No he doesn't.

You are killing the joke, Asaf, you're burying it in a shallow grave to exhume it quickly thereafter and kill it again.

No, I'm not.

That joke was a stillborn.

You are right, and just to top it of the Buddist never recieved any change for this his food either. Because change must come from inside

lol

I love math jokes

When I grew up my best friend was sqrt(-3).

I wonder how many zero bits there are after subtraction

#(a-b) >= #a + #b?

where # counts zeros this time

Fustenberg! What a hero.

@QED No -- consider 3-1. Your conjecture is then 1>=2+1.

ok

are there any measures that work on subtraction?

The co-measure? It returns 0 for the entire space and 1 for null sets :P

i'll never solve this :D

I'll solve your riddle if you write my thesis for me.

@AsafKaragila Don't you think that stuff only goes well when you do it yourself?

You mean solving riddles?

Does anybody want to hear a math joke?

No.

I need a number between 1 and 13.

@AsafKaragila Or writing thesis's.

@Skullpatrol No.

@AsafKaragila Pi/e.

An integer in that range.

Strictly between?

@JonasTeuwen Yeah, that too.

2.

@JonasTeuwen No, not strictly between.

Hmmm, 2 is no good.

1.

One is the loneliest number that you'll ever meet. It is also not very good.

Then why don't you say between 3 and 13?

I don't like 3.

How about 7?

Too hard, pick again.

...!?

9.

Oh, now you're just guessing.

Yes, I don't even know what I am guessing.

You guessed 9.

I need to solve some questions from the homework assignment tomorrow. I have like 13 possible questions to answer, and I'm checking your suggestion against whether or not it would be a good problem to solve in class.

Okay. 13?

Possible, but it's a mild variant on 10 which I'd have to solve for that. I guess I'll just solve 10.

Nice.

wow - we bashed Skull in - no math jokes for us

I'm confused by this. Where does the basis orthogonal to W come in if he wants to project onto W? I'm serious, I remember nothing of linear algebra.

Good night.

Good night!

@Matt He needs to project orthogonally to the subspace he's projecting to -- otherwise anything could be the result.

Good nigt, Jonas.

@Matt But it would be easier for him to normalize his orthogonal basis; then he could get v-proj(v) simply by multiplying the projection direction with its dot product by v.

@HenningMakholm Oh, I think I see. Does the basis in which he expresses his projection matrix have to be orthogonal to W?

Not sure I follow. The projection matrix itself (i.e., the matrix of the projection mapping, I suppose?) is probably expressed in the standard basis.

Oh.

But he's not showing the projection matrix explicitly, so I'm not entirely sure how your question relates to his.

I think I need a brush up of this.

The canonical procedure would be something like: Find an orthogonal basis for W. Extend it to an orthogonal basis B for the entire R^4. Take v, move it to basis B, zero out the coordinates that correspond to basis vectors not in W, move back to the standard basis.

It turns out that you don't actually need to orthogonalize the basis for W, but the new basis vectors from the rest of R^4 must be orthogonal to W; otherwise you don't get a unique result.

I think I should go to sleep : (

Good night folks!

Night.