Dylan Moreland

Does that mean anything

Hm, true.

@jshin47 What doubts do you have?

@PeterTamaroff I seem to recall that Matt was having some issues with it. Maybe I'll dig up that thread.

An idle question for the regulars: do you think that the question limit has changed anything?

Back in a bit, maybe. Good luck with topology!

I'm not disappointed. I don't know how else to answer these questions, though.

But I don't see how any of that interacts.

It's also interesting to note that topologists have uses for modular forms. See this, for example.

Jacob Lurie.

But computing homology of certain manifolds with rational coefficients? That sounds like algebraic topology to me.

You need some other input. Often that has to come from Lie algebra methods or representation theory or blah blah.

But I would say that the purely algebraic machinery you'd find in Hatcher is usually not enough to tell you what you want.

The book by Ken Brown is a good one.

Matt has been doing a lot with the cohomology of arithmetic groups. That's something to look into. That's a wild subject.

Look at what Lurie is doing, for example.

At a high enough level algebraic topology seems indistinguishable from hardcore algebraic geometry to me.

There's still a lot of wiggle room there.

I don't think there are serious uses for general topology.

In a sense, almost everyone uses topology.

@PeterTamaroff It depends on what you mean.

Hard to say why. He looks for insight. That seems like an obvious thing to do but I don't know, he puts it into practice about as well as anyone I've seen.

He's pretty amazing.

Yes.

@PeterTamaroff Probably. I'm a grad student. I work on number theory under the user Matt E.

@PeterTamaroff Ah, good for him then.

@PeterTamaroff Hi Peter. How're things?

[I cannot believe that I logged back in just to figure this out, but:] What happened with Arturo?

Like here, you could pick $y = (y_1, y_2)$ such that $y_1 = x_1$. So with your $d$, $d(x, y) = |x_2 - y_2|$. That's nice and simple.

Definitely. There are lots of those in the ball. But it seems to me that you should write down a nice formula for a certain $y$ which makes things easy to check.

You'll just draw a box instead of a disk around $x$. That's okay.

But the one you've written is strongly equivalent to it, so topologically there's no difference.

I guess most people would use the Euclidean metric.

Oh. It doesn't really matter.

Might help to draw it when $n = 1$, so we're looking at $n + 1 = 2$, the plane.

So I give you an $r > 0$. Give me a $y \in B(x, r)$ such that $y \notin A$.

I think you can. Take a point $x \in A$. To show what you want it's enough to show that no ball centred at $x$ is contained in $A$.

@PeterTamaroff “Empty” is probably not a good idea.

You could probably just reduce this to negating a statement.

Well, look. Let $x$ be a point of $X$. Either there is a neighborhood of $x$ entirely contained in $A$ or $X \setminus A$, or it is the case that all neighborhoods of $x$ intersect both.

Arturo bringing out the autism.

Okay nevermind same old Makoto.

@anon Rap really does make people violent huh

Indeed it does.

Now everything is into $\mathbb R$ and you're very happy.

At any rate, one way to go is to say that a map into the product is continuous iff the components are.

Then you should delete it from the domain.

@BenjaLim You do not want that formula at $0$.

The problem with answering Makoto's questions is that they get punted off the main page realllly fast.

Hello John.