Anything involving neighbourhoods?

No... we just defined $B_{\rho}(x, \epsilon)=\{ y \in X: \rho(x,y)< \epsilon\}$ and said that U is open if for each $x \in U$ there is a $\epsilon>0$ such that $B_{\rho}(x, \epsilon) \subset U$. @DanielFischer

@evinda I meant a characterisation of closed sets in terms of neighbourhoods. Do you know a characterisation of open sets via neighbourhoods?

@JasperLoy Hi there - how are things?

Keep trying, bro!

Not sure about Jonas - met up with him 2 weeks ago, and he was just about to hand in the final version of his PhD thesis, so he would have been very busy indeed

He just stayed for an hour and had a couple of drinks with me and my wife

Very happily retired, thanks - enjoying a lot of hill-walking and music - I have started with piano lessons :)

@DanielFischer We haven't such properties in class... Could you tell me which we have to use?

@JasperLoy That sounds interesting. I did some work on set theory and logic many years ago and liked it.

@JasperLoy I will continue doing my master...

@evinda Yeah, sorry, I overlooked that above. That's good. Now, what is the definition of closed sets you're working with?

I did it as part of a "taught Masters" degree that I started purely to convince people that I was capable of doing a PhD :)

I never finished the masters, but I did get the PhD :)

@JasperLoy I still have just the same old account

I remember getting it recently - let me check ...

Can't find the email - if you email me again at googlemail, I will send the pic

@DanielFischer A set $K \subset \mathbb{R}^n$ is closed if its complement $\mathbb{R}^n \setminus{K}$ is open.

@JasperLoy replied!

@JasperLoy Continuity and differentiation

@BalarkaSen

No, I haven't. @JasperLoy

why is this true ?

@L33ter mhm?

@evinda Good (substitute an arbitrary space for $\mathbb{R}^n$). So we know that $X \setminus \overline{A}$ is open. Using the definition of open sets, if $d(x,A) = 0$, can $x$ be in $X \setminus \overline{A}$?

p : E --> B is a covering map then for each b in B the subspace p^{-1}(b) of E has the discrete topology. For each slivce Valpha is open in E and intersect the set p^{-1}(b) in a single point.

why does it intersect in a single point?

Recall that V_\alpha is homeomorphic to U.

The homeomorphism being precisely $p|_{V_\alpha}$

ohh

oh I see

@JasperLoy roflmao!!!

ok ok good

@JasperLoy Good! My main aim in life at the moment is to climb 214 hills/mountains in the UK :)

I wonder if they have a cobel prize, for comathematicians.

Whoops, I meant comathematicians, for the cobel prize.

@JasperLoy every Tuesday I walk 5-6 hours nonstop over mountainous countryside - I love it :)

@DanielFischer Is the above definition of the open set an iff statement?

@evinda Definitions are always iff.

@JasperLoy Excellent!

Hi @robjohn we keep missing each other

@OldJohn How is your weekend? We just put our Holiday tree up...

@L33ter Can you prove why $p^{-1}(b)$ is discrete?

@JasperLoy Yes, I need to get some sleep soon

Not really hard, just a quicky-checky thingy.

@robjohn fairly quiet here - no tree uet

@JasperLoy Bye for now!

well we proved already singletons are open it is trivial to show that a topology will be discrete if all of its singletons are open

since a topology is closed under union

@robjohn Will drop by again soon - right now, I have to go, I'm afraid

arbitrarily union

right, ok.

Night all.

@DanielFischer night

so there's a generalization of covering spaces where fibers are not discrete.

those fellows are called fiber bundles.

just as a remark.

cool

what are you studying right now?

covering spaces

I need to finish covering spaces and fundmental group of circle today

Make sure you do exercises.

alright

You have an exercise due from me, remember?

yeah

the star convex problem

it is in munkres I noticed

in the fundmental group section

mhm

fundamental group *

and also the abelian fundamental group problem. :)

yeah I want to do that problem

@OldJohn Enjoy the winter weather!

If a continuous function from a disk to the plane is the identity on the boundary, then the disk is a subset of its image right? The argument I was thinking of was to suppose not. Then there is a point x in disk not in image. Then we can take a point d in disk and consider f(d).

There are paths p1 and p2 such that the product of p1, the path from terminal point of p1 to initial point of p2 along the boundary, and the inverse of p2 "contains" the point x. Then get some fundamental group contradiction? Any thoughts? This may not be a good approach...

@user2154420: Your conclusion is correct. Yes, one wants to use the fundamental group. Suppose there were a point in the disc not in the image. By picking an appropriate homeomorphism we may as well assume this point is zero (if you don't like this, just assume it's zero for notational purposes).

Then our assumption would be that we can extend the map $S^1 \hookrightarrow \Bbb R^2 \setminus \{0\}$ to a map from the disc. That is, this map is null-homotopic.

But that's obviously false since this map represents a generator of the fundamental group of the punctured plane.

I was going to go with that approach, but wouldn't that assume that zero is the only point not in f(D)?

All we have is there exists some point not in f(D) for the contradiction. We don't know how many

How does it assume that at all?

Oooohhh... wait

The map doesn't have to be surjective in any way

:)

Which is the definition of the degree of a prime ideal of an algebraic number field?

@anon do you have an idea?

Wait I am getting confused about the last part of the proof... I know that if every map from S^1 to some top space X is null homotopic then this is equivalent to every map extending to D^2 to X and the fundamental group at any point being trivial... so how does this one example show a contradiction?

I guess I don't understand the part "this map represents a generator of the fundamental group of the punctured plane"

Perhaps I don't understand fundamental groups as well as I thought...

@user2154420: For laziness I'm going to call the pinctured plane $X$. What is $\pi_1(X)$?

@anon Do you have an idea why it holds that $\frac{||Ax||}{||x||}= \left \{ A \left( \frac{x}{||x||}\right) \right \}$, given that A is continuous?

Z

Can you tell me a loop that represents $1 \in \Bbb Z$?

S^1?

Yes, but you should understand why. This goes back to your computation that pi_1(X)=Z.

It doesn't actually make sense to say S^1 represents 1, because S^1 is an abstract topological space. You meant the unit circle.

Now the key is: elements of pi_1(Y,y), where Y is some arbitrary space and y is a point in it, are by definition equivalence classes of basepoint preserving maps (S^1,1) \to (Y,y). And one of these is the 0 element precisely if this map extends to a map from D^2.

You should understand why.

So the unit circle represents 1 because the distinct elements of our group are how many times we go about the origin

because there is no way to continuously deform a loop that goes around once into one that goes around twice

kind of like if I had the circle around the origin vs. if I had an infinity symbol, but bent the infinity so both loops go around the origin

Hi @MikeMiller @BalarkaSen

@user2154420 intuitively, no. but the proof is complicated.

Gosh, this is so annoying ...

Hi @PVAL.

Right, the proof involves covering spaces.

?

If you know what $pi_1(S^1)$ is there isn't much to do..

I am throwing cautions to the air, in case @user2154420 is talking about all this intuitively.

@PVAL One can know it but not be comfortable with it, of course...

Also, hello

I think I get it now

So really what's left to prove is if a map is null homotopic on S^1 then it extends to a map on D^2

Thanks!

Cool fact I can't resist talking about: $SL_n(\mathbf Q_p)/SL_n(\mathbf Z_p)$ inherits a geodesic metric space structure coming from the ultrametric on p-adics which make it into a CAT(0) space.

It's in fact a tree.

Hey guys, I am trying to make a nice and simple two-to-one function, is there something invalid about making this two-to-one function $f$ be defined by $f(x) = x^3/x$?

Why is that cool?

Hmm, I meant CAT(k) for some k < 0 up there, I believe.

@MikeMiller Because I have been told that it parallels what happens with SL_n(R)/SL_n(Z), which gets a Riemannian manifold structure from the Killing form on SL_n(R) (don't ask me what that means), so it hints at an analogue of Riemannian manifolds in algebraic varities. Also, since it's CAT(< 0), it means you might be able to do synthetic hyperbolic geometry on p-adic algebraic varities.

That's meaningless to me, sorry. I'll trust that it's interesting to someone.

Sure, it's not very understandable to me too.

Hello peoples.

Morning.

But it's a curious fact. I have no intuition why SL_n(Q_p)/SL_n(Z_p) should be something so nice.

Hi @PedroTamaroff

@Lucas Take any set $S$ and consider $f:S\times \{0,1\}\to S$ that sends $(s,0)$ to $s$ for $(s,1)$ to $s$.

$f:x\mapsto x²$ is two to one except at $0$.

$SL_3(\Bbb R)/SL_3(\Bbb Z)$ is the interior of the trefoil knot complement.

@MikeMiller What's up?