d(x, y)/(1+d(x, y)) IIRC

I think he used nested interval theorem

@Daminark: Sure, or the least upper bound property.

that's the one I know, but I think there's another one in Munkres @Balarka, perturbative was asking about it last week or maybe it was just something related

I remember at the time that he went through Rudin exactly

You can do $\min(1,d(x,y))$ ...

Triangle inequality is probably easier with that one.

Ah, that should work

that sure looks easier to work with

Anyhow, @Daminark, now that I've given you that hint, you should finish immediately.

I guess the kernel of $G * G\to G\times G$ is $[G,G]$ so to find $[[G,G], G]$ one has to look at $[G,G]*G\to[G,G]\times G$.Geometrically if $G$ is a fundamental group then $G*G$ is gonna be the fundamental group of some connected sum of things and the map maps this summy object fundamental group to some product space fundamental group and the kernel of this map is [[G, G],G].

So he was like "Yeah take an open cover and assume that there's no finite subcover. Cut it in half. One of these boxes also has no finite subcover. Cut that in half. Keep going. So the nested interval theorem generalizes to k-cells, then the intersection of these subboxes isn't empty, so consider an open set containing it, assume it's a ball, but the boxes eventually get small enough that they're only inside the ball. Boom"

Rather handwavy, but it is the idea, of course.

That's a true topologist coming up with a proof

He can get a bit handwavy when he's rushing

You should've seen his proof of divergence theorem

Or, "divergence theorem"

(I don't know if it was actually that)

Well, not really. There are worse/better proofs than that

@AliCaglayan Not connected sum. Wedge product.

But back to the orthogonal group, it's closed automatically

Yes that is correct

Make that wedge sum

the connected sum thing has augmented free product stuff in it

Why do you want to understand [[G, G],G]? Do you have nothing better to do with your life?

And if by sleight of hand we start talking about operator norm, it's super bounded

@BalarkaSen I have good reason

Really I am trying to understand nilpotent free groups of calss c

You're working in $\Bbb R^{n^2}$ with the usual metric topology, though, @Daminark, aren't you?

Wait though in finite dimensions all norms are equivlanet so it's not even sleight of hand :P

*equivalent

They are free groups modulo the c+1th term in the lower central series

You can use the usual metric just fine. What do you know about the column vectors of an orthogonal matrix?

Orthonormal

So what does that tell you about the Euclidean length?

Oh wait

Right so it's $\sqrt{n}$

Right.

Looking at spaces they are fundamental group of is not going to help, @Ali. There are millions of spaces a group can be a fundamental group of.

What's the definition?

Learn Gromov's theorem about groups of polynomial growth instead.

Yeah I can study them algebraically using the associated Lie ring and I get all the things I need but I just tried to see if I could geometrically come up with them

there's a funny proof of the Heine-Borel theorem in $\Bbb R$ using real induction, but I think it'd be a pain to generalize

What I know is the we can represent a surface as f (x,y) = ( x, y , f(x,y) for example

ugh, Pete Clark is a big fan of real induction. I think it's a waste of time.

hi @Gridley

Could someone help me with those? I think the first two essentially say that the cardinality of M is at least 1 (meaning that they are topological properties)

hi Ted

I have seen that proof too

I think that's also the first and last time I've seen real induction used somewhere

At the end of my first year calc class we started using Calc on Manifolds, and the proof there was nice

at least 1, Gridley?

It directly invoked suprema to prove that $[a,b]$ is compact

At most sorry

Agreed.

Then you use that the product of compact sets is compact, gets you any box

Then given any closed and bounded set, we stuff it in a box, and then it's compact

So what do you think about c), Gridley?

Ok, I discovered I have a bit of a mess here

It doesn't seem to be saying much. Just that the distance between two points is some epsilon.

To graph a function of n variables, I need n+1 variables right ?

Yes, @Maks.

One will store the value of the function and the rest are the values of the variables

@Gridley: So what do you think the answer is?

Now, when I have a function with two variables

That it is a topological property

And I'm finding a line tangent to a point, there isnt only one line

So we find the plane that contains all lines that are tangent to that point

What characterizes the spaces, @Gridley?

And we call it tangent plane

@Maks: I want you to say tangent to the surface at that point :P

There comes the problem

What's the difference between the surface and the grapg

I'm not quite sure.

I don't care. I'm just saying you can't say tangent to the point.

Just saying that 'there exists' doesn't seem to give a lot of information.

Oh ok

What spaces will that hold in, @Gridley? Think back to what you told me for a) and b).

So the plane tangent to the surface at at that point

Well it means that you have at least two separate points

Right, @Gridley.

So since its a statement about cardinality does that mean it must be metric invariant?

Yup.

What about d)?

Ah okay.

I think 4) is saying that there exists a point in M ( our x) where we can find another point as close as we want to it.

(not counting the point itself)

yep

Do you know the phrase limit point?

Now we have directional derivatives

which tells us how much does a function grow in a specific direction

Vaguely.

The direction is a vector, a unit vector

Well, don't worry about it, @Gridley. So what do you think about d)? Does that depend on the metric or just on the topology?

I'd think topology. It seems to be a property preserved by continuity.

(and hence a homeomorphism)

Continuity alone doesn't preserve it (think of a constant map).

For connectedness of $O(n)$, would it be good to try and prove that $SO(n)$ is a clopen subset?

So... the only difference between the two functions is that one of them is for functions with 3 variables and the other one with two ?

(Not sure if it's true, just wondering if it is, and if that's feasible to try)

OH NO

Its used for level curves

@Daminark: What precisely are you claiming?

Yes, @Daminark. SO(n) is in fact a connected component.

He probably wants to prove O(n) is not connected.

That $O(n)$ is disconnected because it has a proper subset which is closed and open in it

@MartinSleziak here

OK. Proving disconnected is a lot easier than all that ...

hmmm, I a bit stuck o =n 4

on 4*

@Maks: The first is also level surfaces.

The second is only for graphs $z=f(x,y)$.

Oh wait a second

Determinant maps it to a disconnected set

There you go.

I see, I think I get it

@Gridley: You had it right. You just said something wrong along the way (the continuity thing).

If I'm asked to give the equation of the plane tangent to the graphic

I use the second one

The one without the $f_z$, right ?

I'm suggesting you can always write every problem as a level surface and avoid messing it up. That's the exercise I told you to do twice.

Yeah but I cant figure it out :(

Take the derivatives of $F$.

Huh?

My bad hehe

Reread exactly what I typed.

So a property is topological if M,N are metric spaces and there is a homeo f:M -> N then either both spaces have that property or none right?

Right, @Gridley.

write $ z = f(x,y) $ as the level set $F(x,y,z) = f(x,y) - z = 0$

So now how do you fix your question you just asked me?

Like that

So do it.

Well it feels like if you can get arbitrarily close to our 'x' in one metric you'd have to be able to in the other.

but I am not sure how to prove it

Let me re-read level curves/surfaces theory

@Maks. $z=f(x,y)$. Put the damn $z$ in there.

@Gridley: Think by contradiction.

@TedShifrin I think your problem 5.1.5 has a typo?

I don't think so, DogAteMy.

It says $\|A\|\le\sqrt n\|A\|$

No, we are talking about different norms.

But it uses $\|\cdot\|$ for both of them.

One is defined in terms of $Ax$ for unit vectors $x$. The other is defined of thinking the matrix as a vector in Euclidean space.

You've written: $\|A\|\le\sqrt{\sum_{ij}a_{ij}^2}\le\sqrt n\|A\|$

Oh, there's the other norm in the middle. What are you bitching about?

Oh, wait

Derp

It's just an ok notation.

LOL, uh huh.

Pretend I didn't say anything…

OK :)

I guess for some reason I was confusing $\le$ for $\ge$?

How little faith you have in me ... Sigh.

Oh, so the norms are the same.

Sorry

Grr

I'm still drawing a blank on the contradiction. Would you use the 'boundedness' for topologically equivalent metrics? (cd_1 < d_2 < Cd_1)

Could you have an $\epsilon$-ball around $x$ with only $x$ in it in one metric but zillions of points in the other metric?

I guess if it wasn't true then would it be true that a sequence converges wrt one metric but not another.

OK. Why is that a problem?

You'd have different open sets

Be specific.

@TedShifrin ◔_◔ ?? Please do not hate me, but how do I get the normal line to the tangent plane ?

◔_◔

Normal vector to plane?

Yes

Do it.

I dont know how...

You do. I'm quitting.

Are you giving up on teaching ?

I'm done for today.

You'd be able have a sequence such that d(x_k,x) tends to 0 in d1 but not d2.

You said open sets were different. What set is open in one case and not in the other?

@GridleyQuayle there are equivalent metric such that you can't find constants c and C making those inequalities true

@Maks Gradient I think, if it's parametric

Ignore that ...

Gridley, answer my question. I'm leaving.

@AlessandroCodenotti is the implication the other way? @TedShifrin The set containing the sequence in question?

Simple set.

M?

@GridleyQuayle yes (that's just a detail which isn't important for your question by the way, ignore it for the moment)

Huh?

Is that right ?

=0

sorry, I've got a vague intuition about it but I'm really not sure how to pin it down.

x?

The singleton?

So the statement is saying that there is an x in M where we can construct a sequence of y's which tend to that x. If we could construct that sequence with respect to one metric but not another that would imply that the set containing the y's was closed wrt to one metric, but not to the other

Or $\{x\}$ is open in one case and certainly not in the other.

which is a contradiction. That's what I've understood

How is {x} open in one?