How Does Total Boundedness of Product Spaces Relate to Compactness?

[radou]

Homework Statement

I really got a problem with these products.

If Xn is metrizable with dn, and if D(x, y) = suo{di(xi, yi)/i} is the metric which induces the product topology on X = ∏ Xn, show that if Xn is totally bounded for every n (under dn), then X is totally bounded under D. Conclude without using the Tychonoff theorem that a countable product of compact metrizable spaces is compact.

The Attempt at a Solution

Since Xn is totally bounded for every n, this means that for any n and any ε > 0 there exists a finite covering of Xn by ε-balls. Somehow, I must find a finite collection of ε balls (if ε is given) in X which covers X. Any discrete hint is welcome... I had a few ideas, but they don't work.

 

[radou]

Of course, proving the other statement is easy.

If Xi is compact for every i, then it is totally bounded, too. By the previous (yet unproved) result, a countable product X of totally bounded spaces is totally bounded in the metric D. Since X is complete under D (Theorem 43.4.), we conclude (by Theorem 45.1.) that X is compact.

 

[micromass]

OK, I think I've got an idea. I'll try not to give away to much.

As a first hint: can you describe me the ball B((xn)n,1/3) in the producttopology?
In general, can you describe B((xn)n,r)?

 

[radou]

OK, I think I've got an idea. I'll try not to give away to much.

As a first hint: can you describe me the ball B((xn)n,1/3) in the producttopology?
In general, can you describe B((xn)n,r)?

Hm, OK.

If x = (xi) is in (X, D), we have B(x, r) = {y in X : D(x, y) = sup{di(xi, yi)/i} < r}. Here it may be important that if r > 1, B(x, r) = X (because di is the bounded metric on Xi). So cleary, for any ε > 1, the open ball B(x, ε) is a finite open cover for X. It follows that we may assume ε <= 1, right?

 

[micromass]

Hmm, what you said is certainly correct, but it's not really what we're looking for...

What I want is the following: I want to express B((x_n),1/3) in the form

B((x_n)_n,1/3)=\prod_{n\in \mathbb{N}}{U_n}

where the Un are all open sets in Xn. Furthermore, the most important part is that Un=Xn for all (but finitely many) n.

Try to work out what the Un are in this case.

 

[radou]

Hmm, OK...I know this has got something to do with the definition of D, i.e. with the fact that we divide di with i... I'll think about it.

 

[radou]

OK, let x be in X.

B(x, 1/3) = {y in X : sup{di(xi, yi)} < 1/3}.

Now, if y = (yn), I conclude the following:

The values di(xi, yi) for i > 3 really don't matter, since their maximal value is 1/i, which is in this case clearly less than 1/3. Now, let's look at the first three components of y.

d1(x1, y1) should definitely be less than 1/3. d2(x2, y2) should definitely be less than 2/3. d3(x3, y3) should definitely be less than 1.

So, if y lies in B(x, 1/3), the first component should lie in the open ball in X1 given with B1(x1, 1/3), the second (in X2, and the same for X3 later on, of course) in B2(x2, 2/3), and the third in B3(x3, 1).

Hence, unless I'm mistaken, we have B(x, 1/3) = B1 x B2 x B3 x X4 x X5 X ...

This is easy generalized, i.e. for any n, B(x, 1/n) = B1 x B2 x ... x Bn x Xn+1 x ... , where for any i = 1, ..., n, Bi is an open ball in Xi of radius i/n.

If this is correct, just confirm please, without further hints, and I'll proceed.

 

[micromass]

Yes, this is correct!

 

[radou]

Hm, now clearly, if ε is of the form 1/n, where n is a positive integer, I can always find a finite open cover for X with ε-balls... For example, if ε = 1/2, we have B(x, 1/2) = B1 x B2 x X3 x X4 x X5 X ... where B1 is of radius 1/2 and B2 of radius 1. Now, since Bi is totally bounded, for i = 1, 2, for B1 take a finite cover for X1 by open balls of radius 1/2, and for B2 take a finite cover for X2 by open balls of radius 1.. The number of combinations of these elements is finite, and any such open ball in X remains an 1/2-ball. So, instead of B1 and B2 we may have Bi x Bj x ... x Bn x Xn+1 x ..., where i = 1, ... p and j = 1, ... q, and this is a finite collection of 1/2-balls which covers X (p and q are of course the numbers of elements of our finite covering collections for X1 and X2, respectively).

This is basically an unimportant observation. I'm still chasing the idea on how to do this or something similar for any ε > 0. Again, I'll point out that the cases for which ε > 1 are uninteresting, right?

 

[micromass]

Yes, the case where \epsilon&gt;1 is not interesting, in fact, only the small \epsilon is interesting.

So you've proven the case where \epsilon=1/n. I claim that this already proves the general case: if you have epsilon&gt;0, you can choose 1/n&lt;\epsilon. A finite open cover of balls of radius 1/n, will induce a finite open cover of balls of radius \epsilon...

 

[micromass]

Alternatively, you can also prove a formula for B((x_n)_n,\epsilon) in general. This will be very similar to a formula for B((x_n)_n,1/n)...

 

[radou]

Yes, the case where \epsilon&gt;1 is not interesting, in fact, only the small \epsilon is interesting.

So you've proven the case where \epsilon=1/n. I claim that this already proves the general case: if you have epsilon&gt;0, you can choose 1/n&lt;\epsilon. A finite open cover of balls of radius 1/n, will induce a finite open cover of balls of radius \epsilon...

OK, this may seem as a stupid question, but how can such a finite open cover by balls of radius 1/N < ε induce a finite open cover by balls of radius ε? I don't see this immediately... If it's trivial, please let me know, so I'll think about it a bit more!

 

[radou]

Yes, that occurred to me...A formula inspired by a sequence which converges to ε?

 

[micromass]

OK, this may seem as a stupid question, but how can such a finite open cover by balls of radius 1/N < ε induce a finite open cover by balls of radius ε? I don't see this immediately... If it's trivial, please let me know, so I'll think about it a bit more!

Well, if you have an open cover by balls of radius 1/N, then you just enlarge the balls so that they have radius \epsilon. This will still be a cover of X...

 

[micromass]

Yes, that occurred to me...A formula inspired by a sequence which converges to ε?

Hmm, I was thinking of the formula

B((x_n)_n,\epsilon)=B(x_1,\epsilon)\times B(x_2,2\epsilon)\times B(x_3,3\epsilon)\times...

 

[radou]

Well, if you have an open cover by balls of radius 1/N, then you just enlarge the balls so that they have radius \epsilon. This will still be a cover of X...

Yes, I understand the concept, but I need to find out how..

Hmm, I was thinking of the formula

B((x_n)_n,\epsilon)=B(x_1,\epsilon)\times B(x_2,2\epsilon)\times B(x_3,3\epsilon)\times...

OK, I'll think about this a bit more and post later/tomorrow. Actually, a few hours ago, I hated this problem, and now I like it!

 

[micromass]

Yes, I understand the concept, but I need to find out how..

It's not that hard: if \{B(x,1/n)~\vert~x\in E\} is a finite cover of balls of radius 1/N, then \{B(x,\epsilon)~\vert~x\in E\} is a finite cover of balls of radius epsilon... You just replace every ball in the first collection by a larger ball...

OK, I'll think about this a bit more and post later/tomorrow. Actually, a few hours ago, I hated this problem, and now I like it!

Aha, somebody's beginning to like infinite products?

 

[radou]

It's not that hard: if \{B(x,1/n)~\vert~x\in E\} is a finite cover of balls of radius 1/N, then \{B(x,\epsilon)~\vert~x\in E\} is a finite cover of balls of radius epsilon... You just replace every ball in the first collection by a larger ball...

YOu won't believe me right now, but it just occurred to me a few seconds ago! It's utterly trivial! I was thinking too hard, I guess...

 

[micromass]

Haha, I believe you

That's another question solved I guess...

 

[radou]

Haha, I believe you

That's another question solved I guess...

Yes, thanks a lot!

 

[radou]

Btw, another simple intermediate result:

For any rational number r, we can find a finite cover for X by open balls of radius r.

This basically follows from the proved case for radiuses of type 1/n.

If r = m / n, one easily concludes that the first ball in the product has a radius less than m / n, the second less than 2m / n, etc... Of course, until for some i, 2i < n is less than 1, the other cases are always trivially true. And this happens a finite number of times.

Edit: of course, we're talking about rational numbers of type r > 0.

 

[micromass]

Yes, but doesn't that proof generalize to arbitrary real numbers?

 

[radou]

Yes, but doesn't that proof generalize to arbitrary real numbers?

Well, that's where my lack of precise knowledge in analysis lacks a little..

Well, any real number can be approximated with rational numbers, right? i.e. with sequences of rational numbers, more precisely...Perhaps I'm overcomplicating again, but I'd like to see a rigorous proof...

If we construct such a sequence which converges to some real number, then the corresponding "sequence of open covers" (if such term makes sense at all, but actually its important than an open cover in this situation if fully represented with our rational number, or not? is it unique?) "converges" to our open cover with balls of radius ε... Something like that?

 

[radou]

Of course, ε is real, so it is in the closure of the rationals, and there exists a sequence of rationals converging to ε.

Edit: so, basically, that's it, I guess.

Edit 2: when I think better, the "uniqueness" I mentioned somewhere in the upper post isn't really important, ignore it.

 

[micromass]

Yes, all of this should be correct. However, can't it be done much easier.

Take (y_n)_n in B((x_n)_n,r). Then certainly d_n(x_n,y_n)/n\leq r. Thus d_n(x_n,y_n)\leq rn. This shows that

B((x_n)_n,r)=\prod_{n\in \mathbb{N}}{B(x_n,nr)}

 

[radou]

Yes, all of this should be correct. However, can't it be done much easier.

Take (y_n)_n in B((x_n)_n,r). Then certainly d_n(x_n,y_n)/n\leq r. Thus d_n(x_n,y_n)\leq rn. This shows that

B((x_n)_n,r)=\prod_{n\in \mathbb{N}}{B(x_n,nr)}

Ah, yes, of course!

This holds for any n until rn < 1, then we set the other "factors of the product" to equal Xi.

Well, this is obviously the easiest way to prove it, right?

 

[micromass]

Yes, I think this is the easiest proof of this. Although I think your method could work to...

 

[radou]

Yes, I think this is the easiest proof of this. Although I think your method could work to...

OK. Thanks for the n-th time, as n --> ∞!