Uniform topology and local finiteness

[radou]

Homework Statement

For some reason, the uniform topology always causes me problems. So, let's work this through.

Let Rω be given the uniform topology, i.e. the topology induced by the uniform metric, which is defined with d(x, y) = sup{min{|xi - yi|, 1}, i is in ω}.

Given some n, let Bn be the collections of subsets of Rω of the form ∏Ai, where Ai = R for i <=n, and Ai = {0} or Ai = {1}, for i > n. One needs to show that the collection B = U Bn is countably locally finite, but neither countable nor locally finite.

The Attempt at a Solution

First of all, it is obvious that B is not countable, since the collection of sequences of zeros and ones is not countable. Let's leave the local finiteness of B for later.

To see if B is countably locally finite, consider Bn, for some fixed positive integer n. Let x be an element of Rω. We need to show that there exists a neighborhood of x which intersects Bn in finitely many members.

Since we're dealing with a metric space, we can consider open balls around the element x in the uniform metric d. But now I'm a bit confused and would be grateful for a push in the right direction.

Thanks in advance, as always.

 

[micromass]

For some reason, the uniform topology always causes me problems.

Trust me, you're not the only one

To see if B is countably locally finite, consider Bn, for some fixed positive integer n. Let x be an element of Rω. We need to show that there exists a neighborhood of x which intersects Bn in finitely many members.


Since we're dealing with a metric space, we can consider open balls around the element x in the uniform metric d. But now I'm a bit confused and would be grateful for a push in the right direction.

Yes, you are correct. You'll need to work with open balls.

Take an elementen $$(x_n)_n$$ and consider the balls $$B((x_n)_n,\epsilon)$$. How would you check if a sequence $$(y_n)_n$$ is an element of this ball?


Now, as for the actual question. I would consider balls with radius 1/2 (it's just a guess, but I have a feeling it should work with it)...

 

[radou]

Well, clearly, an element y of Rω is an element of the open ball B(x, 1/2) iff for no index i |xi - yi| is greater than 1/2. But x is fixed (i.e. its components, or members of the sequence, whatever you like), and I don't see how to relate it to the 0's and 1's...

 

[radou]

Could we prove this by assuming every neighborhood U of x intersects Bn in infinitely many elements and arrive at a contradiction? (i.e. then the negation of this statement would hold, i.e. "there exists a neighborhood U of x which intersects Bn in finitely many members)

 

[micromass]

Well, clearly, an element y of Rω is an element of the open ball B(x, 1/2) iff for no index i |xi - yi| is greater than 1/2.

OK, this is correct. This might have seemed an obvious question, but it's quite useful.

But x is fixed (i.e. its components, or members of the sequence, whatever you like), and I don't see how to relate it to the 0's and 1's...

[/QUOTE]

OK, so let me dig out the problem a little bit (I'm not sure if that is proper english, but hey...).
Take an arbitrary x, we need to find a ball B(x,epsilon) such that this ball meets finitely many elements in B. I claim that epsilon=1/2 always works (for any x!).

Let's simplify the problem a bit. Let's Take B=B₀. So I only look at the sets $$\prod{\{x_n\}}$$, where the x_n take on the values 0 and 1.

Obviously, we will have to split up in cases (Indeed, the x can be very different). So let's take a look at some of the cases:

x=(x_n)_n=(2,2,2,2,...)
I claim that B(x,1/2) does not intersect any element of B₀. Indeed, assume that (x_n)_n is an element of B(x,1/2) and assume that (x_n)_n does lie in an element of B₀.
Since every y_n is 0 and 1, we see that |x_n-y_n|>1. But this implies that (y_n)_n can not lie in B(x,1/2)...

An analogous argument to the one above, already handles a lot of cases. But what if we take

x=(1,1,1,...)
I claim that B(x,1/2) only intersects B₀ in one element. (and that is the element $$\prod{\{1\}}$$. Assume that B(x,1/2) also intersects an element of the form $$\prod{\{y_n\}}$$. Assume that there is a n such that y_n=0 (if such an n does not exist, then y=x). Then $$|x_n-y_n|=|1-0|=1$$. Thus the $$(y_n)_n$$ does not lie in the ball B(x,1/2).

It's quite difficult to clearly write such arguments down, so I hope you understand this. If you do, then I don't think it's that hard to make a general proof...

 

[micromass]

Could we prove this by assuming every neighborhood U of x intersects Bn in infinitely many elements and arrive at a contradiction? (i.e. then the negation of this statement would hold, i.e. "there exists a neighborhood U of x which intersects Bn in finitely many members)

Hmm, it seems unlikely that the proof could be done this way. But you can always try it...

 

[radou]

OK, before I completely go through this, I don't completely understand with what you mean by B0. This can be a subset of any Bn (i.e. for any n), since B0 entirely (unless I misunderstood) consists of zeroes and ones?

 

[micromass]

OK, before I completely go through this, I don't completely understand with what you mean by B0. This can be a subset of any Bn (i.e. for any n), since B0 entirely (unless I misunderstood) consists of zeroes and ones?

Well, I may have abused notation here and there. I tried not to do it, but it might have happened...

Anyway, with B0 I mean B0 entirely. So, I mean $$B_0=\{\prod\{x_n\}~\vert~x_n=0~\text{or}~x_n=1\}$$. Of course an element in B0 is a set. But I might have identified the set $$\prod\{x_n\}$$ with the sequence $$(x_n)_n$$... I can do this because $$\prod\{x_n\}=\{(x_n)_n\}$$, thus a set is determined by the sequence...

 

[radou]

Hm, this notation confuses me a little (sorry about that)..

According to the problem statement Bn is defined for positive integers. So, for example:

B1 = ∏Ai, such that A1 = R, Ai = {0} or {1} for any i > 1.
B2 = ∏Ai, such that A1 = A2 = R, and Ai = {0} or {1}, for any i > 2.

Elements of B1, for example, are sequences xn such that their first member is any real number, and all the other members equal either 0 or 1. The same for B2, the first two members of xn are any real numbers, and the rest are 0's and 1's.. Perhaps I misunderstood the problem.

 

[micromass]

Hm, this notation confuses me a little (sorry about that)..

According to the problem statement Bn is defined for positive integers. So, for example:

B1 = ∏Ai, such that A1 = R, Ai = {0} or {1} for any i > 1.
B2 = ∏Ai, such that A1 = A2 = R, and Ai = {0} or {1}, for any i > 2.

Elements of B1, for example, are sequences xn such that their first member is any real number, and all the other members equal either 0 or 1. The same for B2, the first two members of xn are any real numbers, and the rest are 0's and 1's.. Perhaps I misunderstood the problem.

Hmm, that's the problem with things like $$\mathbb{R}^\omega$$, the notation is always horrible...

It's not true that $$B_1=\prod A_i$$. If you look closer, then you'll see that B1 is the set which contains all elements of the form $$\prod A_i$$.
Thus $$B_1=\{\prod A_i~\vert~...\}$$. Likewise with B0 and B2.
So elements of B1 are actually subsets of $$\mathbb{R}^\omega$$. So in Munkres words, we could actually say that B1 is some kind of superset...

 

[radou]

Hmm, is it then okay to write the following:

If we consider B1, an element of this collection would (for example) be R x {0} x {0} x {1} x {0} x ... , which is itself a subset of Rω, right? Now, the elements of this subset are sequences described in my post #9 (under B1, but with wrong notation, I see now why), right?

 

[micromass]

Yes, that would be correct.

Now elements in B0 are, for example

{0}x{0}x{1}x{0}x{0}x...

and the only element of this is (0,0,1,0,0,...)

But sometimes (a very wrong abuse of notation) I might identify these two things... Sorry if that happens...

 

[radou]

OK, I know I'm starting to get annoying right now, but doesn't Munkres think of the positive integers as the set {1, 2, ...} i.e. without the element 0? So I assume by B0 you don't mean an element of the collection Bn, but a subset of B1?

Sorry if I'm going into too much details, but I can't start to consider your post #5 until the notation is perfectly clear. :)

 

[micromass]

OK, I know I'm starting to get annoying right now, but doesn't Munkres think of the positive integers as the set {1, 2, ...} i.e. without the element 0? So I assume by B0 you don't mean an element of the collection Bn, but a subset of B1?

Sorry if I'm going into too much details, but I can't start to consider your post #5 until the notation is perfectly clear. :)

Ow, I'm sorry. I'm so used to $$\mathbb{N}=\{0,1,2,3,...\}$$. I didn't realize that Munkres did it the other way...

Anyway, with $$\mathcal{B}_0$$ I mean: sets of the form $$\prod{A_i}$$ such that Ai={0} or Ai={1}.

I know that Munkres' sequence starts with $$\mathcal{B}_1$$, but assume for a moment that it actually starts with $$\mathcal{B}_0$$... (it actually makes no much difference, but I think it's easier to understand)

Technically, $$\mathcal{B}_0$$ is neither an element nor a subset of $$\mathcal{B}_1$$. Although, the elements of $$\mathcal{B}_0$$ are subsets of elements of $$\mathcal{B}_1$$. If you already seen this in Munkres: $$\mathcal{B}_0$$ is a sort of refinement of $$\mathcal{B}_1$$.


If all this is not clear, I will rewrite my post with $$\mathcal{B}_1$$ instead of $$\mathcal{B}_0$$. But I think it's easier the way it is now...

 

[radou]

OK, I get it now, thanks. I'll post again when I solve this.

 

[micromass]

OK, I'm going to sleep now! Good luck!

 

[radou]

OK, I'll try to start with B1 (I'm stubborn, so I'll ignore B0):

B1 is the collections of all subsets of Rω of the form R x {a product of an infinite sequence of 0's and 1's}. Forgive my informal notation, but I think it's clear what I'm trying to say.

Now, let x be in X. Take the open ball B(x, 1/2). I'm trying to show if any B(x, 1/2) intersects B1 in any element. First of all, some element of some subset of B1 is in B(x, 1/2) iff for all xi and yi, |xi - yi| < 1/2, and it seems not only that (since 1/2 could nevertheless be the supremum!), but even a stronger condition, which I don't see how to express in any other way than say, for example, that for every i, |xi - yi| < 1/4. Now, a huge problem is that I don't have a clue how x looks like, i.e. its components.

The sketch for a proof with choosing values for x = (2, 2, 2, ...) etc. is something I don't see how to derive a general proof.

If, for example, I choose x = (1, 1, 1, ...), and look at B(x, 1/2), then y = (1, 1, 1, ...) is an element of R x {1} x {1} x ... which is in B(x, 1/2). So, we see for sure that B(x, 1/2) intersects one element (i.e. subset) of the collection B1. Does it intersect any other one? Well, if we take an element from B1 which contains {0}, the distance equals 1, so no element of B1 containing {0} in the "sequence" (I'm referring to the collection) can be in B(x, 1/2).

Now, what if I choose a random x and consider again B(x, 1/2)? In general, the supremum of the distances of components (definition of the uniforom metric) equals 1, if we choose x with huge components, or at least one huge component. I don't see how I could answer this, since this depends of the choice of x.

 

[micromass]

OK, I'll try to start with B1 (I'm stubborn, so I'll ignore B0):

Stubborness is the sign of a good mathematician

B1 is the collections of all subsets of Rω of the form R x {a product of an infinite sequence of 0's and 1's}. Forgive my informal notation, but I think it's clear what I'm trying to say.

Now, let x be in X. Take the open ball B(x, 1/2). I'm trying to show if any B(x, 1/2) intersects B1 in any element. First of all, some element of some subset of B1 is in B(x, 1/2) iff for all xi and yi, |xi - yi| < 1/2, and it seems not only that (since 1/2 could nevertheless be the supremum!), but even a stronger condition, which I don't see how to express in any other way than say, for example, that for every i, |xi - yi| < 1/4. Now, a huge problem is that I don't have a clue how x looks like, i.e. its components.

Well, maybe you could try the following approach:
Take x arbitrary, there are a few situations that could occur
1) B(x,1/2) doesn't intersect B1 in any element. This is for example the case with x=(2,2,2,...). In this case, there is no problem.

2) B(x,1/2) does intersect B1 in an element. We want to show that it intersects in exactly one element. Assume that $$y=(y_1,y_2,y_3,...)$$ and [tex]z=(z_1,z_2,z_3,...) are both elements of elements of B1 (that is: y1 is arbitrary and the other yn are 0 or 1 and likewise for z). And assume that y and z are also in B(x,1/2).

Now assume that y and z are both coming from a different element in B1. That is, there exist a n>1 such that yn=0 and zn=1 (or the other way around). Thus |yn-zn|=1. But we also have y,z in B(x,1/2). Thus |yn-zn|<=|yn-xn|+|zn-xn|<1/2+1/2=1. Thus |yn-zn|<1. This is a contradiction...

This should do it...

 

[radou]

Hmm, ok... So in general, we could apply this same logic to a general element Bn and arrive at a conclusion that B(x, 1/2) only intersects it in one element, because of the generalized triangle inequality?

 

[micromass]

Yes, the same proof (with probably some minor modifications) could be applied to Bn.

But there's still a "fun" part remaining. Proving that B is not locally finite...

 

[radou]

Yes, the same proof (with probably some minor modifications) could be applied to Bn.

Ah, OK.

But there's still a "fun" part remaining. Proving that B is not locally finite...

Since I don't like this problem, my subconscience probably made me forget that this part exists.

OK, let's assume B = U Bn is locally finite. Then, for any x in X, there exists a neighborhood U of x which intersects finitely many elements of B, i.e. finitely many subsets of Rω of the defined form. I guess I can be more precise and for U take an open ball B(x, ε). I'm hoping to arrive at a contradiction. Let's see.

First of all, this ε should for sure be less or equal to 1, for in the contrary, B(x, ε) would intersect all elements from B, unless I'm mistaken.

Let's take for x the element x = (0, 0, 0, ...) (I'm improvising right now, and I have the feeling this could come in handy). Since, by assumption, B(x, ε) intersects B in finitely many elements, we can choose a sequence from each of this sets Bn that B intersects.

Now, first of all, I'd like to make one thing clear: when I say that B(x, ε) intersects finitely many elements of B = U Bn, does this mean that it intersects finitely many of the Bn elements, or can it theoretically intersect finitely many subsets of one single Bn? i.e., can I look at B as a huge collection where all subcollections are in a sense equal, or do I have to respect a certain collection-set hierarchy?

Edit: actually, I think I can answer this mayelf; since U is a neighborhood, it is a set. So it only makes sense to intersect it with a set. If I try to intersect it with a collection, U should itself need to be in that collection in order for such an operation to defined, so I guess, theoretically, yes, one U could intersect one single Bn in finitely many of its subsets. Corect me if I'm wrong.

 

[micromass]

Since I don't like this problem, my subconscience probably made me forget that this part exists.

OK, let's assume B = U Bn is locally finite. Then, for any x in X, there exists a neighborhood U of x which intersects finitely many elements of B, i.e. finitely many subsets of Rω of the defined form. I guess I can be more precise and for U take an open ball B(x, ε). I'm hoping to arrive at a contradiction. Let's see.

First of all, this ε should for sure be less or equal to 1, for in the contrary, B(x, ε) would intersect all elements from B, unless I'm mistaken.

Let's take for x the element x = (0, 0, 0, ...) (I'm improvising right now, and I have the feeling this could come in handy). Since, by assumption, B(x, ε) intersects B in finitely many elements, we can choose a sequence from each of this sets Bn that B intersects.

The choice of x=(0,0,0,...) is a wonderful choice! But I don't really see how contradiction comes in handy here. Can't you just show that any neighbourhood of x intersects infinitely many elements of B. In fact you can prove something better! You can show that x belongs to infinitely many elements of B! (so there's no need to look for open balls around x).

To see this: can you find an element of B1 such that x is an element of it?
Can you do thesame thing for B2, B3? And in general: for Bn?

Now, first of all, I'd like to make one thing clear: when I say that B(x, ε) intersects finitely many elements of B = U Bn, does this mean that it intersects finitely many of the Bn elements, or can it theoretically intersect finitely many subsets of one single Bn? i.e., can I look at B as a huge collection where all subcollections are in a sense equal, or do I have to respect a certain collection-set hierarchy?

Edit: actually, I think I can answer this mayelf; since U is a neighborhood, it is a set. So it only makes sense to intersect it with a set. If I try to intersect it with a collection, U should itself need to be in that collection in order for such an operation to defined, so I guess, theoretically, yes, one U could intersect one single Bn in finitely many of its subsets. Corect me if I'm wrong.

Yes, you have answered your own question correctly!

 

[radou]

The choice of x=(0,0,0,...) is a wonderful choice! But I don't really see how contradiction comes in handy here. Can't you just show that any neighbourhood of x intersects infinitely many elements of B. In fact you can prove something better! You can show that x belongs to infinitely many elements of B! (so there's no need to look for open balls around x).

To see this: can you find an element of B1 such that x is an element of it?
Can you do thesame thing for B2, B3? And in general: for Bn?

Hmm, well x = (0, 0, 0, ...) is definitely in every element of B! Hence, it follows that every neighborhood if x (we don't even have to think in term of open balls now!) intersects every element of B in x itself, right? So the family can't be locally finite!

 

[micromass]

Wow, wait. I don't see why x=(0,0,0,...) is in every element of B.

Surely, it is not true that x is in [\tex]\mathbb{R}\times {1}\times \{0\}\times \{0\}\times...[/tex]. And that element is an element of B, right?

In fact,you know that Bn is locally finite, so x can only be in a finite number of elements of Bn. So what you said doesn't seem correct to me...

 

[radou]

Wow, wait. I don't see why x=(0,0,0,...) is in every element of B.

Surely, it is not true that x is in [\tex]\mathbb{R}\times {1}\times \{0\}\times \{0\}\times...[/tex]. And that element is an element of B, right?

In fact,you know that Bn is locally finite, so x can only be in a finite number of elements of Bn. So what you said doesn't seem correct to me...

Oh, pardon me, I formulated this badly!

What I wanted to say is: for x = (0, 0, 0, ...) we can always find a subset of any of the collections Bn (i.e. for any n), which contains x. For example, in B1, x is an element of R x {0} x {0} x ...

Hm, but from this we can only conclude that for every n, there exists a subset of Bn such that every neighborhood of x intersects it. So, x is definitely in an infinite number of such subsets. Which means that we can't have a locally finite familly...But I feel have a "set vs. collection" issue here again..

 

[radou]

But isn't this okay? When I think twice... we have shown that any neighborhood of x intersects B in infinitely many sets, right? (even more precise, in x itself, but this doesn't really matter)

 

[micromass]

This seems entirely correct!

I hope you didnt start to hate $$\mathbb{R}^\omega$$ yet
Well yeah, product spaces are imo the hardest part of topology, since there is not much intuitive about them. But you'll get used to them...

 

[radou]

This seems entirely correct!

I hope you didnt start to hate $$\mathbb{R}^\omega$$ yet
Well yeah, product spaces are imo the hardest part of topology, since there is not much intuitive about them. But you'll get used to them...

Yeah, I guess.. Now finally, on to the Nagata-Smirnov metrization theorem!