Do I need to use this hint from Rudin?

[futurebird]

This is an assignment for class so, please don't tell me how to do this problem or give too much away. I just need to know why I need to prove this. Let me explain. Here is the problem from Rudin.

Chapter 2, 25
Prove that every compact metric space K has a countable base, and that K is therefore separable.

Hint: For every positive integer n, there are finitely many neighborhoods, radius 1/n whose union covers K.

Relevant Theorems:

1. Every separable metric space has a countable base.
2. If X is a metric space in which every infinite subset has a limit point then X is separable.
3. If E is an infinite subset of a compact set K, then E has a limit point in K.

My response:
(ignoring the hint)
K is compact, so every infinite subset of K has a limit point. Hence, K is separable. Every separable metric space has a countable base.

This can't be right. Why did Rudin give the "hint?"

 

[konthelion]

Prove that every compact metric space K has a countable base, and that K is therefore separable.

Hint: For every positive integer n, there are finitely many neighborhoods, radius 1/n whose union covers K.

For the hint: since K is compact, then it forms a countable collection,which we claim to be a base, of those neighborhoods of radius 1/n whose union covers K.

 

[futurebird]

For the hint: since K is compact, then it forms a countable collection,which we claim to be a base, of those neighborhoods of radius 1/n whose union covers K.

I see that, but why would one bother to do it that way when you can already see it's true using the theorems.

 

[matt grime]

K is compact, so every infinite subset of K has a limit point.

Counter example: { 1/n : n in N} is an infinite subset of the (compact) interval [0,1] with the obvious metric.

 

[futurebird]

Counter example: { 1/n : n in N} is an infinite subset of the (compact) interval [0,1] with the obvious metric.

That has a limit point, 0. I don't see how this is a counterexample. It may be undefined, but it still has a limit point. (and the limit point is in [0, 1].

 

[matt grime]

Um, 0 is not in the set {1/n : n in N}, surely that is important (not that I've thought long and hard about what you're trying to do)? You've also not invoked the metric structure either, which is another worry - try to think about a compact non-metrizable space. Do such exist? Metamathematically if they don't, why bother to define compactness at all?

 

[futurebird]

Um, 0 is not in the set {1/n : n in N}, surely that is important (not that I've thought long and hard about what you're trying to do)?

It's not important that 0 is not in the set. The limit point of a set need not be in the set.

Let S be a subset of a metric space X. We say that a point x in X is a limit point of S if every open set containing x also contains a point of S other than x itself.​

 

[futurebird]

You've also not invoked the metric structure either, which is another worry - try to think about a compact non-metrizable space. Do such exist? Metamathematically if they don't, why bother to define compactness at all?​

I'm confused about what you mean by this?

 

[matt grime]

It's not important that 0 is not in the set. The limit point of a set need not be in the set.

Thank you, I'm aware of that vacuously true statement. But perhaps you could put in the full statement of the hypothesis that states when a space is separable given some property of subsets having limit points - I can't recall it off the top of my head, and it's generally best to include as much helpful information as possible.

 

[matt grime]

You've also not invoked the metric structure either, which is another worry - try to think about a compact non-metrizable space. Do such exist? Metamathematically if they don't, why bother to define compactness at all?​

I'm confused about what you mean by this?

You have a statement about compact *metric* spaces, and you've attempted a proof only invoking *compactness*. This ought to make you wonder why metricity (to invent a word) was mentioned. I repeat the fact that I'm just looking at the question you've posed and I'm simply asking you the questions that you ought to be asking yourself if you're not sure what's going on - you may have a correct proof from invoking more powerful theorems than you're supposed to use at this stage.

 

[futurebird]

Thank you, I'm aware of that vacuously true statement. But perhaps you could put in the full statement of the hypothesis that states when a space is separable given some property of subsets having limit points - I can't recall it off the top of my head, and it's generally best to include as much helpful information as possible.

I'm still rather new to all of this it's hard to judge what people will know and what's less common.

In response to you other point: I have no idea if one could have compactness without being in a metric space. To have compactness we need and open cover.

So we need open sets, open sets have all points interior. The definition of interior uses the concept of neighborhood and that requires a metric. Right?

 

[futurebird]

I don't know anything about topological spaces, though I'm reading about them on wikipedia now... I don't know if they are relevant here. Isn't a metric space just a "special kind" of topological space?

 

[futurebird]

So I'm not doing anything wrong? Just being lazy? GREAT!