Boundary points of subsets when viewed with the subset topology

[Damidami]

Hi! I have this two related questions:

(1) I was thinking that \mathbb{Q} as a subset of \mathbb{R} is a closed set (all its points are boundary points).

But when I think of \mathbb{Q} not like a subset, but like a topological space (with the inherited subspace topology), are all it's points still boundary points? I think not, as there ir no "exterior" points now. So are the only clopen sets of \mathbb{Q} the empty set and the full \mathbb{Q} set now?

(2) The other question: I saw this example in the book Counterexamples in Analysis:

A function continuous on a closed interval and not bounded there (and therefor, not uniformly continuous there) (Note: the domain is a subset of \mathbb{Q})
f(x) = \frac{1}{x^2 - 2} \ \ \ 0 \leq x \leq 2

It confuses me because of the theorem that a continuous function over a compact set is uniformly continuous. I don't know in what part of the proof it was used that the compact set has to be also a complete space.

Thanks.

 

[micromass]

Hi! I have this two related questions:

(1) I was thinking that \mathbb{Q} as a subset of \mathbb{R} is a closed set (all its points are boundary points).

No, \mathbb{Q} is certainly NOT a closed subset of \mathbb{R}. Sure, all the points of \mathbb{Q} are boundary points, but they are not ALL the boundary points! Indeed, every irrational number is also a boundary point of \mathbb{Q}.

Specifically, the boundary of \mathbb{Q} is \mathbb{R}. And the closure of \mathbb{Q} is also \mathbb{R}.

But when I think of \mathbb{Q} not like a subset, but like a topological space (with the inherited subspace topology), are all it's points still boundary points? I think not, as there ir no "exterior" points now.

Indeed, \mathbb{Q} is closed in itself.

So are the only clopen sets of \mathbb{Q} the empty set and the full \mathbb{Q} set now?

No, certainly not! There are many clopen sets in \mathbb{Q}. For example

]-\pi,\pi[\cap \mathbb{Q}=[-\pi,\pi]\cap \mathbb{Q}

is also clopen.

(2) The other question: I saw this example in the book Counterexamples in Analysis:

A function continuous on a closed interval and not bounded there (and therefor, not uniformly continuous there) (Note: the domain is a subset of \mathbb{Q})
f(x) = \frac{1}{x^2 - 2} \ \ \ 0 \leq x \leq 2

It confuses me because of the theorem that a continuous function over a compact set is uniformly continuous.

First of all, the domain of our function is [0,2]\cap \mathbb{Q}. This is NOT compact. We can see this because compact => complete. Since the above set is not complete, it can also not be compact!

I don't know in what part of the proof it was used that the compact set has to be also a complete space.

A compact set in a metric space is ALWAYS complete!

 

[Damidami]

Thanks a lot!
It's much clear now.

There is something that still I can't understand: The set \mathbb{Q} \cap [0,2] is closed and bounded (as a subset of \mathbb{Q}). So doesn't that mean that it is compact?

I already read that compact implies complete, but on the other hand doesn't the Heine-Borel theorem apply?

Now that I think better, that theorem talks about closed and bounded subsets of \mathbb{R}^n, it's different with \mathbb{Q} because it's not complete, right? It even isn't closed as a subset of \mathbb{R}.

 

[micromass]

Thanks a lot!
It's much clear now.

There is something that still I can't understand: The set \mathbb{Q} \cap [0,2] is closed and bounded (as a subset of \mathbb{Q}). So doesn't that mean that it is compact?

I already read that compact implies complete, but on the other hand doesn't the Heine-Borel theorem apply?

Now that I think better, that theorem talks about closed and bounded subsets of \mathbb{R}^n, it's different with \mathbb{Q} because it's not complete, right? It even isn't closed as a subset of \mathbb{R}.

Indeed, the theorem talks about subsets of \mathbb{R}^n. Your set is not closed and bounded as a subset of \mathbb{R}^n, so we cannot conclude compactness.

 

[Bacle2]

Thanks a lot!
It's much clear now.

There is something that still I can't understand: The set \mathbb{Q} \cap [0,2] is closed and bounded (as a subset of \mathbb{Q}). So doesn't that mean that it is compact?

I already read that compact implies complete, but on the other hand doesn't the Heine-Borel theorem apply?

Now that I think better, that theorem talks about closed and bounded subsets of \mathbb{R}^n, it's different with \mathbb{Q} because it's not complete, right? It even isn't closed as a subset of \mathbb{R}.

You can check that \mathbb{Q} \cap [0,2] is not compact, since, e.g., in any compact metric space , every sequence has a convergent subsequence. Take, then,
any sequence that "wants to" converge to , e.g., √2, and see that it cannot have a
convergent subsequence. And Heine-Borel does not apply, since the set is not closed (subset of ℝⁿ): closed (as Micromass also pointed out) , since \mathbb{Q} \cap [0,2] does not contain --among many other--the limit point √2 ; remember that closed subsets of a set contain all their limit points.