Boundary points of subsets when viewed with the subset topology

[Damidami]

Hi! I have this two related questions:

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

But when I think of [itex]\mathbb{Q}[/itex] 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 [itex]\mathbb{Q}[/itex] the empty set and the full [itex]\mathbb{Q}[/itex] 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 [itex] \mathbb{Q} [/itex])
[itex] f(x) = \frac{1}{x^2 - 2} \ \ \ 0 \leq x \leq 2 [/itex]

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 [itex]\mathbb{Q}[/itex] as a subset of [itex] \mathbb{R} [/itex] is a closed set (all its points are boundary points).

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

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

But when I think of [itex]\mathbb{Q}[/itex] 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, [itex]\mathbb{Q}[/itex] is closed in itself.

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

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

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

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 [itex] \mathbb{Q} [/itex])
[itex] f(x) = \frac{1}{x^2 - 2} \ \ \ 0 \leq x \leq 2 [/itex]

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 [itex][0,2]\cap \mathbb{Q}[/itex]. 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 [itex] \mathbb{Q} \cap [0,2] [/itex] is closed and bounded (as a subset of [itex] \mathbb{Q} [/itex]). 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 [itex] \mathbb{R}^n [/itex], it's different with [itex] \mathbb{Q} [/itex] because it's not complete, right? It even isn't closed as a subset of [itex] \mathbb{R} [/itex].

 

[micromass]

Thanks a lot!
It's much clear now.

There is something that still I can't understand: The set [itex] \mathbb{Q} \cap [0,2] [/itex] is closed and bounded (as a subset of [itex] \mathbb{Q} [/itex]). 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 [itex] \mathbb{R}^n [/itex], it's different with [itex] \mathbb{Q} [/itex] because it's not complete, right? It even isn't closed as a subset of [itex] \mathbb{R} [/itex].

Indeed, the theorem talks about subsets of [itex]\mathbb{R}^n[/itex]. Your set is not closed and bounded as a subset of [itex]\mathbb{R}^n[/itex], 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 [itex] \mathbb{Q} \cap [0,2] [/itex] is closed and bounded (as a subset of [itex] \mathbb{Q} [/itex]). 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 [itex] \mathbb{R}^n [/itex], it's different with [itex] \mathbb{Q} [/itex] because it's not complete, right? It even isn't closed as a subset of [itex] \mathbb{R} [/itex].

You can check that [itex] \mathbb{Q} \cap [0,2] [/itex] 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 [itex] \mathbb{Q} \cap [0,2] [/itex] does not contain --among many other--the limit point √2 ; remember that closed subsets of a set contain all their limit points.