# Boundary points of subsets when viewed with the subset topology

1. Nov 30, 2011

### 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 continous 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 dont know in what part of the proof it was used that the compact set has to be also a complete space.

Thanks.

2. Nov 30, 2011

### micromass

Staff Emeritus
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 $\mathbb{R}$.
Indeed, $\mathbb{Q}$ is closed in itself.

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.

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!!

A compact set in a metric space is ALWAYS complete!!

3. Nov 30, 2011

### 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 $\mathbb{R}$.

4. Nov 30, 2011

### micromass

Staff Emeritus
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.

5. Dec 2, 2011

### Bacle2

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.