# Homework Help: Identify compact subsets

1. May 16, 2010

### complexnumber

1. The problem statement, all variables and given/known data

Identify the compact subsets of $$\mathbb{Q} \cap [0,1]$$ with the relative topology from $$\mathbb{R}$$.

2. Relevant equations

3. The attempt at a solution

Is it all finite subsets of $$\mathbb{Q} \cap [0,1]$$? The relative topology contains single rational points in $$[0,1]$$, hence only finite subsets can have a finite subcover. Is this correct?

2. May 16, 2010

### Landau

Does it? An open set in $\mathbb{Q} \cap [0,1]$ containing a rational q, by definition of subspace topology, contains $(q-\epsilon,q+\epsilon)\cap\mathbb{Q}$ for some epsilon>0. But between q and q+epsilon is another rational. So I guess singletons {q} can't be open: the space is not discrete.

We do have that a discrete space is compact if and only if it is finite.

3. May 17, 2010

### complexnumber

Since $$S$$ only contain closed single rational points, and the definition of subspace topology is $$\tau_S = \{ S \cap U | U \in \tau \}$$, shouldn't the subspace topology only contain single points and not $$(q-\epsilon, q + \epsilon)$$?

4. May 17, 2010

### Landau

But what does your $U\in\tau$ look like? U is an open in R, so contains an open interval. Every open interval contains at least two rationals q1,q2. And S also contains q1 and q2. So $S\cap U$ contains q1 and q2. (of course all restricted to [0,1])
Or am I making a mistake?

5. May 17, 2010

### complexnumber

$$U \in tau$$ can be any open interval in $$\mathbb{R}$$ and contain infinite number of rational numbers, but when it intersects with a set of rational numbers, the resulting set contains infinite number of rational points. You were right. I didn't think what $$U$$ is.

6. May 22, 2011

### jeckt

I'm also doing this question right now - and the answer aren't very clear but I agree with the OP. The only compact sets are finite subsets of $$\mathbb{Q}\cap [0,1]$$ since the relative topology is rational points.

But that seems too simple...

7. May 22, 2011

### Dick

Think about the set {0,1,1/2,1/3,1/4,1/5,...}. Finite, no. Compact?

8. May 22, 2011

### jeckt

Hmmm....I'm not seeing why the set would be compact. Maybe I'm viewing the relative topology wrong. I'm thinking that the relative topology is rational points or sets of rational points in $$[0,1]$$ as the above posters have said.

So the set you said - there wouldn't be a finite subcover the way I'm thinking about the relative topology i don't think...unless I'm not thinking hard enough.

Thanks for the help btw!

9. May 22, 2011

### Dick

A basis for the relative topology in [0,1] intersect Q is open intervals in R intersected with [0,1] intersect Q. It's not rational points or finite subsets of rational points and certainly not ANY subset of rational points. Any nonempty subset of the relative topology has an infinite number of points. If C is a cover of A={0,1,1/2,1/3,1/4,1/5,...} then one of the open sets contains 0. You can also think of this problem in terms of sequential compactness. Is there a subsequence of A that converges to something not in A?

Last edited: May 22, 2011
10. May 23, 2011

### jeckt

Sorry I'm a bit slow but I'm not seeing how open intervals in $$\mathbb{R} \cap [0,1] \cap \mathbb{Q}$$ has an infinite number of points (you mean infinite number of rational points right? if that what you mean then that makes sense to me)

11. May 23, 2011

### Dick

Of course I mean an infinite number of rational points. They are just intervals in Q.

12. May 28, 2011

### jeckt

Been so busy lately haven't been able to give this some thought. Hmmm I get how the topology works now (i think) but the problem becomes i can't think of a subset of $$\mathbb{Q} \cap [0,1]$$ that isn't compact.

Maybe you can head me in the right direction.

And again - thanks for all the help

13. May 28, 2011

### HallsofIvy

Look at the example Dick gave in post #7 but drop "0" from the set:
{1, 1/2, 1/4, 1/5, ...}. For any n, there exist a rational number strictly between 1/n and 1/(n+1) so that, for each of those points, we can find an open interval that contains that point but no others in that set.

Last edited by a moderator: May 29, 2011
14. May 28, 2011

### jeckt

oh wow! Thanks HallsofIvy!! That really clears things up - so any set that contains it's limit points and finite subsets of $$\mathbb{Q}\cap [0,1]$$ are compact. So the set you mention doesn't have its limit point; which is zero.

If the set contained 0 then any cover of the point 0 will contain infinitely many points of the set. Thus with the other points not covered by the interval containing zero, there would be finitely many of them and your argument wouldn't construct a cover with no finite subcover.

Generalising that you can look at a set such as $$\left\{ 1 - \frac{1}{n} \right\}_{n=1}^{\infty }$$. With your argument this set won't be compact but if add the point {1} to the set then it is.

Am I on the right track?

Thanks again btw, I feel like I'm understanding the problem much better now.

15. May 28, 2011

### Dick

You are on the right track with the second example, sure. But the point to the first example is that if the set contains 0 then to cover 0 you need to need to contain an open set in the induced topology which would have the form [0,r) for some number r>0. Go from there. Just saying including 0 means including an infinite number of points is pretty vague.

Last edited: May 29, 2011
16. May 29, 2011

### jeckt

yeah thanks dick. I was just trying to get the ideas down but mathematically I'd have to define it much more rigorously. Other than that I think that's all the compact subsets there are for this question.

Thanks for the great help guys.