Showing that the rationals are not locally compact

[radou]

Homework Statement

This seems suspiciously easy, so I'd like to check my reasoning.

The Attempt at a Solution

I used the following theorem:

If X is a Hausdorff space, then X is locally compact iff given x in X, and a neighborhood U of x, there exists a neighborhood V of x such that Cl(V) is compact and Cl(V) is contained in U.

The rationals are clearly Hausdorff. Assume they are locally compact, so the implication from the theorem holds. But then there are irrational numbers in Cl(V), and such a set cannot possibly be contained in U.

 

[Office_Shredder]

If you treat the rationals as a metric space, it's not embedded in the reals so when you take the closure of a set you don't pick up any real numbers. You actually have to look at what an open set and its closure look like, and show that it's not going to be compact

 

[micromass]

Hmm, it seems that your confused between some notions.

When asking if the rationals are locally compact, they are talking about the topological space of the rationals. The closure in this topological space will only contain rational numbers. We are not talking about the closure in R here, this is a very different closure operator!

There is however a relation between the closure in Q and the closure in R, that is:

$$cl_\mathbb{Q}(A)=cl_\mathbb{R}(A)\cap \mathbb{Q}$$

So, when talking about the topological space $$\mathbb{Q}$$ means that all open sets are in $$\mathbb{Q}$$, all closed sets are in $$\mathbb{Q}$$, and the closure operator yields only sets in $$\mathbb{Q}$$

Hmm, not sure if I explained this the right way... But all I want to say is that you shouldn't confue between the closure in Q and the closure in R. These are different thing.
For example, take $$]0,1[\cap \mathbb{Q}$$. This is an open set of Q. The closure in R of this set is [0,1]. The closure in Q on the other hand is $$[0,1]\cap \mathbb{Q}$$...

 

[radou]

That's exactly the mistake I feared I'd make. :)

OK, since the exercise texts is "Show that the rationals Q are not locally compact.", I assume it referrs to the rationals alone, and not as a subspace of R, right?

 

[micromass]

Yes, you have to take only the rationals. You can't take Q as subspace of R.

But that doesn't mean you can't work with R in any way. Let i:Q--> R be the canonical injection, which is continuous. Then for any compact set K of Q, it must be so that i(K) is compact in R. Thus we DO have that if a set is compact in Q, it must be compact in R. So while your initial argument was incorrect, you can make the connection to R (and you will have to...)

 

[radou]

OK, but before that, what do the open sets in the rationals look like? Which topology do we have here (since we don't have the subspace topology, which was just resolved in the upper posts)? Obviously local compactness is a topological property, so it will in general depend on the topology?

 

[micromass]

We do have the subspace topology on $$\mathbb{Q}$$.

All that I said in my previous posts is that the closure of a rational set in Q, stays in Q, as opposed to take the closure of a rational set in R.

 

[radou]

OK, I helped myself with the "closure issues" with the following theorem:

If Y is a subspace of X, then a set A is closed in Y iff it equals the intersection of a closed of X with Y.

So, we did obtain the structure of the open and closed sets in Q through defining the subspace topology on Q as a subspace of R, right? For example, for some <a, b> in R, the set <a, b>$$\cap$$Q is open in Q and its closure equals [a, b]$$\cap$$Q?

 

[micromass]

Yes!
So you see now that the closure of a set in Q doesn't contain any irrational...

 

[radou]

Yes!
So you see now that the closure of a set in Q doesn't contain any irrational...

OK, that's actually what I envisioned at the beginning, only it was tremendously stupid of me to take closures of sets in Q (as a subspace of R) in R!

Now that this is straightened out, I'll think about it and update a bit later.

 

[micromass]

Good luck!

 

[radou]

By the way, another way to eventually prove this just occurred to me.

If a subset of the rationals is compact, then, since the rationals are metrizable, it must be sequentially compact, too. Can one prove that for any interval of rational numbers one can find a sequence with no convergent subsequence?

 

[radou]

Basically, what one needs to prove is that no closed interval of rational numbers [a, b] is compact.

 

[micromass]

Wouldn't it be easier the way I pointed out?
Since the image of a compact set must be compact, we have that if a set A in $$\mathbb{Q}$$ is compact, then A is compact in $$\mathbb{R}$$.

This implies that $$[a,b]\cap \mathbb{Q}$$ is not compact in Q, since it is not compact in R.

If you want, you can probably prove this through sequential compactness, but you'll have to make a detour through R again. I don't think it's all that obvious to... :uhh:

 

[radou]

Let [a, b] be a interval of rational numbers. This interval contains at least one irrational number c, and it is closed in our subspace. Then the intervals [a, c-1/n] and [c+1/n, b], for some suitable "starting" n (and hence n+1, n+2, etc.), such that c-1/n and c+1/n lie in the interval [a, b], cover [a, b] but have no finite subcover (I found a proof that [0, 1] is not compact in the rationals, so this should be analogous).

So this contradicts the theorem I was using in post #1, and the rationals are not locally compact.

I hope this works and that I didn't write something stupid again, but there always seem to be some issues when I consider the rationals, so I'd like to clear them up.

 

[micromass]

Yes, that seems fine! Nice proof

 

[radou]

Oh, I just saw you post!

Well yes, using the canonical injection seems easy enough, thanks. :)

 

[micromass]

On a related side, I really do think that you can show that the finite sets are the only compact sets in Q...

 

[radou]

OK, thanks a bunch! (by the way, I just eliminated the number 666 from my post counter )

 

[micromass]

Lol, the devil won't like that

 

[radou]

On a related side, I really do think that you can show that the finite sets are the only compact sets in Q...

Well, by the same analogy, we can prove that any open interval of rationals <a, b> isn't compact either, so this only leaves finite sets, right?

 

[micromass]

Yes, you're right

 

[radou]

Btw, this canonical injection is defined like the identity, right? So it maps every rational number to the same one?

 

[micromass]

Yes, the canonical embedding is just $$i:\mathbb{Q}\rightarrow\mathbb{R}:x\rightarrow x$$

 

[radou]

OK, so proving its continuity is trivial, right? i.e. since the indentity map i : R --> R is continuous, if we remove all irrational numbers from the domain, it stays continuous, right?

 

[micromass]

Yes, you just restrict the domain. So the continuity follows from Theorem 18.2, page 107...

 

[radou]

Yes, I just looked it up. OK, thanks!

 

[Valjeanie]

Let [a, b] be a interval of rational numbers. This interval contains at least one irrational number c, and it is closed in our subspace. Then the intervals [a, c-1/n] and [c+1/n, b], for some suitable "starting" n (and hence n+1, n+2, etc.), such that c-1/n and c+1/n lie in the interval [a, b], cover [a, b] but have no finite subcover (I found a proof that [0, 1] is not compact in the rationals, so this should be analogous).

So this contradicts the theorem I was using in post #1, and the rationals are not locally compact.

I hope this works and that I didn't write something stupid again, but there always seem to be some issues when I consider the rationals, so I'd like to clear them up.

I'm confused, is that an open cover, because it seems like all of those sets are closed? Or is this different because we're talking about the closure being compact?