# Locally Compact

1. Nov 29, 2005

### philosophking

I'm having a hard time understanding why the rationals are not locally compact.

The definition of local compactness is that every neighborhood of x in X is contained in some compact subset of X.

But what are the compact subsets of X? I think this is my biggest problem. I know that the compact subsets of the reals are just the closed and bounded intervals, but I am unsure as to what they look like in the space of rationals.

Could i say that in any interval about x, there contains both rationals and irrationals (the set of rationals and irrationals is dense in the reals), so any compact set would have to contain irrationals too... or something :(. I'm very confused. Thanks for the help.

2. Nov 29, 2005

### matt grime

What they (compact sets) look like in the rationals depends on the topology you're giving them. You are after all treating the rationals, Q, as a topological space, with some topology, and trying to show this topological space is locally compact. This will depend on the topology you are placing in Q. Presumably you wish to think about the subspace topology from Q inside R, and a generic nbd of x is then (a,b)nQ for some open set (a,b) in R. Why is this set not contained in some compact subset?

Well, try finding a compact subset of Q with the subspace topology that does not contain a finite number of points (any set with only a finite number of points in it is automatically compact in any topology: given an infinite open cover X_i of the set {y_1,..,y_n} pick one open set containing each of the y_i (possibly with repetition, that is the same open set may cover y_1 and y_4 etc)).

That should be a broad enough hint. Or even more explicitly, take any set in Q that contains infinitely many points. Can you think of any infinite covers of it by open sets? Can you make the open sets in the cover all disjoint?

A compact set *in Q* cannot contain ANY irrationals since it is a subset of Q, by the way.

Last edited: Nov 29, 2005
3. Nov 29, 2005

### philosophking

This is great, thank you so much for the help. It is indeed the subspace topology that I'm working with (I overlooked that very large detail).

Also, wouldn't any compact subset of Q with the subspace topology contain an infinite number of points? Because between any two numbers there is an infinite number of rational numbers.

From there, I guess I would say that I can find an infinite cover of this subset that does not contain a finite subcover, thus proving that it is not compact. And I do this by making the open sets all disjoint, but I will work on this later.

Oh yeah, and to your last point--I wanted to show that a compact set of Q would have to contain irrationals (because doesn't a neighborhood of x if x is in Q contain irrationals, or is this wrong?), which would lead to a contradiction, meaning no compact set contains a neighborhood in the rationals.

4. Nov 29, 2005

### matt grime

I repeat myself, but ANY set in Q compact or otherwise, does not contain a single irrational number BY DEFINITION. We are only dealing with sets that lie inside Q.

The following quote makes no sense, since it is self contradictory:

"Also, wouldn't any compact subset of Q with the subspace topology contain an infinite number of points? Because between any two numbers there is an infinite number of rational numbers.
From there, I guess I would say that I can find an infinite cover of this subset that does not contain a finite subcover."

How can something be both compact and not compact?

5. Nov 30, 2005

### philosophking

yeah, i shouldn't have said "compact subset of Q". I should have just said "any subset of Q". sorry for the confusion!

6. Nov 17, 2010

### Ice Vox

That's not true at all... there is in fact an infinite subset of the rationals which is compact under the standard topology...

Consider the set {1, 1/2, 1/3, 1/4, ...} U {0}

This set is compact because any open set around 0 contains all but a finite number points in the set. I understand that that is not directly applicable to this situation because the open sets are of the form (a,b) n Q but it is a problem I ran into when trying to make this proof work.