Separable Hausdorff Spaces: Example with Non-Separable Subspace

[tylerc1991]

Homework Statement

Give an example of a separable Hausdorff space (X,T) that has a subspace (A,T_A) that is not separable.

Homework Equations

A separable space is one that has a countable dense subset.

The Attempt at a Solution

Let (X,T) (i.e. the separable Hausdorff space) be the real number line with the usual topology. Now let the subspace (A,T_A) be the irrational numbers. The topology generated will in essence be the discrete topology. Since the irrational numbers are not countable they do not have a countable dense subset. If this is just rubbish can someone give me a kick in the right direction? Thank you very much!

 

[mathwonk]

why is the induced topology discrete? that is what scrapes my molecules on hearing it.

 

[radou]

Hm, an example doesn't not occur to me right now, but this won't work, since for a countable dense subset of the irrationals, you can take the the set {$$\frac{m\sqrt(2)}{n}$$ : m, n are integers}.

 

[tylerc1991]

why is the induced topology discrete? that is what scrapes my molecules on hearing it.

True, since the irrationals are dense, the induced topology is not discrete. But the irrationals are not countable, so they don't have a countable dense subset?

Hm, an example doesn't not occur to me right now, but this won't work, since for a countable dense subset of the irrationals, you can take the the set {$$\frac{m\sqrt(2)}{n}$$ : m, n are integers}.

There has to be just one integer relation in the set builder notation for a countable set correct? and what about e or pi?

 

[micromass]

Hmm, it took me some time to come up with this. Do you know what ordinal numbers are?

If you do, then you could take $$\omega_1$$, the first uncountable ordinal. This induces a Hausdorff topological space which is not seperable. On the other hand, the space $$\omega_1+1$$ is seperable and contains $$\omega_1$$.

If the above is chinese to you, then I'll look for another example. As a hint though: the book "Counterexamples of topology" by Steen and Seebach contains all these fun examples. I really recommend it if you're serious about topology.

 

[micromass]

True, since the irrationals are dense, the induced topology is not discrete. But the irrationals are not countable, so they don't have a countable dense subset?

The irrational DO have a countable dense subset. In fact, all the subsets of $$\mathbb{R}$$ have a countable dense subset, it is only a bit tricky to find it.
In case of the irrationals: $$\{\pi+q~\vert~q\in \mathbb{Q}\}$$ is a countable dense subset.

 

[tylerc1991]

The irrational DO have a countable dense subset. In fact, all the subsets of $$\mathbb{R}$$ have a countable dense subset, it is only a bit tricky to find it.
In case of the irrationals: $$\{\pi+q~\vert~q\in \mathbb{Q}\}$$ is a countable dense subset.

But there is no correspondence between the irrational numbers and the integers. I thought that is what made the irrational numbers irrational, and hence uncountable?

But if this is in fact false, going back to the original question, this seems much more difficult. If every subset of the reals is countable then how am I going to answer the question?

 

[micromass]

But there is no correspondence between the irrational numbers and the integers. I thought that is what made the irrational numbers irrational, and hence uncountable?

Yes, the irrational numbers are uncountable this is true. BUT it does have a countable dense subset! Separability doesn't demand the countability of the whole space, but only of a dense subset of it.

But if this is in fact false, going back to the original question, this seems much more difficult. If every subset of the reals is countable then how am I going to answer the question?

This question is indeed a bit hard. Coming up with a suitable counterexample is the trick. I take it that you have never heard of ordinals before. So we will try something else. Let's begin with constructing the space A. We will extend A later to X. But A has to be not separable and Hausdorff. Can you think of a space that satisfies this?

 

[tylerc1991]

Yes, the irrational numbers are uncountable this is true. BUT it does have a countable dense subset! Separability doesn't demand the countability of the whole space, but only of a dense subset of it.



This question is indeed a bit hard. Coming up with a suitable counterexample is the trick. I take it that you have never heard of ordinals before. So we will try something else. Let's begin with constructing the space A. We will extend A later to X. But A has to be not separable and Hausdorff. Can you think of a space that satisfies this?

A doesn't necessarily have to be Hausdorff, just not separable. All of the spaces we have dealt with up to this point have been separable as far as I know. What about a singleton like {0}? Is this separable?

 

[micromass]

Well, if X has to be Hausdorff, then every subspace of X is Hausdorff. So we do want A to be Hausdorff, otherwise our A will be useless in our construction of the counterexample.

Yes, {0} is separable. In fact, all finite and countable sets are separable. So you'll have to come up with an uncountable set to get a non-separable space... So, what uncountable topological spaces do you know?

 

[tylerc1991]

Well, if X has to be Hausdorff, then every subspace of X is Hausdorff. So we do want A to be Hausdorff, otherwise our A will be useless in our construction of the counterexample.

Yes, {0} is separable. In fact, all finite and countable sets are separable. So you'll have to come up with an uncountable set to get a non-separable space... So, what uncountable topological spaces do you know?

Well the real numbers are uncountable, but they are separable. I'm pretty sure all metric spaces are separable, so it's probably something that I've never heard of, which makes the question that much more frustrating.

 

[micromass]

No, not all metric spaces are separable. And the topology that I'm thinking of is a metric space. And it is certainly a topology that you know.

OK, can you perhaps list every topological space that you know? And can you add if they are separable or not?

 

[tylerc1991]

No, not all metric spaces are separable. And the topology that I'm thinking of is a metric space. And it is certainly a topology that you know.

OK, can you perhaps list every topological space that you know? And can you add if they are separable or not?

R^n with usual topology - separable
Natural numbers with discrete topology - separable
Rational numbers with topology generated by usual topology on reals - separable
Irrational numbers with topology generated by usual topology on reals - separable
countless number of finite topologies - all separable

lower limit topology on reals - don't think this is metrizable
finite complement topology on reals - ""
etc.

 

[micromass]

Yes, go on! We're looking for a very basic kind of topology. Probably one of the first topologies that you've heard of.

 

[tylerc1991]

The trivial topology isn't metrizable, but the discrete topology on the reals is metrizable. But this has a countable local basis at each point, and hence makes the metric space second countable and hence separable, correct?

 

[micromass]

The trivial topology isn't metrizable, but the discrete topology on the reals is metrizable. But this has a countable local basis at each point, and hence makes the metric space second countable and hence separable, correct?

Yes, the discrete topology has a countable local base, but this only makes it first countable. Second countable spaces have a countable base (not just a countable local base).

In fact, the discrete topology was exactly the one I was looking for. Can you show that the discrete topology is non-separable (when it is uncountable).

 

[tylerc1991]

Yes, the discrete topology has a countable local base, but this only makes it first countable. Second countable spaces have a countable base (not just a countable local base).

In fact, the discrete topology was exactly the one I was looking for. Can you show that the discrete topology is non-separable (when it is uncountable).

Possibly, but the bigger question is this: what is the superset going to be now? if the discrete topology on the reals is some induced topology from a superset, I'm not allowed to say that we have some usual topology with the real numbers anymore, right? Rephrased, now that we have found (or it is yet to be proved) that (A,T_A) is not separable, how then am I going to go backwards to find a superset that IS separable?

 

[micromass]

Well, that's something you'll have to figure out. You'll need to adjoin some points to A and put a topology on it such that it is separable.

Now, have you seen ways to adjoin only one point to A and put a natural topology on that?

 

[tylerc1991]

Well, that's something you'll have to figure out. You'll need to adjoin some points to A and put a topology on it such that it is separable.

Now, have you seen ways to adjoin only one point to A and put a natural topology on that?

I have not. This question actually comes fairly early in the book. I will think more about it and will talk to the prof. 1 on 1 about it I suppose, thank you very much for your time and help!