# A non-separable set

1. Jun 23, 2008

### Kreizhn

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

Let (X,d) be a metric space, A a closed subset of X. Suppose we've found an uncountable subset $U \subseteq A$ such that $\forall x,y \in U, \; d(x,y) \geq c$ for some positive constant c. Show that A is not separable

2. Relevant equations
A is separable if it contains a countably dense subset

3. The attempt at a solution

For the sake of contradiction assume that A is separable. That is, there exists a countable set S such that cl(S) = A, or $\forall a \in A \; \forall r>0, \; B(a;r) \cap S \neq \emptyset$

Now my idea from here is as follows. Since U is uncountable and S is necessarily countable, we can choose $x \in U\setminus _S$. Then $\forall r>0 \; B(x;r) \cap S \neq \emptyset$ by assumption. Now I want to show a contradiction using the fact that $\forall x,y \in U , \; d(x,y) \geq c$. I've been thinking about how to find a suitable r such that $B(x;r) \subseteq U\setminus_S$ which would cause $B(x;r) \cap S = \emptyset$ and give a contradiction. However, I can't find a way to guarantee that a sufficiently small r exists. Hence that idea doesn't seem to be leading anywhere, and I'm not sure what to try next.

2. Jun 23, 2008

### Dick

Take r=c/4. B(u,c/4) for each u in U must contain at least one point in the countable dense subset S but only one point in U. But no s in S is contained in more than one of the balls (why?). Can you use this to construct an injective map from an uncountable set into a countable set? What's wrong with that? I'm a little disturbed because I haven't used the fact that A is closed. But not that much, because if U is a subset of X then it's always contained in a closed set. Namely X. Have I missed something subtle?

3. Jul 1, 2008

### Kreizhn

I know the reply is a little late but I've been a bit busy.

I like the idea, but I'm not too sure about the injective map from an uncountable set to a countable one. I assume we want to map U -> S injectively, but we've only shown that every s is contained uniquely to a ball around an element in u, there's no guarantee that the number of elements in $B(u; \frac{c}{4} ) \cap S$ is precisely one, and so I can't think of a way of creating a well-defined map.

I'm wondering if perhaps we shouldn't go in the other direction; i.e. show there's a surjective map from S to U which would also give a contradiction. We can do this since every s in S can be uniquely identified with a certain u in U, and since $B(u; \frac{c}{4} ) \cap S \neq \emptyset$ then for every u, an "inverse" would exist (though again not guaranteed to be a singleton set).

Am I missing something obvious that guarantees injectivity? Is there a reason why the surjectivity wouldn't work? Indeed, if the surjective argument works, I could (although not necessary) create the stronger bijective map.

4. Jul 1, 2008

### Dick

It's certainly true you have to pick an s for each u. But that's what the axiom of choice is for. It's injective because if s is in B(u1,c/4), then it's not also in B(u2,c/4), so there's no chance of two elements of U mapping to the same element of S. The surjection ought to work as well. Take the subset S' of the elements of S that are contained in the balls. Now map them to U in the obvious way. Either way you get the contradiction card(S)>=card(U), I would say the choice of injecting one way or surjecting the other way are two different ways of stating the same argument.