Open sets in a separable metric space

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September 19th 2009, 10:45 AM
tttcomrader

Open sets in a separable metric space

Suppose that $(X, \rho )$ is a separable metric space. Then there exists a sequence $\{ x_n \}$ such that it is dense in X. Prove that any open set $U \subset X$ is a union of elements from the set $B = \{ B(x_n,r) : r \in \mathbb {Q} \}$ .

My idea so far.

So since X is separable, I know that every nonempty open subset U contains at least one element of the sequence $\{ x_n \}$ .

So say that $U_1$ is an nonempty open subset in X, then there should exist at least one element from the sequence, say $x_m, \ m \in \mathbb {N}$ , that is contained in U. And since U is open, we can find some $\epsilon _1 > 0$ such that $B(x_m, \epsilon _1) \subset U$ .

Also, I believe that I can pick a point, say $y \in U$ , such that $x_m \in B(y, \delta ) ^c$ for some small $\delta > 0$ . Then I should be able to find another point $x_i \in B(y, \delta )$ since it is open. Then I can get another $B(x_i , \epsilon _2 )$ ...

But how should I proceed to get $U \subset \bigcup B(x_n,r)$ ?

Thank you!
September 19th 2009, 12:57 PM
Opalg

Quote:

Originally Posted by tttcomrader

Suppose that $(X, \rho )$ is a separable metric space. Then there exists a sequence $\{ x_n \}$ such that it is dense in X. Prove that any open set $U \subset X$ is a union of elements from the set $B = \{ B(x_n,r) : r \in \mathbb {Q} \}$ .

My idea so far.

So since X is separable, I know that every nonempty open subset U contains at least one element of the sequence $\{ x_n \}$ .

So say that $U_1$ is an nonempty open subset in X, then there should exist at least one element from the sequence, say $x_m, \ m \in \mathbb {N}$ , that is contained in U. And since U is open, we can find some $\epsilon _1 > 0$ such that $B(x_m, \epsilon _1) \subset U$ .

Also, I believe that I can pick a point, say $y \in U$ , such that $x_m \in B(y, \delta ) ^c$ for some small $\delta > 0$ . Then I should be able to find another point $x_i \in B(y, \delta )$ since it is open. Then I can get another $B(x_i , \epsilon _2 )$ ...

But how should I proceed to get $U \subset \bigcup B(x_n,r)$ ?

Let $x\in U$ . Since U is open, there is a neighbourhood of x contained in U. So there exists a rational number r such that $B(x,r)\subset U.$

Next, the sequence $\{x_n\}$ is dense in X, so there exists m such that $\rho(x_m,x)<r/2$ . Then $x\in B(x_m,r/2)\subset B(x,r)\subset U.$

Thus every element of U lies in a set in $B$ that is contained in U. The union of these sets (for $x\in U$ ) is therefore the whole of U.
September 19th 2009, 01:04 PM
Plato

Quote:

Originally Posted by tttcomrader

Suppose that $(X, \rho )$ is a separable metric space. Then there exists a sequence $\{ x_n \}$ such that it is dense in X. Prove that any open set $U \subset X$ is a union of elements from the set $B = \{ B(x_n,r) : r \in \mathbb {Q} \}$ .

I know that you want comment on you have done.
But frankly I do not follow what you are trying to do.

But this follows at once from a standard theorem.
A separable metric space has a countable basis.
If you can prove that then you will have done this problem.
September 24th 2009, 09:22 AM
tttcomrader

Quote:

Originally Posted by Opalg

Let $x\in U$ . Since U is open, there is a neighbourhood of x contained in U. So there exists a rational number r such that $B(x,r)\subset U.$

Next, the sequence $\{x_n\}$ is dense in X, so there exists m such that $\rho(x_m,x)<r/2$ . Then $x\in B(x_m,r/2)\subset B(x,r)\subset U.$

Thus every element of U lies in a set in $B$ that is contained in U. The union of these sets (for $x\in U$ ) is therefore the whole of U.

I follow the proof, but how do I know that $B(x_m,r/2)\subset B(x,r)$ ? Thanks.

So if I pick an element $y \in B(x,r)$ , I have $\rho (x,y) < r$ , and since $x\in B(x_m, \frac {r}{2} )$ , I then have $\rho (x,x_m) < \frac {r}{2} <r$ .

Pick $y \in B(x,r)$ , then I have $\rho (y,x)<r$ , is that how I should proceed?
September 24th 2009, 10:05 AM
Opalg

Quote:

Originally Posted by tttcomrader

I follow the proof, but how do I know that $B(x_m,r/2)\subset B(x,r)$ ? Thanks.

So if I pick an element $y \in B(x,r)$ , I have $\rho (x,y) < r$ , and since $x\in B(x_m, \frac {r}{2} )$ , I then have $\rho (x,x_m) < \frac {r}{2} <r$ .

Pick $y \in B(x,r)$ , then I have $\rho (y,x)<r$ , is that how I should proceed?

To show that $B(x_m,r/2)\subset B(x,r)$ , you need to pick $y\in B(x_m,r/2)$ . Then (triangle inequality!) $\rho(y,x)\leqslant\rho(y,x_m) + \rho(x_m,x) < \tfrac r2 + \tfrac r2 = r$ , and so $y\in B(x,r)$ .
September 24th 2009, 01:03 PM
Moo

Just a remark :

Quote:

So since X is separable, I know that every nonempty open subset U contains at least one element of the sequence $\{ x_n \}$ .

This is actually because $\{x_n\}$ is dense in E.