Open sets in a separable metric space

• September 19th 2009, 10:45 AM
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). 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
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). 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} .

Pick $y \in B(x,r)$, then I have $\rho (y,x), 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} .

Pick $y \in B(x,r)$, then I have $\rho (y,x), 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.