# Open sets in a separable metric space

• Sep 19th 2009, 10:45 AM
Open sets in a separable metric space
Suppose that $\displaystyle (X, \rho )$ is a separable metric space. Then there exists a sequence $\displaystyle \{ x_n \}$ such that it is dense in X. Prove that any open set $\displaystyle U \subset X$ is a union of elements from the set $\displaystyle B = \{ B(x_n,r) : r \in \mathbb {Q} \}$.

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 $\displaystyle \{ x_n \}$.

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

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

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

Thank you!
• Sep 19th 2009, 12:57 PM
Opalg
Quote:

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

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 $\displaystyle \{ x_n \}$.

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

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

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

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

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

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

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

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.
• Sep 24th 2009, 09:22 AM
Quote:

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

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

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

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

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

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

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

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

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

To show that $\displaystyle B(x_m,r/2)\subset B(x,r)$, you need to pick $\displaystyle y\in B(x_m,r/2)$. Then (triangle inequality!) $\displaystyle \rho(y,x)\leqslant\rho(y,x_m) + \rho(x_m,x) < \tfrac r2 + \tfrac r2 = r$, and so $\displaystyle y\in B(x,r)$.
• Sep 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 $\displaystyle \{ x_n \}$.
This is actually because $\displaystyle \{x_n\}$ is dense in E.