# sigma-algebra in separable space

• Oct 2nd 2013, 11:23 PM
drabi
sigma-algebra in separable space
Hi,
I don't know how to prove this exercise, actually I don't even know how to start (Thinking):

Let $(E,d)$ is separable metric space. Prove that its borel $\sigma$-algebra $B(E)$ is generated by system $U(x,\epsilon) = \left\{y \in E: d(y,x)<\epsilon\right\}$.

I would appreciate any hint.
• Oct 3rd 2013, 02:42 AM
Plato
Re: sigma-algebra in separable space
Quote:

Originally Posted by drabi
Let $(E,d)$ is separable metric space. Prove that its borel $\sigma$-algebra $B(E)$ is generated by system $U(x,\epsilon) = \left\{y \in E: d(y,x)<\epsilon\right\}$.
I would appreciate any hint.

This notation uses vocabulary that is hard for me to follow. I think it means that you are to show that the set of balls. $\mathcal{U}(x;\epsilon)$ forms a basis for the metric topology.

The countable part comes from the given separable. So perhaps you need to post some more about the terms here. Does "its borel $\sigma$-algebra $B(E)$" simply mean the metric generated topology?

Do you have a definitions of basis?
• Oct 5th 2013, 06:47 AM
drabi
Re: sigma-algebra in separable space
Hi,

I'm sorry for the vocabulary, I'll try to explain:
borel $\sigma$-algebra $\mathcal{B}(E)$ means $\sigma$-algebra generated by open subsets of $E$.

So I'm supposed to show, that it's ok to take only the sets of open balls $\mathcal{U}(x,\epsilon)$ instead of all open subsets.

It is clear now?
• Oct 5th 2013, 07:36 AM
Plato
Re: sigma-algebra in separable space
Quote:

Originally Posted by drabi
borel $\sigma$-algebra $\mathcal{B}(E)$ means $\sigma$-algebra generated by open subsets of $E$.
So I'm supposed to show, that it's ok to take only the sets of open balls $\mathcal{U}(x,\epsilon)$ instead of all open subsets.

The fact that $E$ is separable means that it contains a countable dense subset $\mathfrak{D}$.

If $\mathcal{O}$ is an open set in $E$ then for each $x\in\mathcal{O}$ then $\exists\delta_x$ such that $\mathcal{U}(x,\delta_x)\subset\mathcal{O}$.

Now let $\epsilon_x=\delta_x/2$. Then $(\exists t_x\in\mathfrak{D})[d(x,t_x)<\epsilon_x]$.
Thus we see that $x\in\mathcal{U}(t_x,\epsilon_x)]\subset\mathcal{O}$.

$\mathcal{O}=\bigcup\limits_{x \in \mathcal{O}} \mathcal{U}(t_x,\epsilon_x)$

I have done far too much for you. So you fill in details.
• Oct 5th 2013, 08:46 AM
SlipEternal
Re: sigma-algebra in separable space
Let $\mathcal{O}(E)$ be the set of all open sets of $E$. Let $\mathcal{P}(E)$ be the power set of $E$.
Let $\mathcal{B}(E) = \left\{B \subseteq \mathcal{P}(E) \mid B \mbox{ is a }\sigma\mbox{-algebra and }\mathcal{O}(E) \subseteq B \right\}$

Then by definition, we know the Borel $\sigma$-algebra $B(E) = \bigcap_{B \in \mathcal{B}(E)} B$.

Next, let $U(E)$ be the $\sigma$-algebra generated by the sets $\mathcal{U}(x;\epsilon)$. In other words, it is the intersection of all $\sigma$-algebras containing the sets $\mathcal{U}(x;\epsilon)$.

Now, we want to show that $U(E) = B(E)$. How do you show that two sets are equal? You show that $U(E) \subseteq B(E)$ and $B(E) \subseteq U(E)$.

For any $\epsilon>0$ and any $x\in E$, $\mathcal{U}(x;\epsilon)$ is an open set of $E$, so it is an element of $\mathcal{O}(E)$. Since $\mathcal{U}(x;\epsilon) \in \mathcal{O}(E) \subseteq B(E)$, and $U(E)$ is the smallest $\sigma$-algebra containing all sets $\mathcal{U}(x;\epsilon)$, it must be that $U(E) \subseteq B(E)$.

Next, how do we show that $B(E) \subseteq U(E)$? We take an arbitrary open set $O \in \mathcal{O}(E)$. Then we need to show that $O \in U(E)$. If this is true for any arbitrary $O$, then since $B(E)$ is the smallest $\sigma$-algebra containing $\mathcal{O}(E)$, it must be that $B(E) \subseteq U(E)$.

So, question: What exactly is the system $\mathcal{U}(x;\epsilon)$? Is $x\in E$ fixed? Is $\epsilon>0$ fixed? Are neither fixed? If neither is fixed, then it is easy to find a union of sets of the form $\mathcal{U}(x;\epsilon)$ that will contain every point of $O\in \mathcal{O}(E)$ and you don't even need complements. If one is fixed, but not another, this becomes a little more difficult. If both are fixed, then you most definitely do not get $U(E)=B(E)$.

Edit: If $x\in E$ is fixed, but $\epsilon>0$ is arbitrary, then you do not get $U(E) = B(E)$ either. Alternately, if $x\in E$ is arbitrary, but $\epsilon>0$ is fixed, you might still be able to show $U(E) = B(E)$... I would have to think about that.

Edit 2: If $\epsilon>0$ is fixed, but $x\in E$ is arbitrary, then if for all $y \in E$, $\{y\} = \left( \left\{ x \in \mathcal{U}(z;\epsilon) \mid z\in E \mbox{ and }y \notin \mathcal{U}(z;\epsilon) \right\} \right)^c$, then you can show $U(E) = B(E)$.