Disprove $\sigma$ (singletons in R) = Borel field

1. Mar 1, 2013

Lily@pie

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

${ E }_{ 1 }:=\{ \left( -\infty ,x \right) :x\in \Re \}$

${ E }_{ 2 }:=\{ \left\{ x \right\} :x\in \Re \}$

Prove $\sigma \left( { E }_{ 1 } \right) =\sigma \left( { E }_{ 2 } \right)$

2. Relevant equations

I know ${ E }_{ 1 }$ generates the Borel field

(i.e.) $\sigma \left( { E }_{ 1 } \right)=B(ℝ)$

3. The attempt at a solution

I know this is wrong as $\sigma \left( { E }_{ 2 } \right) \varsubsetneqq \sigma \left( { E }_{ 1 } \right)$

Hence my attempt is assume $\sigma \left( { E }_{ 1 } \right) \subseteq \sigma \left( { E }_{ 2 } \right)$

Let $F \in σ(E_{1})$ be a non-empty set

As $F \in E_{1} \Rightarrow F \in σ(E_{1})$

When $F \in E_{1}$, $F = (-\infty, x)$ for some $x \in ℝ$

I want to prove that for some $x \in ℝ$ $F = (-\infty, x) \notin E_{2}$, which means $F = (-\infty, x) \notin σ(E_{2})$.
It is kind of obvious to me as $-∞ \notin ℝ$, and $x \in ℝ$ implies that $(-\infty, x)$ cannot be generated by a countable union of singletons.

Hence $\sigma \left( { E }_{ 1 } \right) \subseteq \sigma \left( { E }_{ 2 } \right)$ must be wrong.

But I'm not sure if this statement is strong enough.

Thanks!

2. Mar 1, 2013

Dick

No, that's not good enough, I hope you know why. Let G be the set consisting of all sets X such that either X is countable or $X^C$ is countable. Can you show that G is a sigma algebra? Now can you show that's exactly the sigma algebra generated by the singletons?

3. Mar 1, 2013

Lily@pie

I could prove that G is a sigma algebra.

However, I'm not quite sure in proving G=σ(singletons)

Can I say $G \subseteq$ {{x}:x\in ℝ}?
Then $G=σ(G) \subseteq$ σ({{x}:x\in ℝ}).

I could also show {{x}:x\in ℝ} $\subseteq σ(G)$

Hence, G=σ({{x}:x\in ℝ}.

Therefore to show $σ( E_{1})$ not subset of G,

Let $E\in E_{1}$ be (-∞,x] for some real number x.

since (-∞,x] and (x,∞) are not countable, $E \notin$ G =σ({{x}:x\in ℝ}) .

Am I heading to the correct direction?

Thanks a lot :)

4. Mar 1, 2013

Dick

Given what follows, I'd be interested in hearing the general idea of how you proved G is a sigma algebra. Because the middle part sounds strange. You clearly can't say that G is a subset of the singletons, it's not. There are certainly elements of G that aren't singletons. Just pick an element of G and show that it is in the sigma algebra generated by the singletons. But the end part sounds good.

Last edited: Mar 1, 2013
5. Mar 2, 2013

Lily@pie

To prove that G is a sigma algebra, I just showed that G satisfy the 3 axioms of sigma algebra.

I admit the middle part is strange as I am not sure.

I've tried the following to prove G = σ({{x}:x$\in$ ℝ})

1st, I will prove that $G \subset$ σ({{x}:x$\in$ ℝ}).

Since an element in G is either countable to the complement is countable, it can be written as a countable union of singletons. Hence the element will be in the sigma field.

However, I really don't know how to show that σ({{x}:x$\in$ ℝ}) $\subset G$. Can I say that an element in sigma(singleton) can be singleton or the complement of singleton?

6. Mar 2, 2013

Dick

Maybe you just aren't making complete statements. It's surely not true that if a set has a countable complement then it's a "countable union of singletons", is it? For the second part, let's call S=σ({{x}:x$\in$ ℝ}). You know that every element of G is in S. And G is a sigma algebra. And the singletons are in G. S is defined to be the SMALLEST sigma algebra containing the singletons. So?

Last edited: Mar 2, 2013
7. Mar 2, 2013

Lily@pie

I get the second part now.

This is what I was thinking for the first part,

Let A={{x}:x$\in$ ℝ}
For the first part, I am trying to prove that $G \subseteq σ(A)$

If $F \in G$, F is countable or the complement is countable,

If F is countable, we can write F as $\stackrel{\bigcup}{x\in F}${x}. F is now a countable union of elements in σ(A), so $F \in σ(A)$.

If FC is countable, $F^{C}=\stackrel{\bigcup}{x\in F^{C}}${x}. FC is now a countable union of elements in σ(A), so $F^{C} \in σ(A)$ and this implies $(F^{C})^{C}=F \in σ(A)$.

(i.e) $G \subseteq σ(A)$

Am I heading the right direction? Thanks heaps!

8. Mar 2, 2013

Dick

Yes, you are. Now if you are confident you have proved that G is a sigma algebra using the axioms, you should be done. Like I said, if you want to explain how that worked, I could look at that part too.