Disprove [itex]\sigma[/itex] (singletons in R) = Borel field

[Lily@pie]

Homework Statement

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

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

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

Homework Equations

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

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

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!

 

[Dick]

Homework Statement

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

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

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

Homework Equations

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

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

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!

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?

 

[Lily@pie]

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?

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 :)

 

[Dick]

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 :)

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.

 

[Lily@pie]

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.

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?

 

[Dick]

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?

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?

 

[Lily@pie]

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?

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 F^C is countable, $F^{C}=\stackrel{\bigcup}{x\in F^{C}}${x}. F^C 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!

 

[Dick]

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 F^C is countable, $F^{C}=\stackrel{\bigcup}{x\in F^{C}}${x}. F^C 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!

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.