- #1
Lily@pie
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Homework Statement
[itex]{ E }_{ 1 }:=\{ \left( -\infty ,x \right) :x\in \Re \} [/itex]
[itex]{ E }_{ 2 }:=\{ \left\{ x \right\} :x\in \Re \} [/itex]
Prove [itex]\sigma \left( { E }_{ 1 } \right) =\sigma \left( { E }_{ 2 } \right) [/itex]
Homework Equations
I know [itex]{ E }_{ 1 }[/itex] generates the Borel field
(i.e.) [itex]\sigma \left( { E }_{ 1 } \right)=B(ℝ)[/itex]
The Attempt at a Solution
I know this is wrong as [itex]\sigma \left( { E }_{ 2 } \right) \varsubsetneqq \sigma \left( { E }_{ 1 } \right) [/itex]
Hence my attempt is assume [itex]\sigma \left( { E }_{ 1 } \right) \subseteq \sigma \left( { E }_{ 2 } \right) [/itex]
Let [itex] F \in σ(E_{1})[/itex] be a non-empty set
As [itex]F \in E_{1} \Rightarrow F \in σ(E_{1}) [/itex]
When [itex]F \in E_{1}[/itex], [itex]F = (-\infty, x) [/itex] for some [itex]x \in ℝ[/itex]
I want to prove that for some [itex]x \in ℝ[/itex] [itex]F = (-\infty, x) \notin E_{2} [/itex], which means [itex]F = (-\infty, x) \notin σ(E_{2}) [/itex].
It is kind of obvious to me as [itex]-∞ \notin ℝ[/itex], and [itex]x \in ℝ[/itex] implies that [itex](-\infty, x)[/itex] cannot be generated by a countable union of singletons.
Hence [itex]\sigma \left( { E }_{ 1 } \right) \subseteq \sigma \left( { E }_{ 2 } \right) [/itex] must be wrong.
But I'm not sure if this statement is strong enough.
Thanks!