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

  1. Mar 1, 2013 #1
    1. The problem statement, all variables and given/known data

    [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]

    2. Relevant equations

    I know [itex]{ E }_{ 1 }[/itex] generates the Borel field

    (i.e.) [itex]\sigma \left( { E }_{ 1 } \right)=B(ℝ)[/itex]

    3. 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. :confused:

  2. jcsd
  3. Mar 1, 2013 #2


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    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?
  4. Mar 1, 2013 #3
    I could prove that G is a sigma algebra.

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

    Can I say [itex]G \subseteq [/itex] {{x}:x\in ℝ}?
    Then [itex] G=σ(G) \subseteq[/itex] σ({{x}:x\in ℝ}).

    I could also show {{x}:x\in ℝ} [itex] \subseteq σ(G)[/itex]

    Hence, G=σ({{x}:x\in ℝ}.

    Therefore to show [itex]σ( E_{1})[/itex] not subset of G,

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

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

    Am I heading to the correct direction?

    Thanks a lot :)
  5. Mar 1, 2013 #4


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    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
  6. Mar 2, 2013 #5
    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[itex]\in[/itex] ℝ})

    1st, I will prove that [itex]G \subset[/itex] σ({{x}:x[itex]\in[/itex] 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[itex]\in[/itex] ℝ}) [itex]\subset G[/itex]. Can I say that an element in sigma(singleton) can be singleton or the complement of singleton?
  7. Mar 2, 2013 #6


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    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[itex]\in[/itex] 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
  8. Mar 2, 2013 #7
    I get the second part now.

    This is what I was thinking for the first part,

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

    If [itex]F \in G[/itex], F is countable or the complement is countable,

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

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

    (i.e) [itex]G \subseteq σ(A)[/itex]

    Am I heading the right direction? Thanks heaps!
  9. Mar 2, 2013 #8


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    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.
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