Assume $\,\mathcal{F}$ be a nonempty collection of subsets of $\Omega$. $\,\mathcal{F}$ is called a $\sigma$-Algebra whenever

  1. if $A\in\mathcal{F}$ then $A^c\in\mathcal{F}$, and
  2. if $A_1,A_2,...\in\mathcal{F}$ then $\bigcup_{n=1}^{\infty}A_n\in\mathcal{F}$

Is there a countably infinite Sigma-Algebra? Why?I want to prove it by definition (without real analysis).

  • 1
    $\begingroup$ As not all of the quants here are experts in measure theory I would add from a blog: "there are no σ-algebras that are infinitely countable. This means that any σ-algebra S is either finite (and is therefore just an algebra) or very 'BIG' in cardinality, in the sense that it is uncountable." taken from here: yaronhadad.com/… $\endgroup$ – Ric Jul 6 '15 at 7:59
  • $\begingroup$ @Richard Whatever you want ! $\endgroup$ – user16651 Jul 6 '15 at 8:12
  • 2
    $\begingroup$ See here: math.stackexchange.com/questions/320035/… $\endgroup$ – Ulysses Jul 6 '15 at 12:21

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