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There is the exercise:

Does there exist an infinite $\sigma$-algebra which has only countably many members?

Intuitively I think that the answer is no but I've really trouble to proof it.

enter image description here This is one proof But I have few questions about it. I don't understand this part in this proof: " If not, then every other measurable set of $X$ must intersect both $E$ and $E^c$ nontrivially."

I also don't understand why is $F_n$ a set of size $n$ measurable sets? for example $F_2$ has three members . by this logic $F_n$ should have $n+1$ members. why do we need to make $n$ sufficiently large? Any help would be appreciated.

Asaf Karagila
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JohnNash
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1 Answers1

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  • Answer to your main question: "every other measurable set of $X$ must intersect both $E$ and $E^c$ nontrivially" must be understood as: "every nonempty $E'\in\mathfrak M$ distinct from $E$ and $E^c$ must satisfy $E'\cap E\ne \varnothing$ and $E'\cap E^c\ne\varnothing$" i.e. "[...] must satisfy $A\not\subset E^c$ and $A\not\subset E$".
    Since $\varnothing\ne E\ne X$, this is the exact negation of "$E$ or $E^c$ contains a nonempty measurable $E'$ properly contained in it", i.e. of "there exists a nonempty $E'\in\mathfrak M$ such that $E'\subsetneq E$ or $E'\subsetneq E^c$".
  • I agree with you that $F_n$ is of size $n+1$ rather than $n$. It is just a blunder, I guess.
  • But the rest of this solution seems to me nonsense hence I wouldn't trust the file where you found it: the "set $F$ of measurables sets $E$ such that $E$ appears in all $F_n$ for sufficiently large $n$" is unquestionably "well defined" but may be empty and anyway, the remaining sentences have neither head nor tail to me.

Now, may If $S$ is an infinite $\sigma$ algebra on $X$ then $S$ is not countable help?

Anne Bauval
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