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The maximal sigma algebra on a set is its power set.

When the set is countable, its maximal sigma algebra can be generated by all singleton subsets, i.e. subsets each consisting of exactly one element.

Conversely, if the maximal sigma algebra on a set can be generated by all singleton subsets, must the set be countable?

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up vote 12 down vote accepted

Assume $X$ is uncountable. The countable-cocountable $\sigma$-algebra, that is to say the $\sigma$-algebra consisting of the sets which are countable or have countable complement, contains the singletons, so the $\sigma$-algebra generated by the singletons can't be the entire power set (note that $X$ contains two disjoint uncountable sets).

It is an easy exercise to prove that the singletons generate the countable-cocountable $\sigma$-algebra on any set. If the set happens to be countable then the countable-cocountable $\sigma$-algebra coincides with the power set.


Edit: For people interested in choice-specific issues, our resident connoisseur Asaf recommends perusal of the following links:

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It might be worth noting that I'm using the axiom of choice here to assert that an uncountable set $X$ contains an uncountable set with uncountable complement (so that the countable-cocountable $\sigma$-algebra is actually distinct from the power set). –  t.b. Jan 9 '12 at 5:23
    
Thanks! How can the axiom of choice be used to prove that an uncountable set X contains an uncountable set with uncountable complement? –  Tim Jan 9 '12 at 6:49
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Let $\omega_1$ be the first uncountable ordinal. By the axiom of choice there's an injection $g: \omega_1 \to X$. There's also a bijection $f: \omega_1 \times \omega_1 \to \omega_1$. Then $h = gf: \omega_1 \times \omega_1 \to X$ is injective and thus gives me an uncountable family of disjoint uncountable sets in $X$. For example $h(\omega_1 \times \{0\})$ and $h(\omega_1 \times \{1\})$ are both uncountable and disjoint subsets of $X$. –  t.b. Jan 9 '12 at 7:03
    
You need that countable unions of countable sets are countable- which is rather harmless and follows from the axiom of choice for countable sets. –  Michael Greinecker Jan 9 '12 at 7:06
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Please no more choice comments to this answer this is veering too far off-topic! –  t.b. Jan 9 '12 at 7:12
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