Let $X$ be an uncountable Tychonoff space. Must there exist a non-Borel subset of $X$?
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By Borel set, do you mean a set which is in the smallest $\sigma$-algebra containing all the open sets? – Asaf Karagila Aug 18 '14 at 10:08
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That is correct – pre-kidney Aug 18 '14 at 10:10
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Maybe it is the trivial counterexample, but take any uncountable discrete metric space. Every set is open, so every set is Borel.
Asaf Karagila
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This is an addition to Asaf Karagila's correct answer.
The standard argument which works for $\mathbb{R}$ is like this: Let $\mathcal{O}(X)$ denote set of all open subsets of topological space $X$, $\mathcal{B}(X)$ set of all Borel subsets of $X$. Also let $σ(\mathcal{A}, \mathcal{S})$ denote the $σ$-algebra generated by the system $\mathcal{A}$ in $σ$-algebra $\mathcal{S}$. So we have $\mathcal{B}(X) = σ(\mathcal{O}(X), \mathcal{P}(X))$.
Since generating a $σ$-algebra means adding countable unions and intersections, it holds that $|σ(\mathcal{A}, \mathcal{S})| ≤ $ |A| + 2^ω $|\mathcal{A}|^ω$. So if $X$ is a topological space such that $|\mathcal{O}(X)|^ω < |\mathcal{P}(X)|$, we have that $|\mathcal{B}(X)| ≤ |\mathcal{O}(X)|^ω < |\mathcal{P}(X)|$. So there is a non-Borel subset of $X$.
Also note that $|\mathcal{O}(X)| ≤ 2^{nw(X)}$, whete $nw(X)$ is network weight of a topological space $X$. So if $X$ satisfies $2^{nw(X)} ≤ |X|$, then we have $|\mathcal{O}(X)|^ω ≤ (2^{nw(X)})^ω = 2^{nw(X)} ≤ |X| < |\mathcal{P}(X)|$. So second countable space of size at least continuum (e.g. $\mathbb{R}$) has a non-Borel subset, as Tomek Kania says in his answer.
user87690
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2Your cardinal arithmetic doesn't take into consideration the case where the system $\cal A$ has a cardinal of countable cofinality (or cases where Singular Cardinal Hypothesis fails). – Asaf Karagila Aug 18 '14 at 12:35
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Sometimes trivial counter-examples are very good and Asaf actually provided one. In general, as Asaf, pointed out the answer is no. However, it is a standard yet challenging exercise on transfinite induction to show that if your space $X$ is additionally second countable then the cardinality of the family of Borel sets is continuum. So the answer is yes for spaces which are second countable and have cardinality continuum.
Note that there are even compact Hausdorff spaces for which every subset is Borel – just take your favourite countable successor ordinal with the order topology. More generally, take any countable Tychonoff space.
Tomasz Kania
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