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Second countability by itself doesn't restrict the cardinality of a topological space, since every set with the trivial topology is a second countable space, but it seems natural to ask whether second countability restricts the cardinality of the Borel $\sigma$-algebra of the space.

Can the cardinality of the Borel $\sigma$-algebra of a second countable space be arbitrarily big?
If this is the case is there a simple construction for a second countable space with an arbitrarily big Borel $\sigma$-algebra?

Martin Sleziak
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Alessandro Codenotti
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2 Answers2

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More generally, suppose $A$ is a collection of subsets of a set $X$ and let $B$ be the $\sigma$-algebra generated by $A$. We can construct $B$ inductively as follows. We define an increasing sequence $B_\alpha$ of collections of subsets of $X$, for each $\alpha<\omega_1$. Let $B_0=A$, and given $B_\alpha$, let $B_{\alpha+1}$ be the set of complements, countable unions, and countable intersections of elements of $B_\alpha$. If $\alpha$ is a limit ordinal, define $B_\alpha=\bigcup_{\beta<\alpha} B_\beta$. Then I claim that $$B=\bigcup_{\alpha<\omega_1} B_\alpha.$$ Indeed, clearly by induction $B_\alpha\subseteq B$ for all $\alpha$. On the other hand, $\bigcup_{\alpha<\omega_1}B_\alpha$ is a $\sigma$-algebra, since given countably many elements of it they are all contained in some $B_\alpha$, and then their union and intersection (and the complement of any one of them) is contained in $B_{\alpha+1}$. Since $B$ is by definition the smallest $\sigma$-algebra containing $A$, $B\subseteq\bigcup_{\alpha<\omega_1}B_\alpha$.

We can now use this to bound the cardinality of $B$. Notice that $|B_1|\leq |A|^{\aleph_0}$ (assuming $|A|>1$). Since $$(|A|^{\aleph_0})^{\aleph_0}=|A|^{\aleph_0^2}=|A|^{\aleph_0^2}=|A|^{\aleph_0},$$ it follows by induction that $|B_\alpha|\leq |A|^{\aleph_0}$ for all $\alpha$. Thus $$|B|\leq \aleph_1\cdot |A|^{\aleph_0}=|A|^{\aleph_0}.$$

Finally, let us apply this to your question. If $X$ is a second-countable space, we can take $A$ to be a countable basis. Every open set is a countable union of elements of $A$ and is thus in $B$, so $B$ will be the Borel algebra of $X$. We thus conclude that the Borel algebra has cardinality at most $$|A|^{\aleph_0}=2^{\aleph_0}.$$ (Of course, this upper bound is easy to achieve, for instance for $X=\mathbb{R}$.)

Eric Wofsey
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Second countable space $X$ has a countable base

$$\mathcal{B} = \{U_i\}_{i=1}^{\infty}$$

Since every open set $U$ is of the form

$$U = \bigcup\mathcal{B}^{'}\mbox{ where } \mathcal{B}^{'}\subseteq\mathcal{B}$$

then it is clear that $X$ has at most $2^{\aleph_0}$ open sets and thus at most $2^{\aleph_0}$ closed sets.

Let's consider three cases:

1) $X$ is $T_1$. Then every point is a closed set. In particular there are at most $2^{\aleph_0}$ points and since Borel $\sigma$-algebra is a subset of a power set, then the size of Borel $\sigma$-algebra is at most $2^{2^{\aleph_0}}$.

2) $X/\sim$ is $T_1$ where

$$x\sim y\ \mbox{ if and only if }\ \bar{x}=\bar{y}$$

where on the right side we have closures. Consider a map

$$F:\mathcal{P}(Y)\to\mathcal{P}(X)$$ $$F(A) = \bigcup_{a\in A}a$$

Note that $A$ is a set of equivalence classes which are sets themselves and thus the definition makes sense. This function is "1-1" between closed sets and it obviously preserves countable unions and intersections. Thus it is "1-1" between Borel $\sigma$-algebras so it follows from 1) $X$ has at most $2^{2^{\aleph_0}}$ Borel sets.

3) $X$ does not satisfy 2). TODO.

freakish
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