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For any topological space $(X,\tau)$, the Borel $\sigma$-algebra $\mathcal{B}$ is the $\sigma$-algebra generated by the open sets. In other words, it is the intersection of all $\sigma$-algebras on $X$ which contain $\tau$.

For a metric space $(X,d)$, I want to prove the following alternative characterization.

The family of Borel sets of a metric space $(X,d)$ is the smallest class $\mathcal{S}$ of subsets of $X$ with the properties

  1. If $E_1,E_2,E_3,\ldots$ belong to $\mathcal{S}$, then so too does $\bigcup_{n=1}^{\infty}E_n$
  2. If $E_1,E_2,E_3,\ldots$ belong to $\mathcal{S}$, then so too does $\bigcap_{n=1}^{\infty}E_n$
  3. $\mathcal{S}$ contains all the closed sets in $X$.

(Bruckner, Bruckner, and Thomson, Real Analysis, Theorem 3.3; proof is an exercise)

My questions:

  • I have completed most of the proof, but I have not succeeded at proving that $\mathcal{S}$ is closed under taking complements. A hint would be great.
  • My proof relies heavily on $X$ being a metric space. Can the result be generalized to, say, a Hausdorff space, or is there a counterexample?

Proof attempt:

$\mathcal{B}$ is a class of subsets of $X$ satisfying the three indicated properties. Since $\mathcal{S}$ is the smallest class of subsets of $X$ with these properties, we have the containment $\mathcal{S} \subseteq \mathcal{B}$.

To show that $\mathcal{B} \subseteq \mathcal{S}$, it suffices to show that $\mathcal{S}$ is a $\sigma$-algebra containing the open subsets of $X$, because $\mathcal{B}$ is the smallest such $\sigma$-algebra.

Let $U \subseteq X$ be any open set. Then $U^c$ is closed.

Consider the function $\rho : X \to \mathbb{R}$ defined by $\rho(x) = d(x,U^c)$. Then $\rho$ is continuous. Indeed, if $x,y \in X$ and $z \in U^c$, then $d(x,z) \leq d(x,y) + d(y,z)$; taking the infimum over $z \in U^c$, we have $\rho(x) \leq d(x,y) + \rho(y)$. Similarly, $\rho(y) \leq d(x,y) + \rho(x)$. Combining these inequalities, we have $$|\rho(x) - \rho(y)| \leq d(x,y)$$ so in fact, the continuity is uniform.

Note also that $\rho(x) = 0$ if and only if $x \in U^c$, because $U^c$ is closed. Therefore, $$\begin{aligned} U &= \{x \in X : \rho(x) > 0\} \\ &= \bigcup_{n=1}^{\infty} \{x \in X : \rho(x) \geq 1/n\} \\ &= \bigcup_{n=1}^{\infty} \rho^{-1}([1/n, \infty)) \\ \end{aligned}$$ This exhibits $U$ as a countable union of closed sets, so $U \in \mathcal{S}$.

To complete the proof, we need to verify that $\mathcal{S}$ is a $\sigma$-algebra. Since $\mathcal{S}$ contains $X$ and is closed under countable unions, we just need to show that it is closed under taking complements. Note that so far, we have only used properties 1 and 3 of $\mathcal{S}$, so presumably we will need property 2 here.

Let $A \in \mathcal{S}$. I want to show that $A^c \in \mathcal{S}$. Unfortunately, I can't use the same method as above because $\rho(x) = 0$ is true for any limit point of $A$, whether or not it is in $A$.

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  • $\begingroup$ HINT Any open subset of a metric space is an $F_\sigma$ set. $\endgroup$ – DanielWainfleet Sep 3 '15 at 0:08
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    $\begingroup$ For a non-Hausdorff counterexample, consider the cofinite topology on an uncountable set. $\endgroup$ – Nate Eldredge Sep 3 '15 at 0:11
  • $\begingroup$ Generally for non-metric spaces (and specifically for Hausdorff locally compact spaces), I have found that Borel sets are the wrong object to look at. The Baire sets are more appropriate. en.wikipedia.org/wiki/Baire_set $\endgroup$ – Stephen Montgomery-Smith Sep 3 '15 at 13:13
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For the answer to the first question: let $\mathcal S' = \{A \in \mathcal S : A^C \in \mathcal S\}$. See that $\mathcal S'$ contains all open sets, that $\mathcal S'$ is closed under taking complements, and that $\mathcal S'$ is closed under countable unions.

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  • $\begingroup$ Very nice! Therefore, $\mathcal{S}'$ is a $\sigma$-algebra containing the open sets, so we have the containments $\mathcal{B} \subseteq \mathcal{S}' \subseteq \mathcal{S}$. I always love these kinds of arguments when I encounter them; I need to get better at generating them myself. $\endgroup$ – Bungo Sep 2 '15 at 23:59
  • $\begingroup$ Wow, that feels like cheating. $\endgroup$ – Nate Eldredge Sep 3 '15 at 0:11
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For a non-metrizable but locally compact Hausdorff counterexample, consider $\omega_1$ with its order topology. I will show that $\mathcal{S}$ does not even contain all the open sets.

Let $\mathcal{B}$ be the collection of all sets $B \subset \omega_1$ such that either $B$ is countable or $B$ contains a club set. (We consider "countable" to include "finite".)

I claim $\mathcal{B}$ contains all the closed sets and is closed under countable union and intersection. It will follow that $\mathcal{S} \subset \mathcal{B}$.

Suppose $F \subset \omega_1$ is closed. If $F$ is countable, then $F \in \mathcal{B}$ by definition. If $F$ is uncountable, then it is unbounded, so it is a club, and thus $F \in \mathcal{B}$ again.

Now suppose $A_1, A_2, \dots \in \mathcal{B}$.

If all $A_n$ are countable, then $A = \bigcup_n A_n$ is also countable. Otherwise, some $A_n$ contains a club, so $A$ also contains a club. In either case $A \in \mathcal{B}$.

If some $A_n$ is countable, then $A' = \bigcap_n A_n$ is also countable. Otherwise, each $A_n$ contains a club $F_n$. A countable intersection of clubs is a club, so $F = \bigcap_n F_n$ is a club and contained in $A'$. In either case $A' \in \mathcal{B}$.

This shows $\mathcal{S} \subset \mathcal{B}$. Now let $U$ be the set of all successor ordinals. I claim $U$ is open and $U \notin \mathcal{B}$. To see $U$ is open, simply note that each successor ordinal is an isolated point.

Now $U$ is unbounded, hence uncountable. And every club contains a limit ordinal. (If $C$ is a club, since it is unbounded we can find an increasing sequence $\alpha_1 < \alpha_2 < \dots$ in $C$. But then $\alpha = \sup_n \alpha_n$ is in $C$, because $C$ is closed, and $\alpha$ must be a limit ordinal.) So $U$ does not contain a club. Hence $U \notin \mathcal{B}$, and we have constructed an open set which is not in $\mathcal{S}$.

To turn this into a compact Hausdorff example, consider $\omega_1 + 1$ and let $\mathcal{B}' = \{B \subset \omega_1 + 1 : B \cap \omega_1 \in \mathcal{B}\}$.

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  • $\begingroup$ Thanks! I have some work to do in order to understand this counterexample. Or, apparently, to understand any counterexample: after asking this question, it occurred to me that I couldn't name any example of a non-metrizable Hausdorff space, with or without the property under discussion here. After a bit of searching, I still haven't found any that I am in a position to fully understand :-) $\endgroup$ – Bungo Sep 3 '15 at 1:20
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    $\begingroup$ @Bungo: $\omega_1$ is a good one to start with. The other one that comes up a lot is uncountable product space. $\endgroup$ – Nate Eldredge Sep 3 '15 at 1:34
  • $\begingroup$ Thanks again for this example. I did some background reading over the past week and now I understand it almost completely. (I still need to digest the "countable intersection of clubs is a club" proof, though.) It occurs to me that if we did not already know that $\omega_1$ with the order topology was non-metrizable, we do now, because it's a counterexample to the theorem in the problem statement. $\endgroup$ – Bungo Sep 15 '15 at 17:27

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