How union and intersection implies complement?

Question

Defining the family of Borel sets to be the smallest σ–algebra that contains all the closed sets in a metric space X, I want to show that it is equivalent to the following definition :

The only part remained to show is that if conditions 1. and 2. of the Theorem 3.3 holds then for any set $E\in B$, also $X\backslash E\in B$. I know being closed under union and complement implies being closed under intersection, but how to prove being closed under union and intersection implies being closed under complement?

Other way to finish my proof is to show that in a metric space every open set is an $F_σ$ ; this is easy for $\mathbb{R}$ but how to prove that for a general metric space?

Answer 1

You can prove instead that each closed set is a $G_\delta$. Let $\langle X,d\rangle$ be a metric space, and let $F\subseteq X$ be closed. For each $n\in\Bbb N$ let $U_n=\bigcup_{x\in F}B(x,2^{-n})$; clearly each $U_n$ is open, and $F\subseteq\bigcap_{n\in\Bbb N}U_n$. Suppose that $x\in\bigcap_{n\in\Bbb N}U_n$; then for each $n\in\Bbb N$ there is an $x_n\in F$ such that $d(x,x_n)<2^{-n}$. Clearly the sequence $\langle x_n:n\in\Bbb N\rangle$ converges to $x$, and $F$ is closed, so $x\in F$. Thus, $F=\bigcap_{n\in\Bbb N}U_n$, and $F$ is therefore a $G_\delta$.

Since every closed set in $X$ is a $G_\delta$, every open set must be an $F_\sigma$.

Answer 2

This is actually a rather clever little proof.

Consider the set $C = \{S \in B : X \setminus S \in B\}$. I claim that this set satisfies the properties

1. If $E_1, E_2, E_3, ... $ belong to $C$ then $\bigcup\limits_{i = 1}^\infty E_i \in C$
2. If $E_1, E_2, E_3, ... $ belong to $C$ then $\bigcap\limits_{i = 1}^\infty E_i \in C$
3. Every closed set is in $C$.

Let's begin with 1. Suppose $E_1, E_2, E_3, ...$ belong to $C$. Then $E_1, E_2, E_3, ...$ belong to $B$. Then $\bigcup\limits_{i = 1}^\infty E_i \in B$. Now consider $X \setminus \bigcup\limits_{i = 1}^\infty E_i = \bigcap\limits_{i = 1}^\infty (X \setminus E_i)$. Since each $X \setminus E_i$ is in $B$, so too is $\bigcap\limits_{i = 1}^\infty (X \setminus E_i)$.

Showing 2 is almost identical to showing 1, so we will skip it for brevity.

Now we must show that every closed set is in $C$. It suffices to show that every open set is in $B$. Consider an open set $U$, and consider the sets $K_n$, indexed by $n \in \mathbb{N}$, defined by $K_n = \{x | \forall y \in X \setminus U, d(x, y) \geq 1/n\} = \bigcap\limits_{y \in X \setminus U} \{x | d(x, y) \geq 1/n\}$. Then clearly, $K_n$ is a closed set for all $n$. Then for all $n$, we have $K_n \in B$. Then in particular, we have $\bigcup\limits_{n = 1}^\infty K_n \in B$.

I claim that $U = \bigcup\limits_{n = 1}^\infty K_n \in B$. For suppose that $x \in U$. Then take $\delta$ such that $B_\delta(x) \subseteq U$. Take $n$ such that $1/n < \delta$. Then $x \in K_n$.

And suppose that $x \in K_n$. Suppose that $x$ is not in $U$. Then we have $d(x, x) = 0 < 1/n$; contradiction. So $x \in U$.

Since $C$ satisfies 1, 2, and 3, we see that $B \subseteq C$. Then $B$ is closed under complementation.