10.2 Analytic Sets
Definition 10.2.1 (Analytic Set).label Let $X$ be a Polish space and $A \subset X$, then $A$ is analytic if there exists a Polish space $Z$ and $f \in C(Z; X)$ such that $f(Z) = A$.
Proposition 10.2.2.label Let $X$ be a Polish space, then:
- (1)
For each family $\seq{A_n}\subset X$ of analytic sets, $\bigcup_{n \in \natp}A_{n}$ is analytic.
- (2)
For each family $\seq{A_n}\subset X$ of analytic sets, $\bigcap_{n \in \natp}A_{n}$ is analytic.
- (3)
For any $A \in \cb_{X}$, $A$ is analytic.
Proof, [Proposition 8.2.1-8.2.3, Coh13]. For each $n \in \natp$, let $Z_{n}$ be a Polish space and $f_{n} \in C(Z_{n}; X)$ such that $A_{n} = f_{n}(Z_{n})$.
(1): By Proposition 10.1.2, $Z = \bigsqcup_{n \in \natp}Z_{n}$ is a Polish space. Let $f \in C(Z; X)$ be the gluing of $\seq{f_n}$, then $\bigcup_{n \in \natp}A_{n} = f(Z)$.
(2): By Proposition 10.1.2, $Z = \prod_{n \in \natp}Z_{n}$ is a Polish space. For each $m, n \in \natp$, $\bracs{f_m \circ \pi_m = f_n \circ \pi_n}$ is closed. Thus
is closed, and Polish by Proposition 10.1.2. Hence $\bigcap_{n \in \natp}A_{n} = f_{1} \circ \pi_{1}(\Delta)$ is also analytic.
(3): By Proposition 10.1.2, every open and closed subset of $X$ is analytic. Thus (1), (2), and Lemma 22.2.2 imply that every element of $\cb_{X}$ is analytic.$\square$
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