TITLE: Borel $\sigma$ algebra on a topological subspace.
QUESTION [11 upvotes]: Let $T$ be a topological space, with Borel $\sigma$-algebra $B(T)$ (generated by the open sets of $T$). If $S\in B(T)$, then the set $C:=\{A\subset S:A\in B(T)\}$ is a $\sigma$-algebra of $S$. 
My question is, if I also generated the Borel $\sigma$-algebra $B(S)$ treating $S$ as a topological subspace, with the inherited topology from $T$, is it true that $B(S)=C$?

REPLY [14 votes]: Note that if $Y$ is any subspace of $T$, then $B(Y) = \{ A \cap Y : A \in B(T) \}$.

As $\{ A \cap Y : A \in B(T) \}$ clearly contains all open subsets of $Y$, and is itself a $\sigma$-algebra on $Y$, then $B(Y) \subseteq \{ A \cap Y : A \in B(T) \}$.
As the inclusion map $i : Y \to T$ is continuous, then $i^{-1} [ A ]$ is a Borel subset of $Y$ for each Borel $A \subseteq T$, but $i^{-1} [ A ] = A \cap Y$, and so $\{ A \cap Y : A \in B(T) \} \subseteq B(Y)$.

If $S \subseteq T$ is Borel, then $A \cap S$ is a Borel subset of $T$ for all Borel $A \subseteq T$, and therefore $\{ A \cap S : A \in B(T) \} = \{ A \in B(T) : A \subseteq S \}$.