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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 $B(S)=C$?

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    $\begingroup$ Have you tried to prove it? $\endgroup$ Oct 28 '12 at 16:06
  • $\begingroup$ I tried but could only prove that $B(S)\subset C$. $\endgroup$
    – Spook
    Oct 28 '12 at 16:13
  • $\begingroup$ This follows from this question and its answers with the generators being the open sets and $f:S\to T$ given by $f(x)=x$. Note that the assumption $S\in B(T)$ is not necessary. $\endgroup$ Oct 28 '12 at 18:13
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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 $\{ A \cap S : A \in B(T) \} = \{ A \in B(T) : A \subseteq S \}$.

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    $\begingroup$ The trick using the inclusion map is neat! :) $\endgroup$
    – rehband
    Nov 24 '14 at 13:22
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To prove that $C\subset B(S)$ We prove that $\{A|A\cap S \in B(S)\}$ is a $\sigma$-algebra and that it contains all opens subsets of $T$. Hence it contains $B(T)$ and the result follows.

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