Show that if $X,Y$ are topological spaces and $f:X\to Y$ is a continous function , then $f$ is a Borel measurable.
Any help what am I supposed to prove here ?
My attempt:
$\mathcal{B}(X) $ is the open subsets of $X$ , $\mathcal{B}(Y)$ is the open subsets of $Y$.
I want to prove that $f^{-1}(U) \in\mathcal{B}(X)$ where $U\in \mathcal{B}(Y) $
$f$ is a continous function implies that $f^{-1}(U)$ is open $\forall $ open $U\subset Y$.
Hence $f^{-1}(U) \in\mathcal{B}(X)$ $\forall U\in \mathcal{B}(Y) $ .
$f$ is borel-measure.
Is my proof correct ?
I think you need to use the definitions of your concepts more clearly. Outline:
Let $\mathcal{O}(X)$ be the open sets of $X$, let $\mathcal{O}(Y)$ be the open sets of $Y$. Then let $\mathcal{B}(X)$ be the $\sigma$-algebra on $X$ generated by $\mathcal{O}(X)$, let $\mathcal{B}(Y)$ be the $\sigma$-algebra on $Y$ generated by $\mathcal{O}(Y)$.
We assume (a) $f : X \to Y$ is continuous, that is: for every $U \in \mathcal{O}(Y)$ we have $f^{-1}(U) \in \mathcal{O}(X)$.
We want to show (b) for every $E \in \mathcal{B}(Y)$ we have $f^{-1}(E) \in \mathcal{B}(X)$.
Do this in steps. (c) for every $U \in \mathcal{O}(Y)$ we have $f^{-1}(U) \in \mathcal{B}(X)$. [From (a) and definition of $\mathcal{B}(X)$]
Let $$ \mathcal{F} = \big\{E \subseteq Y : f^{-1}(E) \in \mathcal{B}(X)\big\} $$ Show (d.1) $\mathcal{F} \supseteq \mathcal{O}(Y)$ [From (c)]
(d.2) $\mathcal{F}$ is a $\sigma$-algebra [Use properties of $f^{-1}$; this is the heart of the proof.]
Deduce (d) $\mathcal{F} \supseteq \mathcal{B}(Y)$. [From (d.1) and (d.2) and definition of $\mathcal{B}(Y)$.]
Conclude (b).