$\{ x\in X: f \text{ is continuous at } x\}$ is borel

Question

Let $X, Y$ be two metric spaces and $f:X\to Y$ a function. Prove that $A:= \{ x\in X: f$ is continuous at x$\}$ is a borel subset of X. I'm stuck here for good. I'm trying to use an equivalent definition of continuity that uses inverse images but I cant adapt the definition I know to point continuity. Any ideas or hints? I also tried writing A as $\displaystyle{\bigcup_{m, n \in \mathbb{N}} A_{n,m}}$, where $A_{n,m} = \{ x \in X: f(D(x, \frac{1}{m}))\subset D(f(x), \frac{1}{n})\}$, no luck though.

Answer 1

A big hint: define $O_n=\{ x \in X : \exists \delta > 0 \: \forall y,z \in B_\delta(x) \: d_Y(f(y),f(z))<1/n \}$. Show that $O_n$ is open and that $A=\bigcap_{n=1}^\infty O_n$.