On Complete Separable metrizable spaces, do the compact sets generate the Borel $\sigma$-algebra?

Question

On Complete Separable metrizable spaces, do the compact sets generate the Borel $\sigma$-algebra? I don't think this is true. Can anyone help me with a counterexample?

Answer 1

Any separable infinite-dimensional Banach space is a counterexample, for example $L_1$. The compact sets are meagre so they generate a sub-$\sigma$-algebra of the $\sigma$-algebra of sets that are either meagre or comeagre. But the Borel $\sigma$-algebra contains the unit ball, which is neither meagre nor comeagre.

Answer 2

Any separable space $X$ which is $not \ \sigma-$compact is a counterexample. To see this, let $\mathscr S$ be the $\sigma-$ring of all $\sigma-$bounded Borel sets and then define $\mathscr S^c:=\left \{ E^{c}\in \mathscr P(X): E\in \mathscr S\right \}.$ Then $\mathscr S\cup \mathscr S^c$ is the $\sigma-$algebra generated by $\mathscr S $ which contains the Borel $\sigma-$algebra generated by the compact sets, but is not all of $\mathscr B(X).$