Let $(X,\|\cdot\|_{X})$ be a separable Hilbert space and let $(Y,\|\cdot\|_{Y})$ be a Banach space endowed with the Borel $\sigma$-field. Let $f:X\to Y$. Is it true that:
- if $f$ is continuous, then $f(A)$ is Borel for all open sets $A$;
- if $f$ is Lipschitz, then $f(A)$ is Borel for all open sets $A$.
Clearly (1) $\implies$ (2), so these are two statements of different strengths. If $\dim(X)<+\infty$, then (1) is trivial. In fact, any open set can be written as the countable union of compact sets, and compactness is preserved under continuity.
But what about the case $\dim(X)=+\infty$?
My attempt was: let $\{e_{i}\}_{i=1}^{+\infty}$ be an orthonormal basis for $X$. Since $X$ is separable, it suffices to prove that (1) or (2) hold whenever $A$ is an open ball (and, ultimately, one can focus directly on the unit ball). Furthermore, the statement holds for all open sets iff it does for closed sets. So let $A$ be the closed unit ball. Then, $A=\overline{\cup_{n=1}^{+\infty}K_{n}}$, where $$K_{n}=\left\{\sum_{i=1}^{n}\alpha_{i}e_{i}\;\;\text{with}\;\;\sum_{i=1}^{n}|\alpha_{i}|^{2}\le1\right\}$$ are compact. However, here I get stuck (and I'm not even sure this is the correct way to go...).