Is $C^\infty(U)$ a Borel subset of $L^2(U)$ when $U$ is a compact subset of $\mathbb{R}^n$?

Asked Apr 23 '23 at 13:59

Active Apr 23 '23 at 13:59

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OK, I was totally wrong in the last question : Is a subspace of a separable Hilbert space always a Borel set?

Now, let me be more specific. Let $U$ be a compact subset of $\mathbb{R}^n$ and think of $L^2(U)$ with its inner product topology. In this setting, $C^\infty(U)$ is clearly a vector subspace of $L^2(U)$.

But, is $C^\infty(U)$ an element of the Borel $\sigma-$algebra generated by the inner product topology of $L^2(U)$?

This specific question seems to deserve a positive answer, I believe...but I still cannot find a rigorous justification. Could anyone help me?

asked Apr 23 '23 at 13:59

Keith

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How exactly do you define $C^\infty(U)$ for U compact? – PhoemueX Apr 23 '23 at 17:22
OK, I was not specific enough. As a set, it is the union of all $C^\infty(U')$ where $U' \subset U$ is open in $\mathbb{R}^n$. – Keith Apr 23 '23 at 17:44

Is $C^\infty(U)$ a Borel subset of $L^2(U)$ when $U$ is a compact subset of $\mathbb{R}^n$?

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