Tychonoff vs. Hilbert Let $(\mathscr H_n,\langle\cdot,\cdot\rangle_n)_{n\in\mathbb N}$ be a sequence of Hilbert spaces. Let $$\mathscr H\equiv\bigoplus_{n\in\mathbb N}\mathscr H_n\equiv\left\{(h_n)_{n=1}^{\infty}\,\Bigg|\,h_n\in\mathscr H_n\,\forall n\in\mathbb N,\,\sum_{n\in\mathbb N}\|h_n\|_n^2<\infty\right\}$$ denote their direct sum, equipped with the inner product $$((h_n)_{n\in\mathbb N},(g_n)_{n\in\mathbb N})\mapsto\sum_{n\in\mathbb N}\langle h_n,g_n\rangle_n.$$ The norm $$(h_n)_{n\in\mathbb N}\mapsto\sqrt{\sum_{n\in\mathbb N}\|h_n\|_n^2}$$ naturally defines a metric topology on $\mathscr H$.
I have encountered the following dilemma. If $C_n\subseteq\mathscr H_n$ is a compact set for all $n\in\mathbb N$ such that $\prod_{n\in\mathbb N} C_n\subseteq \mathscr H$, then I conjecture that $\prod_{n\in\mathbb N} C_n$ may not be compact in $\mathscr H$, seemingly in defiance of Tychonoff's theorem, because the metric topology on $\mathscr H$ may be different from the product topology corresponding to the metric topologies on $(\mathscr H_n)_{n\in\mathbb N}$.
Is this conjecture correct? (I don't need a rigorous proof, I just wonder whether this is the case.)

If a product of compact sets lies within the direct sum, it will always be compact.
The space in question is metrisable, so it's enough to check sequential compactness, and thanks to completeness, it's not hard to do that using the standard diagonal argument.
This is not a consequence of Tychonoff's theorem, however, as Hilbert topology is much finer than the product topology.
A similar argument will not work if you replace $\prod C_n$ with $\bigoplus C_n$, for example if you take $C_n=[0,1]$ in $\mathscr H_n={\bf R}$ (so that $\mathscr H=\ell^2$), the direct sum will be very far from compact.
