Let $a_n \to a$ and $b_n \to b$ in Hilbert spaces $A$ and $B$ respectively. Denote $S_1 = \{a, a_1, a_2,...\}$ and $S_2=\{b,b_1, b_2, ...\}.$
We know $S_1 \times S_2$ is compact in $A \times B$. Is it true that $X:=\{(a,b), (a_1,b_1), (a_2, b_2), ...\}$ is compact in $A \times B$?
I think so since if two sequences converge then so do their products with the same indices. Is there a better way of writing this?
The comments give a great means for showing that $X$ is compact. It can also be shown directly from the definition without much difficulty.
Since $a_n\to a$ and $b_n\to b$, any open neighborhood of $(a,b)$ in $A\times B$ contains all but finitely many members of $X$. Thus if $\mathcal U=\{U_\alpha\}$ is an open cover of $X$, there is some $\alpha$ such that $(a,b)\in U_\alpha$, and thus $(a_n,b_n)\in U_\alpha$ for all but finitely many $n$. The remainder of the $X$ is covered by finitely many members of $\mathcal U$, and therefore $X$ is compact.