§ Direct sum of topological vector spaces

• In vector spaces, direct sum (also direct product) needs projection functors $\pi_1, \pi2_: V \to X, Y$such that $X \times Y = V$.
• In topological vector spaces, these projections also need to be continuous which is a massive thing to ask for.

§ Direct sum need not be closed.

• Let $X$ be a Hilbert space with schrauder basis (topological basis) $e[i]$
• Consider subspaces spanned by the basis $A[k] \equiv e[2k]$, and $B[k] \equiv e[2k] + e[2k+1]/(k+1)$. So $A \equiv span(A[k])$, $B \equiv span(B[k])$.
• Clearly, $A, B$ are subspaces, $A, B$ are closed.
• See that the closure of $A + B$ is the full space, since it contains the hamel basis $e[i]$.
• However, also see that the vector $z \equiv sum_k e[2k]/(k+1)$ is not in $A + B$. If we tried writing it as $a + b$, then we woulnd need $b \equiv \sum_k B[k]$. But this sum does not converge.
• This means that $(A + B)$ is not closed. If it were closed, it would contain the full space (because it's dense).
• Thus, we have an example of the direct sum of two closed subspaces which is not closed, because it is dense.