I thought that this would be a simpe question, and placed it [here][1] at the Mathematics Stackexchange. Now have to elevate it to Mathoverflow.
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LANGUAGE

TVS = topological vector space. Any subspace of a TVS is a TVS in the induced-topology sense.       

DEFINITION

For TVS spaces ${\mathbb{V}}_1\subset\mathbb V$, a TVS subspace ${\mathbb{V}}_2\subset\mathbb V$ is a *topological complement* of ${\mathbb{V}}_1$ in $\mathbb V$, if $\mathbb V$ is their direct sum both algebraically and topologically. This implies that
 $$
 {\mathbb{V}}_1\oplus {\mathbb{V}}_2\,=\,{\mathbb{V}}~\,,
 $$
 and the following addition map is a homeomorphism:
 $$
 {\mathbb{V}}_1\times{\mathbb{V}}_2\,\longrightarrow\,{\mathbb{V}}~~,\qquad
 \left(\, v\in{{\mathbb V}}_1\,,~\,v^{\,\prime}\in{{\mathbb V}}_2\,\right)\,\longmapsto~ (v+v^{\,\prime})\in\mathbb V~~.\\
 $$     

QUESTION A

For a *COUNTABLY* infinite splitting 
$$
{\mathbb V}=\bigoplus_{i\in\cal I}{\mathbb V}_i
$$
to be not only algebraic but also topological, would it be sufficient to impose the condition that a natural map
$$
\bigoplus \mathbb{V}_i\longrightarrow\mathbb{V}
$$ 
exists and is a homeomorphism?

Here $\bigoplus \mathbb{V}_i$ is a set of all families $\left(\mathbb{v}_i\right)_{i\in\cal I}$ with $v_i\in {\mathbb V}_i$ and only finitely many non-zero $v_i$. 
         $~~~~~\\$  

QUESTION B

Would this work also for an uncountable sum (direct integral) of subspaces?

Stated alternatively, can we always be sure that there always exists a necessary measure on the set of these subspaces?

If there is no general answer to question B, can this question be answered for Hilbert spaces?$~~~~~\\$           


KIND REQUEST

This question is of interest mainly to physicists. Could you please make your answer sufficiently detailed and, if possible, understandable to a layman? Thank you!



  [1]: https://math.stackexchange.com/questions/4371932/splitting-of-a-topological-vector-space-tvs-into-an-a-countable-sum-and-b