A year ago, I asked this question here at Mathematics Stackexchange, but no one there managed to answer it. So I am elevating it to MathOverflow.

This question had emerged as an offshoot of a bigger topic discussed here.
All linear vector spaces are over $\mathbb C$, and TVS is standing for Topological Vector Space.

DEFINITION

TVS subspaces $\left\{{\mathbb{V}}_i\subseteq{\mathbb{V}}:\;i\in {\cal{I}}\right\}$ of a TVS space ${\mathbb{V}}$ are in direct sum if the subspace they span
(1) is TVS, i.e. is closed;
(2) is their direct sum $\oplus {\mathbb{V}}_{{j}}\,$:
$$ {\mathbb{V}}_i \cap \left( \oplus {\mathbb{V}}_{\textstyle{_j}} \right)_{ \stackrel{}{\stackrel{_{j\in\cal I}}{\textstyle {_{j\neq i}}}} }\,=\,\{\vec{0}\} \quad\mbox{for}\;\forall\; i\in {\cal{I}}\;\;; $$ The latter item implies that any family of vectors $\left\{ v_i:\;v_i\in{\mathbb{V}}_i,\;i\in {\cal{I}} \right\}$, which are zero for all but finitely many values of $i$, is linearly independent.
$\,$

In a TVS space $\mathbb V$, consider a closed TVS subspace ${\mathbb{W}}\subset\mathbb V$ and the set $\cal D$ of all closed TVS subspaces in direct sum with $\mathbb W$: $$ {\cal D}\,=\,\left\{ {\mathbb{U}}:\;\;\;{\mathbb{U}}\subset{\mathbb{V}}\,,\;\;\; {\mathbb{U}}\;\mbox{closed}\,, \;\;\;{\mathbb{U}}\cap{\mathbb{W}}=\vec{ 0} \right\}\;\,. $$ The set $\cal D$ is partially ordered by inclusion. For an infinite chain ${\cal C}\subset\cal D$, the increasing union $\bigcup_{\cal C}\mathbb U$ is not necessarily closed. Denote is closure with overbar: $$ \tilde{\mathbb{U}}_{\cal C}\equiv\overline{\bigcup_{\cal C}\mathbb U}\,\;. $$

QUESTION 1.

While the union $\bigcup_{\cal C}\mathbb U$ is certainly in direct sum with $\mathbb W$, will this be true for its closure $\tilde{\mathbb{U}}_{\cal C}\,$?

QUESTION 2.

Suppose that a topological representation $A(G)$ of a group $G$ is acting in $\mathbb V$, and assume that all our spaces $\mathbb W$ and $\mathbb U$ are not only closed but also invariant, i.e. contain topological subrepresentations of $A(G)$.

While the union $\bigcup_{\cal C}\mathbb U$ is invariant, will its closure $\tilde{\mathbb{U}}_{\cal C}$ be invariant also?

A year ago, I asked this question here at Mathematics Stackexchange, but no one there managed to answer it. So I am elevating it to MathOverflow.

This question had emerged as an offshoot of a bigger topic discussed here.
All linear vector spaces are over $\mathbb C$, and TVS is standing for Topological Vector Space.

DEFINITION

TVS subspaces $\left\{{\mathbb{V}}_i\subseteq{\mathbb{V}}:\;i\in {\cal{I}}\right\}$ of a TVS space ${\mathbb{V}}$ are in direct sum if the subspace they span
(1) is TVS, i.e. is closed;
(2) is their direct sum $\oplus {\mathbb{V}}_{{j}}\,$:
$$ {\mathbb{V}}_i \cap \left( \oplus {\mathbb{V}}_{\textstyle{_j}} \right)_{ \stackrel{}{\stackrel{_{j\in\cal I}}{\textstyle {_{j\neq i}}}} }\,=\,\{\vec{0}\} \quad\mbox{for}\;\forall\; i\in {\cal{I}}\;\;; $$ The latter item implies that any family of vectors $\left\{ v_i:\;v_i\in{\mathbb{V}}_i,\;i\in {\cal{I}} \right\}$, which are zero for all but finitely many values of $i$, is linearly independent.
$\,$

In a TVS space $\mathbb V$, consider a closed TVS subspace ${\mathbb{W}}\subset\mathbb V$ and the set $\cal D$ of all closed TVS subspaces in direct sum with $\mathbb W$: $$ {\cal D}\,=\,\left\{ {\mathbb{U}}:\;\;\;{\mathbb{U}}\subset{\mathbb{V}}\,,\;\;\; {\mathbb{U}}\;\mbox{closed}\,, \;\;\;{\mathbb{U}}\cap{\mathbb{W}}=\vec{ 0} \right\}\;\,. $$ The set $\cal D$ is partially ordered by inclusion. For an infinite chain ${\cal C}\subset\cal D$, the increasing union $\bigcup_{\cal C}\mathbb U$ is not necessarily closed. Denote is closure with overbar: $$ \tilde{\mathbb{U}}_{\cal C}\equiv\overline{\bigcup_{\cal C}\mathbb U}\,\;. $$

QUESTION 1.

While the union $\bigcup_{\cal C}\mathbb U$ is certainly in direct sum with $\mathbb W$, will this be true for its closure $\tilde{\mathbb{U}}_{\cal C}\,$?

QUESTION 2.

Suppose that a topological representation $A(G)$ of a Lie group $G$ is acting in $\mathbb V$, and assume that all our spaces $\mathbb W$ and $\mathbb U$ are not only closed but also invariant, i.e. contain topological subrepresentations of $A(G)$.

While the union $\bigcup_{\cal C}\mathbb U$ is invariant, will its closure $\tilde{\mathbb{U}}_{\cal C}$ be invariant also?