Closures, intersections and sums of subspaces in Hilbert space Let $X$ be a real Hilbert space, and let $U$ and $V$ be two closed linear subspaces of $X$. Is it true that
$$\overline{(U+V) \cap (U^\perp+V^\perp)} = \overline{U+V} \cap \overline{U^\perp + V^\perp}\quad?$$
The left-hand side is clearly a subset of the right-hand side, but the opposite inclusion stumps me. (The result is clearly true if $X$ is finite-dimensional because all subspaces are automatically closed.)
I checked my trusted Functional Analysis book, math.stackexchange, as well as Halmos' A Hilbert Space Problem book but couldn't find anything. This should be known! Please provide a reference or thought. Thanks!
 A: I think the claim is true.
We will show the more general claim
$$
\overline{(U+A)\cap (U^\perp + B)}
= \overline{U+A} \cap \overline{U^\perp + B}
$$
for (arbitrary) subsets $A,B\subset X$ and a closed subspace $U\subset X$.
Again, it is easy to see that the left-hand side is contained in the
right-hand side.
Let $x$ be an element in the right-hand side.
Then there exist sequences
$\{u_n\}_{n\in\mathbb N}\subset U$,
$\{a_n\}_{n\in\mathbb N}\subset A$,
$\{v_n\}_{n\in\mathbb N}\subset U^\perp$,
$\{b_n\}_{n\in\mathbb N}\subset B$
with $u_n+a_n\to x$ and $v_n+b_n\to x$.
Let us denote the orthogonal projections onto a closed subspace $W$ by $P_W$.
Then, applying the operators $P_{U^\perp}$ and $P_U$ to the convergences above
yields
$$
P_{u^\perp}(u_n) + P_{U^\perp}(a_n) = P_{U^\perp}(a_n)\to P_{U^\perp}(x)
$$
and
$$
P_U(v_n) + P_U(b_n) = P_U(b_n)\to P_U(x).
$$
Addition yields
$$
z_n := P_U(b_n) + P_{U^\perp}(a_n) \to P_{U^\perp}(x)+ P_U(x) = x.
$$
We also have
$$ z_n = P_U(b_n)+P_{U^\perp}(a_n) = P_U(b_n)+ a_n - P_U(a_n) \in U+A $$
and
$$ z_n = P_U(b_n)+P_{U^\perp}(a_n) 
= b_n - P_{U^\perp}(b_n)+ P_{U^\perp}(a_n) \in U^\perp+B $$
and therefore $z_n \in (U+A)\cap (U^\perp+B)$.
Together with $z_n\to x$ this implies that $x$ is in the left-hand side.
A: It is an evident fact that in general topology the intersection of two subsets $A$ and $B$ of some topological space $X$ is included in the intersection of their closures: $\overline{A\cap B}\subset \overline{A}\cap \overline{B}$ always holds by definition of the closure.
But the opposite inclusion is not generally true.
Of course if both subsets are themselves closed, i.e., if $A=\overline{A}$ and $B=\overline{B}$ the opposite inclusion hence the equality holds. Since
$\overline{A}\cap \overline{B}=A\cap B$, by taking the closure on both sides
on gets $\overline{\overline{A}\cap \overline{B}}=\overline{A\cap B}$.
Since the intersection of two closed sets is closed,
$\overline{\overline{A}\cap \overline{B}}=\overline{A}\cap \overline{B}$.
So one gets
$\overline{A}\cap \overline{B}=\overline{A\cap B}$.
This is a very general set up of topological spaces whether or not their topology is induced by a inner scalar product.
If in your Hilbert space setup $U$ and $V$ were in direct sum then the converse inclusion would hold since if $X=A\oplus A^\perp$  is the orthogonal sum of linear subspaces $A$, $A^\perp$, then both $A$ and $A^\perp$ are closed in $X$. This closure property would then hold for $U$, $U^\perp$, $V$,$V^\perp$ as well as for $A=U\oplus V$ and $B=A^\perp=U^\perp\oplus V^\perp$.
Unfortunately as you gently pointed me out , max_zorn, $U+V$ is a not a direct sum and one can not infer from the fact that $U$ (resp. $U^\perp$) and $V$ (resp $V^\perp$) are closed that their sum $U+V$ (resp. $U^\perp+V^\perp$) is also closed.
So the question remains fully open ...
