Clousure of sum of two closed Hilbert subspaces with equivalent basis Let $H$ be an Hilbert space over $\mathbb{C}$
Let $V$ and $U$ be two closed subspaces of $H$
Let $\{v_m\}_{m \in \mathbb{N}} \in V$ be a basis for $V$
Let $\{u_m\}_{m \in \mathbb{N}} \in U$ be a basis for $U$
We know that $\{v_m\}_{m \in \mathbb{N}}$ is equivalent to  $\{u_m\}_{m \in \mathbb{N}}$ that is: there exists a topological isomorphism $T : V \to U$ such that $\forall m \in \mathbb{N}: T(v_m) = u_m$
My question is if is it true that: $V+U$ is closed
Thanks
 A: I suppose you are talking about orthonormal bases. No infinite dimensional Hilbert space has  a countable Hamel basis. 
Any two separable infinite dimensional Hilbert spaces are isometrically isomorphic so your assumption of existence of a topological isomorphism is very weak. Any well known example of two infinite dimensional separable Hilbert subspaces whose sum is not closed would be  a counterexample. 
See Direct sum of two closed subspaces of Banach space is not closed
A: No. The sum need not be closed. Consider the example given here. We start with an orthonormal basis $\{e_n\}_{n=1}^{\infty}$ in $\ell_2$ (say) and define two subspaces as follows:
$$M={\rm span}\{e_{2n}\}_{n=1}^{\infty},\quad N={\rm span}\{e_{2n}+\frac{e_{2n+1}}{n+1}\}_{n=1}^{\infty}$$
A general vector in $M$ has nonzero entries only at even indices, $(0,a_1,0,a_2,\dots)$,
and a general vector in $N$ has the form 
$$(0,a_1,\frac{a_1}{2},a_2,\frac{a_2}{3},a_3,\frac{a_3}{3}\dots)$$
That $M+N$ is not closed was proved in the first accepted answer in the link above. So it remains to show that there exists a topological isomorphism from $M$ to $N$ carrying the basis of $M$ to the basis of $N$. Define $T:M\to N$ by the obvious choice: 
$$T(e_{2n})=e_{2n}+\frac{e_{2n+1}}{n+1}$$
Then a general vector  $(0,a_1,0,a_2,\dots)$ in $M$ is transformed by $T$ into the general vector $x=(0,a_1,\frac{a_1}{2},a_2,\frac{a_2}{3},a_3,\frac{a_3}{3}\dots)$ in $N$.
The norm of the first of $x$ is clearly $\|x\|=(\sum_{n=1}^{\infty}a_n^2)^{1/2}$, whereas the norm of the $Tx$ satisfies
$$\|x\|\leq \|Tx\|=\left(\sum_{n=1}^{\infty}a_n^2+\sum_{n=1}^{\infty}\frac{a_n^2}{(n+1)^2}\right)^{1/2}\leq 2\|x\|$$
Since $\|x\|\leq\|Tx\|\leq 2\|x\|$, we see that $T$ is a topological isomorphism.
