If $H$ is a Hilbert space, $\varnothing \neq A \subseteq H$, $T: H\rightarrow H$ is a linear bounded operator with two properties, then A is closed. Let $H$ be a Hilbert space, $A$ a non empty subset of $H$ and $T:H\rightarrow H$ a linear bounded operator with the following properties:

*

*$T(H) \subseteq A$

*$x-T(x) \perp A \ \ \  \forall x \in H$
I need to prove that $A$ is a closed subspace of $H$. 
I considered a sequence $(a_n)$ in $A$ such that $a_n \rightarrow a \in H$. Since $T$ is bounded, it is also continuous. So $T(a_n) \rightarrow T(a)\in T(H) \subseteq A$. 
So, how would I prove that $a \in A$, knowing that $T(a) \in A$ (if it's even needed)? 
Any help would be greatly appreciated. 
Thanks in advance.

You don't really need to use that $T$ is linear nor that it is bounded; just that it maps $H$ into $A$ and that $I-T$ maps into $A^\perp$. Given any $x\in H$, you can write
$$
x=Tx+(x-Tx)\in A+A^\perp. 
$$
This shows that $H=A+A^\perp$. Now you can see that $A$ is closed in several ways.
For instance, if $\{a_n\}\subset A$ is Cauchy, let $x=\lim_na_n$. We have $x=a+b$, with $a\in A$ and $b\in A^\perp$. So $x-a\in A^\perp$. Now, as $x-a=\lim_n a_n-a$ and $a_n-a\in A$ for all $n$,
$$
\|x-a\|^2=\langle x-a,x-a\rangle=\lim_n\langle x-a,a_n-a\rangle=0. 
$$
Thus $x=a\in A$ and $A$ is closed.
A second way is to show that $A=A^{\perp\perp}$, which makes it closed. Indeed, if $x\in A^{\perp\perp}$, then writing $x=a+b$ with $a\in A$ and $b\in A^\perp$, we have
$$
0=\langle x,b\rangle=\|b\|^2+\langle a,b\rangle=\|b\|^2. 
$$
So $b=0$ and then $x=a\in A$.
As mentioned in the comments, both arguments show that $A$ is a subspace.
