Sequence of unitary operators is unitary 
If $T_n$ is a sequence of unitary linearbounded operators and $T_n\to T$ in the norm of $B(H,H)$ where $H$ is a Hilbert Space ,is it true that $T$ is linear bounded and unitary.

Steps:


*

*I have shown that $T$ is bounded.

*In order for $T$ to be unitary $T$ must be bijective.But I am unable to show that sequence of bijective linear bounded operators is bijective.


Please help.If the second criterion is fulfilled is the result true in general.
Any hints will be useful.
 A: It is well known that $B(H)$, the space of bounded operators on a Hilbert space $H$, with the natural operator composition, the operator-norm and the adjoint operation is a $C^*$-algebra. In particular, the operations are continuous with respect to the norm. This means that if $T_n\to T$ then $T_n^{*}\to T^*$ and also
$T^{*}_nT_n\to T^{*}T$. If $T_n$ is unitary for each $n$, then $T_n^{*}T_n=Id$ for each $n$, and so $T^*T=Id$ by continuity. Similarly, $TT^*=Id$. In particular, if $y\in H$, then $T(T^*y)=y$, so $T$ is surjective, and it is also injective, because it is an isometriy: $||Tx||=||x||$ for all $x$.
A: You want to prove that the unitary operators are closed. An operator $U$ is unitary if and only if $U^{\dagger} U = U U^{\dagger} = I$, and these are closed conditions, so the conclusion is immediate. 
It is not true that a convergent sequence of invertible operators converges to an invertible operator; you can already write down a counterexample in the finite-dimensional case, where $GL_n$ is dense in $M_n$. Invertibility is an open rather than a closed condition. 
