The set of all self-adjoint operators are closed in Hilbert space.
Let $T \in B(H)$ where $H$ is the Hilbert space and $B(H)$ denotes the set of linear operators from $H \mapsto H$.
We are working with the operator norm.
Adjoint of an operator $T$ is denoted by $T^*$.
An operator $T$ is said to self-adjoint if $T = T^*$ where $T^*$ is the adjoint of operator $T$.
To prove the above statement let us consider a sequence of self - adjoint operators $\{T_{n}\}$ converging to a operator $T$ , now if we prove that the operator $T$ is self-adjoint then we would prove that the set of all self-adjoint operators is closed.
So $T_{n} \rightarrow T$
To prove that $T$ is self-adjoint
Proof -
Let us consider $\|T -T^*\| = \|T - T_{n} + T_{n} - T_{n}^* + T_{n}^* - T^*\|$
By triangle inequality we have $\|T -T^*\| \leq \|T -T_{n}\| + \|T_{n} - T_{n}^*\| + \|T_{n}^* - T^*\|$
So we have now that $T_{n} = T_{n}^*$ as $T_{n}$ are self-adjoint operators so the middle term become zero.
so we are left with $\|T -T^*\|\leq \|T -T_{n}\| + \|T_{n}^* - T^*\|$.
Now the term $\|T - T_{n}\| \rightarrow 0$ but how to deal with the second term $\|T_{n}^* - T^*\| = \|T_{n} - T^*\|$?
As if we showed the RHS to be zero then $T= T^*$ implying $T$ is self-adjoint and hence the set of all self-adjoint operators are closed!