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Let $A$ be a bounded operator on complex Hilbert space $H$ that is not self-adjoint. Let $\epsilon>0$ and define $R_{\epsilon}=\{T\in B(H):||A-T||<\epsilon\}$.

How do I prove that there exists an $\epsilon$ such that $R_{\epsilon}$ contains no self-adjoint bounded operators on $H$?

What I thought:
We need to show that for $T\in R_{\epsilon}$ we have $T\neq T^*$.
We know that the map $f:B(H)\rightarrow B(H)$ with $f(T)=T-T^*$ is continuous. So there exists an $\delta>0$ such that for all $\epsilon>0$ we have for $||T_1-T_2||<\delta$ that $$||T_1-T_1^*-T_2+T_2^*||<\epsilon.$$ Further I know that $f$ has kernel the self-adjoint operators, but how can I use all this for the proof?

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You have all of the ingredients. Consider the map $f:B(H)\to \mathbb R$ defined by. $$f(T)=\|T-T^*\|.$$ This map is continuous, and $f^{-1}(\{0\})$ is the collection of self-adjoints. Since $\{0\}$ is closed in $\mathbb R$, the collection of self-adjoints is closed in $B(H)$, hence the collection of non self-adjoints is open in $B(H)$. So if $T\in B(H)$ is not self-adjoint, it is an interior point in the collection of non self-adjoints, and there is some $\varepsilon>0$ such that $R_\varepsilon$ does not contain any self-adjoints.

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  • $\begingroup$ Could you maybe explain how the result follows more explicitly? $\endgroup$ – user402675 Dec 29 '16 at 14:00

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