Proof. Since 1. follows from 2. by adjunction of an identity, it suffices to prove 2. In general, we have $\spec{A}{x}\suf \spec{B}{x}$.

Suppose that $x$ is self-adjoint. Let $\lam\nin\spec{A}{x}$ then by Proposition 6.4.4, $\spec{B}{x}\suf\R$ so to show that $\lam\nin\spec{B}{x}$ we may assume WLOG that $\lam\in\R$. For any $\e>0$, $\lam_{\e}=\lam+i\e\nin\spec{B}{x}$ so $(x-\lam_{\e})\inv\in B$. By continuity of inverse in $G(A)$, $(x-\lam_{\e})\inv\to (x-\lam)\inv\in G(A)$ as $\e\to 0$. Since $B$ is closed, $(x-\lam)\inv\in B$. Thus $\lam\nin\spec{B}{x}$ and $\spec{A}{x}=\spec{B}{x}$ for a self-adjoint $x\in B$.

Now if $x\in B$ is invertible in $A$ then $x^{*}x$ is invertible in $A$ so is in $B$ by self-adjointness of $x^{*}x$. Hence $x$ is left invertible in $B$. Similarly considering $xx^{*}$ shows that $x$ is right invertible in $B$ also. Therefore, $x$ is invertible in $B$. We apply this argument for $x-\lam,\lam\in\com$ to conclude 2.$\square$