It is well known that the spectrum is continuous as function of operator. More precisely, let $\mathcal{H}$ be separable Hilbert space and $\mathcal{B}(\mathcal{H})$ the Banach algebra of linear operators acting on $\mathcal{H}$, then one has
$$(\forall A,B\in\mathcal{B}(\mathcal{H}))(\forall \epsilon>0)(\exists \delta >0 )( \|A-B\|<\delta \Rightarrow \mbox{dist}(\sigma(A),\sigma(B))<\epsilon).$$


My first question is: Is it possible that this continuity is even uniform, i.e.,


$$(\forall \epsilon>0)(\exists \delta >0 )(\forall A,B\in\mathcal{B}(\mathcal{H}))( \|A-B\|<\delta \Rightarrow \mbox{dist}(\sigma(A),\sigma(B))<\epsilon)?$$


I do not expect the affirmative answer, however, I can not prove it. For that reason, I add a second question which, as I hope, could have the affirmative answer.


Let $\mathcal{S}\subset\mathcal{B}(\mathcal{H})$ such that
$$\sup\_{A\in\mathcal{S}}\|A\|<\infty,$$
is then true that
$$(\forall \epsilon>0)(\exists \delta >0 )(\forall A,B\in\mathcal{S})( \|A-B\|<\delta \Rightarrow \mbox{dist}(\sigma(A),\sigma(B))<\epsilon)?$$


Thanks!


**Edit:** Nik Weaver pointed out that even the first statement denoted as "well-known" is not true. I also realized that I did not clarify the notion of distance here. Let me reformulate the statement and provide a verification. Where is the mistake in the following verification?


1. Restatement:


$$(\forall A,B\in\mathcal{B}(\mathcal{H}))(\forall \epsilon>0)(\exists \delta >0 )( \|A-B\|<\delta \Rightarrow \sigma(B)\subset\mathcal{U}\_{\epsilon}(\sigma(A)))$$
Here $\mathcal{U}\_{\epsilon}(M)$ denotes the $\epsilon$-neighborhood of a set $M\subset\mathbb{C}$.


2. "Verification": Let $\epsilon>0$ be given. And define $\delta>0$ by
$$\delta^{-1}:=\max\_{z\notin\mathcal{U}\_{\epsilon}(\sigma(A))}\|(A-z)^{-1}\|.$$
Note that the above maximum exists and is finite since resolvent operator $(A-z)^{-1}$ is an analytic function on the resolvent set $\rho(A)$ and bounded in a neighborhood of $\infty$ for one has
$$ (A-z)^{-1}\sim-\frac{1}{z}, \quad \mbox{ as } z\to\infty,$$
by the von Neumann series argument.


Now, if $\|A-B\|<\delta$ and $z\notin\mathcal{U}\_{\epsilon}(\sigma(A))$, then 
$$ \|(A-B)(A-z)^{-1}\|<1$$
and hence the resolvent operator
$$ (B-z)^{-1}=(A-z)^{-1}(1-(A-B)(A-z)^{-1})^{-1}$$
exists as a bounded operator. Consequently, $z\in\rho(B)$ and the statement follows.



Uniform continuity of spectrum as function of operator
The "well-known" fact is, of course, false. According to a theorem of C. Apostol and B. Morrel (On uniform approximation of operators by simple models, *Indiana Univ. Math. J.* **26** (1977), 427–442), if $K$ is a nonempty compact subset of $\mathbb{C}$ and $A$ is a normal operator such that $\sigma(A)$ is connected and $K \subseteq \sigma(A)$, then $A$ is approximated in norm by operators whose spectrum is contained in $K$.


