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?

- 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 $M\subset\mathbb{C}$.

- "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.