The set $\mathbb{C} \setminus \sigma(x)$ is connected. Suppose that $A$ is a commutative Banach algebra with unit $e$ and the polynomials in $x \in A$ are dense in $A$. Show that the set $\mathbb{C} \setminus \sigma(x)$ is connected, where $\sigma(x)$ denotes the spectrum of $x$. I can show that $\lambda \notin \sigma(x)$ implies $P_n(x) \to (x-\lambda e)^{-1}$ for a sequence of polynomials $\{P_n(x)\}$ in $A$. However, I cannot prove that $P_n(z) \to (z-\lambda e)^{-1}$ uniformly for $z \in \sigma(x)$ without using the Gelfand transform. Any good suggestion?
 A: $
\newcommand{\C}{{\mathbb{C}}}
\newcommand{\N}{{\mathbb{N}}}
\newcommand{\inv}{^{-1}}
$
Suppose by contradiction that $\C\setminus \sigma (x)$ is disconnected,  so it admits at least one bounded connected
component,  say $\Omega $.
Picking any point $\lambda  \in \Omega $, we have that $x-\lambda $ is invertible and we may find  a sequence $\{p_n\}_n$ of polynomials
such that
$$
  \lim_{n\to \infty }\|p_n(x)-(x-\lambda )\inv\|=0.
  $$
Observing that the  Gelfand transform
$$
  A\to C(\sigma (x))
  $$
is contractive and
sends $x$ to the identity function on $\sigma (x)$,
we deduce that
$$
  \lim_{n\to \infty }\sup_{z\in \sigma (x)}|p_n(z)-(z-\lambda )\inv|=0.
  $$
Notice that
for every $n\in \N$ and every $z\in \sigma (x)$, one has that
$$
  |(z-\lambda )p_n(z)-1| = |z-\lambda ||p_n(z)-(z-\lambda )\inv|\leq  \alpha \beta _n,
  \tag{1}
  $$
where
$$
  \alpha =\sup_{z\in \sigma (x)}|z-\lambda |,
  $$
and
$$
  \beta _n=\sup_{z\in \sigma (x)}|p_n(z)-(z-\lambda )\inv|.
  $$
Observing that the boundary of $\Omega $ is contained in $\sigma (x)$, we deduce from the maximum principle that (1) also holds
for every $z\in  \Omega $, and hence
$$
  |p_n(z)-(z-\lambda )\inv|\leq  {\alpha \beta _n\over |z-\lambda |}, \quad\forall z\in \Omega \setminus\{\lambda \}.
  $$
Letting $\gamma (t)=\lambda +re^{it}$, for $t\in [0,2\pi ]$, with $r$ sufficiently small so that $\gamma $ lies in $\Omega $, we than have that
$$
  2\pi i=
  \left|\int_\gamma  p_n(z)-(z-\lambda )\inv \,dz\right| \leq
  {\alpha \beta _n\over r}\text{Length}(\gamma ) =2\pi \alpha \beta _n.
  $$
Since $\beta _n\to 0$, as $n\to \infty $, this is a contradiction.
