Let $A$ be a Banach algebra and $x\in A$. Let $\sigma(x)$ be its spectrum, and $\rho(x)$ be the complement of the spectrum, the resolvent.
Let $K\subseteq \mathbb{C}$ be such that $\delta(K)\subseteq \sigma(x)$. I want to show then that $K\subseteq \sigma(x)\cup$ bounded component of $\rho(x)$. Here $\delta(K)$ is the boundary of the set $K$. In my case, $K=\sigma(a)$ for some $a\in A$, so it is possible that we need to assume that $K$ is compact.
This seems obvious in the following sense: any interior point of $K$ is 'bounded' by its boundary, which in turn lies in $\sigma(x)$, so it can only belong to the bounded component of $\rho(x)$. However, I am unable to prove this rigorously, and would be most grateful for some help!
Also, what is the specific property of $\sigma(x)$ which gives us this result? Is it the compactness, or just the closed-ness, or just the boundedness? In other words, in the statement of the claim, can $\sigma(x)$ be replaced by any compact, or closed, or bounded set?
Edit: It seems that it is the compactness of $K$ which is more important. Is the following argument correct?
We already have that $\delta(K)\subseteq \sigma(X)$. Suppose $K^{\circ}$ is not contained in $\sigma(X)$. Let $U$ be a maximal open subset of $K^{\circ}$ such that $U\subseteq \rho(x)$. $U$ must be bounded. Then $\delta(U)\subseteq \delta(K)\subseteq \sigma(x)$. Hence $U$ must be contained in the bounded component of $\rho(x)$.
Thank you in advance.
Let $K$ and $L$ be two compact set in $\mathbb{C}$ such that $\partial K\subset L$. Then $K\subset L \cup$ bounded component of $L^c$. Note that $L^c$ has unique unbounded component and it is also pathconnected. Let us call that unbounded component as $P$.
Case-1: $\partial K=K$. Then we are done.
Case-2: $\partial K\subset K$. Let $z\in (K-\partial K)\cap P$. Here $z$ is an interior point. Let $U_z$ be maximal open set containing $z$ in $(K-\partial K)\cap P$. Moreover $\partial U_z\cap P\subset \partial K$. Since $K$ is compact, hence bounded. So we can take $y\in P$ at sufficiently large distance. Since $z,y$ is in $P$ we can joint these two by a continuous path which does not intersect $L$. But as $U_z$ is a open neighbourhood containing $z$, every continuous path will intersect with it's boundary. Which is a contradiction to the fact $\partial K\subset L$. So $z$ can not be lie in $P$. Therefore if $z\in K-\partial K$, then $z$ is in bounded components of $L^c$.
Note: We will take $K=\sigma(a)$ and $L=\sigma (x)$.
Note: I have tried to give justification. If there is any wrong argument, let me know please.