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Yes Deveno, that is correct. I think I have worked it out. $K^{c}=(R-K)+C$ where '+' is union. Since K is bounded it is not equal to R so $K^c$ is connected. Is that correct?Perhaps he means:
Prove that if $K \subseteq \{a+bi \in \Bbb C: b = 0\}$ is compact, that $\Bbb C - K$ is connected.
Just a guess.
Show that $X=\mathbb{C} \setminus K$ is path-connected, i.e. for any two points in this set there exists a path contained in this set connecting the two points. Note, this proof has nothing to do with $K$ being compact, rather that $K\not = \mathbb{R}$.Yes Deveno, that is correct. I think I have worked it out. $K^{c}=(R-K)+C$ where '+' is union. Since K is bounded it is not equal to R so $K^c$ is connected. Is that correct?
i've just realised those sets are not disjoint. Why do books act like things are 'obvious'. Specifiaclly Rudins functional analysis http://59clc.files.wordpress.com/2012/08/functional-analysis-_-rudin-2th.pdf page 256-7, no explanation of corollarysYes Deveno, that is correct. I think I have worked it out. $K^{c}=(R-K)+C$ where '+' is union. Since K is bounded it is not equal to R so $K^c$ is connected. Is that correct?
Now for some context for the next question. Let A be a B^* banach subalgebra of a B^* banach algebra B. I have proved that the boundary of ${\sigma}_{A}(x)$ lies in the boundary of ${\sigma}_{B}(x)$. From this, I should be able to deduce that if ${\rho}_{B}x$ is connected, then ${\sigma}_{A}(x)$=${\sigma}_{B}(x)$, but how?
I'm having a hard time trying to prove that if A is a compact subset of R, then C\A is connected. Here's what I tried. Assume C\A is disconnected so C\A=E+F, where E,F are disjoint closed sets. Then, C=E+F+A. A is closed and disjoint from E,F so this implies C is disconnected - a contradiction. Look's fine but this argument should break down when A=R, but since R is closed, I can't see where it does.I do not know what the notation means, so I am just guessing by his use of words. He says that $\sigma_A (x) = \sigma_B(x) \cup \bigcup_i C_i$ where $C_i$ are the bounded components of the complement of $\sigma_B(x)$.
What is a "component of the complement" of $\sigma_B(x)$. It means you take the complement of $\sigma_B(x)$, whatever that means, I am assuming he is taking the complement in the complex numbers, so the complement of $\sigma_B(x)$ is the set $\mathbb{C}\setminus \sigma_B(x)$. This set is not necessarily connected, but it can be written as a union of disjoint sets each of which is connected, those individual sets that comprise the union are called the components (of the complement). If the complement is connected it means it only has one component - itself, otherwise it will have many, possibly infinitely many, components. Thus, if $\sigma_B(x)$ is such that its complement is connected, as in the theorem, it has no bounded components (as it only has one component, itself, which is unbounded) and so by the theorem $\sigma_A(x) = \sigma_B(x) \cup \emptyset$, and from here your corollary follows.