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[SOLVED] compact subset of R
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ThePerfectHacker
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- Jan 26, 2012
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Define "separate".
- Feb 15, 2012
- 1,967
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.
Prove that if $K \subseteq \{a+bi \in \Bbb C: b = 0\}$ is compact, that $\Bbb C - K$ is connected.
Just a guess.
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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?
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?
ThePerfectHacker
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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 $K\not = \mathbb{R}$.
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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 corollarys
ThePerfectHacker
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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.
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.
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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.
- Feb 15, 2012
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Path-connected implies connected (the converse isn't true, just for future reference).
Since $K$ is compact, it is bounded, and thus contained in some interval $[-r,r]$.
Choose $a \in \Bbb R$ so that $a > r$.
We can break it down into several cases:
Case 1: $z,w$ both lie in the same half-plane (with respect to the real axis) and neither is real.
Then if $\gamma:[0,1] \to \Bbb C$ is $\gamma(t) = tz + (1 - t)w$ we have:
$\gamma \cap K = \emptyset$.
Case 2: $z,w$ lie in different half-planes and neither is real. Assume $z$ is in the lower half-plane (we can always switch the roles of $z$ and $w$).
Define $\gamma:[0,1] \to \Bbb C$ by:
$\gamma(t) = 3t(a - i) + (1 - 3t)z$ for $t \in [0,\frac{1}{3}]$
$\gamma(t) = (3t - 1)(a + i) + (2 - 3t)(a - i)$ for $t \in (\frac{1}{3},\frac{2}{3})$
$\gamma(t) = (3t - 2)w + (3 - 3t)(a + i)$ for $t \in [\frac{2}{3},1]$
It should be cleat that $\gamma$ is continuous, and $\gamma \cap K = \emptyset$.
Case 3: one of $z,w$ is real. Use the same path as case 1.
Case 4: both of $z,w$ are real. This is actually the "trickiest" case.
Case 4a: either $z,w < \inf(K)$ or $z,w > \sup(K)$. In both cases use the path of case 1. (Equality is not possible, since $K$ is compact, and thus closed so that $\sup(K) \in K$ and similarly for $\inf(K)$).
Case 4b: either $z < \inf(K), w > \sup(K)$ or vice versa. Again, without loss of generality, assume $z < \inf(K)$. Define $\gamma$ like so:
$\gamma(t) = 3t(z - i) + (1 - 3t)z$ for $t \in [0,\frac{1}{3}]$
$\gamma(t) = (3t - 1)(w - i) + (2 - 3t)(z - i)$ for $t \in (\frac{1}{3},\frac{2}{3})$
$\gamma(t) = (3t - 2)w + (3 - 3t)(w - i)$ for $t \in [\frac{2}{3},1]$
This is the problem I see with your proof: E and F are supposed to be disjoint OPEN sets.
So you should assume that $\Bbb C \setminus K = K^c = E \cup F$ with E,F open and disjoint, in the relative topology. I don't see an easy proof along these lines, since $K$ is not open.
Since $K$ is compact, it is bounded, and thus contained in some interval $[-r,r]$.
Choose $a \in \Bbb R$ so that $a > r$.
We can break it down into several cases:
Case 1: $z,w$ both lie in the same half-plane (with respect to the real axis) and neither is real.
Then if $\gamma:[0,1] \to \Bbb C$ is $\gamma(t) = tz + (1 - t)w$ we have:
$\gamma \cap K = \emptyset$.
Case 2: $z,w$ lie in different half-planes and neither is real. Assume $z$ is in the lower half-plane (we can always switch the roles of $z$ and $w$).
Define $\gamma:[0,1] \to \Bbb C$ by:
$\gamma(t) = 3t(a - i) + (1 - 3t)z$ for $t \in [0,\frac{1}{3}]$
$\gamma(t) = (3t - 1)(a + i) + (2 - 3t)(a - i)$ for $t \in (\frac{1}{3},\frac{2}{3})$
$\gamma(t) = (3t - 2)w + (3 - 3t)(a + i)$ for $t \in [\frac{2}{3},1]$
It should be cleat that $\gamma$ is continuous, and $\gamma \cap K = \emptyset$.
Case 3: one of $z,w$ is real. Use the same path as case 1.
Case 4: both of $z,w$ are real. This is actually the "trickiest" case.
Case 4a: either $z,w < \inf(K)$ or $z,w > \sup(K)$. In both cases use the path of case 1. (Equality is not possible, since $K$ is compact, and thus closed so that $\sup(K) \in K$ and similarly for $\inf(K)$).
Case 4b: either $z < \inf(K), w > \sup(K)$ or vice versa. Again, without loss of generality, assume $z < \inf(K)$. Define $\gamma$ like so:
$\gamma(t) = 3t(z - i) + (1 - 3t)z$ for $t \in [0,\frac{1}{3}]$
$\gamma(t) = (3t - 1)(w - i) + (2 - 3t)(z - i)$ for $t \in (\frac{1}{3},\frac{2}{3})$
$\gamma(t) = (3t - 2)w + (3 - 3t)(w - i)$ for $t \in [\frac{2}{3},1]$
This is the problem I see with your proof: E and F are supposed to be disjoint OPEN sets.
So you should assume that $\Bbb C \setminus K = K^c = E \cup F$ with E,F open and disjoint, in the relative topology. I don't see an easy proof along these lines, since $K$ is not open.
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- Feb 15, 2012
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I think the relative topology is tripping you up here: E and F may be closed in the relative topology, but that does not mean they are STILL closed under the usual topology for the complex numbers.
For example, the interval $(0,1]$ is a closed subset of $(0,2)$ being the intersection of the closed set $[0,1]$ with $(0,2)$, but it is NOT a closed subset of the reals.
In fact we see that $E \cup F$ cannot be closed in $\Bbb C$ for that would imply that for every $k \in K$, $E \cup F$ would have to include the limit point:
\(\displaystyle \lim_{n \to \infty} a_n\)
where $a_n = k + \dfrac{i}{n}$
which is, of course $k$ (so $E \cup F$ doesn't include $k$, which is one of its boundary points).
Which means $\Bbb C = K \cup (E \cup F)$ is not a union of disjoint closed sets.
For example, the interval $(0,1]$ is a closed subset of $(0,2)$ being the intersection of the closed set $[0,1]$ with $(0,2)$, but it is NOT a closed subset of the reals.
In fact we see that $E \cup F$ cannot be closed in $\Bbb C$ for that would imply that for every $k \in K$, $E \cup F$ would have to include the limit point:
\(\displaystyle \lim_{n \to \infty} a_n\)
where $a_n = k + \dfrac{i}{n}$
which is, of course $k$ (so $E \cup F$ doesn't include $k$, which is one of its boundary points).
Which means $\Bbb C = K \cup (E \cup F)$ is not a union of disjoint closed sets.
ThePerfectHacker
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- Jan 26, 2012
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Compactness is not needed here. All you need is for $K$ to be a proper subset of $\mathbb{R}$. Then show you can find a path between any two points in $\mathbb{C}\setminus K$ which is easy to see if you visualize it.
Here is a general exercise from Munkres: Suppose $X,Y$ are connected spaces and $A,B$ proper subsets of $X$ and $Y$, then $(X\times Y) \setminus (A\times B)$ is connected. This is your question essentially, were you think of $\mathbb{C}^2$ as $\mathbb{R}\times \mathbb{R}$, take $X=Y=\mathbb{R}$, with $A= K$ and $B=\{ 0 \}$.
Here is a general exercise from Munkres: Suppose $X,Y$ are connected spaces and $A,B$ proper subsets of $X$ and $Y$, then $(X\times Y) \setminus (A\times B)$ is connected. This is your question essentially, were you think of $\mathbb{C}^2$ as $\mathbb{R}\times \mathbb{R}$, take $X=Y=\mathbb{R}$, with $A= K$ and $B=\{ 0 \}$.