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I tried to prove that If $\omega$ is the complement of a compact set, then $\omega$ has only one unbounded component.

I know that the complement of a large disc containing the compact set is unbounded and connected. I see the prove saying that if $\omega$ is not connect, then it must have another component in the disc. But thus it is bounded. I just have a question of the definition of being "bounded." Why $\omega$ having at least one component bounded by the disc, another unbounded, being a bounded set?

Andrés E. Caicedo
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John He
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  • What’s your ambient space? The real plane? – k.stm Jul 22 '19 at 09:42
  • $\omega$ is in complex plan. – John He Jul 22 '19 at 09:43
  • Your English makes it hard to follow your reasoning. So, here’s my analysis anyway: Yes – $ω$ has exactly one unbounded connected component, hence – yes – if $ω$ is not connected, at least one (namely all but one) of its connected components must be bounded. And in this case, you then ask “Why [does] $ω$ [have] at least one component bounded by the disc [and] another unbounded one, being a bounded set?” What are you refering to as a “bounded set”? Is it $ω$? But $ω$ is not bounded, where are you taking this from? – k.stm Jul 22 '19 at 09:50
  • I was confused by the solution I get from my teacher. He says the complement of the large disc is contained in $\omega$, If omega is not connect, it must have another component in the disc, which proves the theorem. – John He Jul 22 '19 at 09:57
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    Yes, and that’s perfectly fine and indeed proves the assertion. What’s the problem? – k.stm Jul 22 '19 at 10:00
  • Never mind, I understood It wrongly. – John He Jul 22 '19 at 10:01

1 Answers1

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Let $C$ be an unbounded component of $\omega$. Since $\omega^{c}$ is bounded $z \in \omega$ whenever $|z|$ is sufficiently large, say $|z|>R$. Hence there exist a point $z$ common to $C$ and $\{z:|z|>R\}$. This last set is connected so $C \cup \{z:|z|>R\}$ is connected. [If two connected sets have a point in common then their union is connected]. Now maximality of components implies that $C=C \cup \{z:|z|>R\}$. Hence any two unbounded components have a point in common. Since any two components are either equal or disjoint it follows that there is only one unbounded component.

Kavi Rama Murthy
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