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Let $\gamma^*$ be the image of a simple loop $\gamma$ in $\mathbb{C}$. On page 203 of Rudin's Real and Complex Analysis, while introducing the winding number, he gives the following reason for why the complement of $\gamma^*$ must have only one unbounded component. (This is before any mention of the Jordan Curve Theorem.)

Note that $\gamma^*$ is compact, hence $\gamma^*$ lies in a bounded disk $D$ whose complement $D^c$ is connected; thus $D^c$ lies in some component of $\Omega$. This shows that $\Omega$ has precisely one unbounded component.

I want to make sure I understand his reasoning.

Note that $\gamma^*$ is compact, hence $\gamma^* \subset D$ (it lies in a bounded disk, this is from the topology of $\mathbb{C}$? or some form of Lebesgue number lemma?) whose complement $D^c$ is connected (since the complement is now path-connected?); thus $D^c$ lies in some component of $\Omega$ (how do we know $D^c$ isn't the component (maximal connected subset?). This shows that $\Omega$ has precisely one unbounded component (since $D^c$ is also unbounded).

J.-E. Pin
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Lemon
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  • Relevant https://math.stackexchange.com/questions/133290/complement-of-a-bounded-set-b-in-mathbbrn-has-exactly-one-unbounded-co?rq=1 – Sumanta Sep 10 '20 at 15:53

2 Answers2

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For a rundown on your reasoning, read after the horizontal rule below. First, let me prove what he mentions.

Following up on Rudin's notation, to see why $\Omega$ is the unique component which can be unbounded, suppose that there is another, say $\Omega'$. Since $\Omega'$ is unbounded, there is some $x$ which belongs to $\Omega'$ and also $D^c$. Therefore, $\Omega'$ and $\Omega$ have a point in common. Since they are connected components, they must coincide.

Note that nowhere in this argument do we conclude that there is only one other component, and neither does Rudin claim that. This is part of the Jordan Curve Theorem, as you imply.

Note that $\gamma^*$ is compact, hence $\gamma^* \subset D$.

Yes, it lies in a bounded disk. This is true in general for any compact set inside a metric space $M$, and there is no need for Lebesgue number lemma: given a compact set $K$ and a point $p \in M$, there is an open (or closed) ball centered on $p$ big enough that contains $K$. Just consider the collection of open balls of radius $n$ centered at $x$ and apply the definition to see why this holds. (And take the closure of the maximum of the resulting radii after you apply it, if you want to have a closed ball in the end.)

whose complement $D^c$ is connected (since the complement is now path-connected?)

Yes, that is a valid way to justify it.

thus $D^c$ lies in some component of $\Omega$ (how do we know $D^c$ isn't the component (maximal connected subset?).

It could be the entire component. (Just take $\gamma$ to be a circle, and $D$ to be the disk determined by it.) So we don't know that, but it is irrelevant.

This shows $\Omega$ has precisely one unbounded component (since $D^c$ is also unbounded).

Unfortunately, nothing that you mentioned makes this conclusion directly valid. (At least, if what Rudin mentioned doesn't make it valid, then what you didn't also does not.)

Aloizio Macedo
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  • In the last statement "This shows $\Omega$ has 1 unbounded comp" is from Rudin, but the parenthesis is my reasoning that he left out. Because i thought since $D^c$ is unbounded, whatever component containing it is also unbounded. As to how I know there is exactly 1? I guess I used Jordan Curve theorem accidentally even though i said this probably came before the theorem. – Lemon Sep 30 '19 at 03:50
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$\gamma[[0,1]] \subseteq D$ where $D$ is some closed disk.

It's easy to see (e.g. by path-connectedness, as you say) that $\Bbb C\setminus D$ is connected. And as $\gamma^\ast=\Bbb C\setminus \gamma[[0,1]] \supseteq \Bbb C \setminus D$, and the latter set is unbounded and connected, there is at least one unbounded component of $\gamma^\ast$. That is all that Rudin's reasoning gives.

To get the at most (or exactly) one, consider the component of $\infty$ in $\Bbb C \setminus \gamma[[0,1]]$ in the Riemann sphere $\Bbb C^\ast$.

Henno Brandsma
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  • So for the most part i am right except the first statement which the disk just came from the usual topology of $\mathbb{C}$ – Lemon Sep 30 '19 at 03:52
  • @Hawk, yes, the disk comes from boundedness in $\Bbb C$ and its complement is an open neighbourhood of $\infty$. – Henno Brandsma Sep 30 '19 at 05:44