# Prove $S^2 - C$ has $2$ connected components whenever $C \simeq S^1$

I am trying to prove the following, which is a problem set question from a course on differential geometry and topology that I am currently taking:

Suppose $$C \subseteq S^2$$ is homeomorphic to $$S^1$$. Prove that $$S^2 - C$$ has two connected comoponents.

It makes sense, if I draw a closed curve that does not cross itself on a spherical object, the curve will separate two connected regions on the spherical surface.

I feel that I have made some progress, but I am stuck. The symbol $$\simeq$$ stands for "is homeomorphic to" in this context.

My attempt:

Since $$S^2$$ is not homeomorphic to $$S^1$$, we have that $$S^2$$ is not homeomorphic to $$C$$. Thus, $$S^2 \neq C$$, so $$S^2 - C$$ is non-empty. Choose $$p \in S^2 - C$$.

Then $$S^2 - \{p\}$$ is homeomorphic to $$\mathbb{R}^2$$ via the stereographic projection $$\sigma_p : S^2 - \{p\} \to \mathbb{R}^2$$. The restriction of $$\sigma_p$$ to $$C$$ is a homeomorphism between $$C$$ and $$\sigma_p(C)$$. Hence, $$\sigma_p(C) \simeq S^1$$. I.e. $$\sigma_p(C)$$ is a Jordan curve.

Applying the Jordan Schoenflies theorem to $$\sigma_p(C)$$, we have that $$\mathbb{R}^2 - \sigma_p(C)$$ has two connected components, with each component having boundary $$\sigma_p(C)$$.

But $$\mathbb{R}^2 - \sigma_p(C) = \sigma_p((S^2 - \{p\}) - C)$$, and $$(S^2 - \{p\}) - C \simeq \sigma_p((S^2 - \{p\}) - C),$$ by the restriction of $$\sigma_p$$ to $$(S^2 - \{p\}) - C.$$

Since the number of connected components is preserved under homeomorphism, we deduce that $$(S^2 - \{p\}) - C$$ has two connected components, call them $$X$$ and $$Y$$. We have

$$(S^2 - \{p\}) - C = X \sqcup Y,$$ where $$\sqcup$$ denotes a disjoint union.

This is about as far as I can get.

We have $$\{p\} \sqcup ((S^2 - \{p\}) - C) = S^2 - C$$, and I believe that either $$X \sqcup \{p\}$$ and $$Y$$ are the connected components of $$S^2 - C$$, or $$Y \sqcup \{p\}$$ and $$X$$ are the connected components of $$S^2 - C$$. I'm not sure how to prove this though.

Any ideas on how to proceed with this line of reasoning? It would be great to see other approaches too!

## 1 Answer

The key is to identify which of the two connected components should be completed to contain $$p$$. To do this, let $$U_p\subset S^2$$ be a neighborhood of $$p$$ disjoint from $$C$$; such a neighborhood exists because $$C$$ is compact. Then $$\sigma_p$$ will map $$U_p\setminus\{p\}$$ into one of the two connected components of $$\mathbb{R}^2\setminus\sigma_p(C)$$, and therefore $$U_p\setminus\{p\}$$ lies in one of the two connected components $$X$$ and $$Y$$ forming $$(S^2\setminus\{p\})\setminus C$$.

Without loss of generality let us say $$U_p\setminus\{p\} \subset X$$. Clearly $$S^2\setminus C$$ is the union of $$X\cup\{p\}$$ and $$Y$$. The union is disjoint, because $$X$$ and $$Y$$ were disjoint and $$p$$ is not an element of either $$X$$ or $$Y$$. All that remains is to check that $$X\cup\{p\}$$ is connected. For this, let $$X\cup\{p\} = A\cup B$$, where $$A$$ and $$B$$ are disjoint and open. Suppose without loss of generality that $$p\in A$$. Then $$X = (A\setminus\{p\})\cup B$$. Thus $$X$$ is a union of disjoint open sets. But $$X$$ is nonempty and connected, so one of these sets must be $$X$$ and the other empty. In fact it must be that $$B$$ is empty; for if $$B=X$$, then $$A = \{p\}$$, which is not open. We conclude that whenever $$X\cup\{p\} = A\cup B$$ with $$A,B$$ disjoint and open, one of the two must be empty and the other must be $$X\cup\{p\}$$. Therefore $$X\cup\{p\}$$ is connected and proof is complete.

• If $U_p$ is disjoint from $C$, then we have $\sigma_p(U_p - \{p\}) \cap \sigma_p(C) = \emptyset$, but this alone doesn't necessarily mean that $\sigma_p(U_p - \{p\})$ has an empty intersection with one of the connected components of $\mathbb{R}^2 - \sigma_p(C)$. Could this be fixed by letting $U_p$ be connected? There would be some additional proof needed then. – user657854 Nov 13 '19 at 3:17
• Fair point. Sure, it's enough for $U_p$ to be connected, as then it can only lie in a single connected component. Arranging this isn't hard: embed $S^2$ into $\mathbb{R}^3$, give it the topology induced by the Euclidean metric, and take $U_p$ to be a small punctured disc around $p$. – Gyu Eun Lee Nov 13 '19 at 11:52