Let \(A\) be a compact contractible subspace of \(S^2\). Show that \(A\) does not separate \(S^2\).
Solution
Let \(A\subset S^2\) be compact and contractible. Suppose, for contradiction, that \(A\) separates \(S^2\). Choose points \(a,b\) in two distinct components of \(S^2-A\).
Since \(A\) is contractible, the inclusion \[ i:A\hookrightarrow S^2-\{a,b\} \] is nullhomotopic: contract \(A\) to any point \(a_0\in A\), and regard the contraction as taking place inside \(A\subset S^2-\{a,b\}\).
Identify \[ S^2-\{a,b\}\cong S^1\times\mathbb R. \] The Borsuk lemma from this section says that if a compact subset of \(S^2\) separates two points \(a,b\), then its inclusion in the twice-punctured sphere cannot be nullhomotopic; equivalently, a compact set with nullhomotopic inclusion into \(S^2-\{a,b\}\) cannot separate \(a\) from \(b\). This contradicts the preceding paragraph.
Hence \(S^2-A\) is connected: a compact contractible subset of \(S^2\) does not separate the sphere.