Let \(D\) be a subspace of \(S^2\) homeomorphic to the topologist’s sine curve \(\overline{S}\). (See §24.) Show that \(D\) does not separate \(S^2\). [Hint: Let \(h: \overline{S} \to D\) be the homeomorphism. Given \(0 < c < 1\), let \(\overline{S}_c\) equal the intersection of \(\overline{S}\) with the set \(\ts{(x, y) \mid x \leq c}\). Show that given \(a, b \in S^2 - D\), there is, for some value of \(c\), a path in \(S^2 - h(\overline{S}_c)\) from \(a\) to \(b\). Conclude that there is a path in \(S^2 - D\) from \(a\) to \(b\).]
Let \(C\) be a subspace of \(S^2\) homeomorphic to the closed topologist’s sine curve. Show that \(C\) separates \(S^2\) into precisely two components, of which \(C\) is the common boundary. [Hint: Let \(h\) be the homeomorphism of the closed topologist’s sine curve with \(C\). Let \(C_0 = h(0 \times [-1, 1])\). Show first, using the argument of Theorem 63.4, that each point of \(C - C_0\) lies in the boundary of each component of \(S^2 - C\).]
Solution
Write \[ \overline S=\bigl\{(x,\sin(1/x)):0<x\le1\bigr\}\cup(\{0\}\times[-1,1]). \]
(a) We first show that the standard \(\overline S\subset S^2\) does not separate the sphere. Its complement in \(\mathbb R^2\) is path connected. Indeed, if \(0<x\le1\), a point above the graph can be joined vertically to the line \(y=2\), while a point below the graph can be joined vertically to \(y=-2\); neither segment crosses the graph. Points with \(x\le0\) or \(x>1\) can likewise be joined to one of these two horizontal lines while avoiding the vertical limiting segment. The two lines \(y=2\) and \(y=-2\) are joined through the half-plane \(x>1\), which misses \(\overline S\). Thus all points of \(\mathbb R^2-\overline S\) lie in one path component. Adding the point at infinity does not change connectedness, so \(S^2-\overline S\) is connected.
Now let \(h:\overline S\to D\subset S^2\) be the given homeomorphism. The Borsuk lemma from §62 says that an injective continuous image of a compact nonseparating subset of \(S^2\) is again nonseparating. Hence \(D=h(\overline S)\) does not separate \(S^2\). This proves (a).
(b) Let \(A\) denote the standard closed topologist’s sine curve. It is obtained from \(\overline S\) by adjoining the broken-line arc \(P\) from \((0,-1)\) to \((1,\sin1)\), chosen so that \[ P\cap\overline S=\{(0,-1),(1,\sin1)\}. \] Let \(h:A\to C\) be the given homeomorphism and put \[ D=h(\overline S),\qquad Q=h(P). \] By part (a), \(D\) is closed and connected and \(S^2-D\) has one component. The arc-attachment result proved earlier in this section says that adjoining an arc whose intersection with a closed connected set consists exactly of the two endpoints increases the number of complementary components by one. Hence \[ S^2-C=S^2-(D\cup Q) \] has exactly two components; call them \(U\) and \(V\).
It remains to prove that \(C\) is the boundary of both. Let \[ C_0=h(\{0\}\times[-1,1]). \] Every point of \(C-C_0\) is locally an ordinary point of an embedded arc. The local argument in the proof of Theorem 63.4 applies there: a sufficiently small disk meets \(C\) in a single crosscut, and its two sides lie in the two distinct components \(U\) and \(V\). Consequently \[ C-C_0\subset \partial U\cap\partial V. \] Every point of \(C_0\) is a limit of points of the oscillating graph, hence a limit of points of \(C-C_0\). Since boundaries are closed, \[ C_0\subset\partial U\cap\partial V. \] Thus \(C\subset\partial U\cap\partial V\).
Conversely, \(U\) and \(V\) are components of the open set \(S^2-C\). Since open subsets of the locally connected sphere have open components, both \(U\) and \(V\) are open in \(S^2\). Hence no point of \(S^2-C\) can lie in the boundary of either component, so \[ \partial U\cup\partial V\subset C. \] Therefore \[ \boxed{\partial U=\partial V=C}, \] and \(C\) separates \(S^2\) into precisely two components with common boundary \(C\).