Show that the one-point compactification of \(S_\Omega\) is homeomorphic with \(\overline{S}_\Omega\).
Solution
Goal: Prove that the one-point compactification \(S_\Omega^*\) of the minimal uncountable well-ordered set \(S_\Omega = [0, \Omega)\) is homeomorphic to \(\overline{S}_\Omega = [0, \Omega]\) equipped with the order topology.
1. Properties of \(\overline{S}_\Omega\): 1. \(\overline{S}_\Omega = [0, \Omega]\) in the order topology is compact and Hausdorff.2. The subspace topology on \(S_\Omega = [0, \Omega) \subset \overline{S}_\Omega\) is identical to the intrinsic order topology on \(S_\Omega\).3. The remainder \(\overline{S}_\Omega \setminus S_\Omega = \{\Omega\}\) is a single point.
1.1. Every well-ordered set with a largest element is compact in the order topology (Munkres Theorem 27.1), and every order topology is Hausdorff.Thus \(\overline{S}_\Omega\) is compact Hausdorff.
1.2. Since \(S_\Omega\) is an initial segment (convex set) in \(\overline{S}_\Omega\), the subspace topology and the order topology on \(S_\Omega\) coincide.
2. Compact subsets of \(S_\Omega\): A closed subset \(C \subseteq S_\Omega\) is compact in \(S_\Omega\) if and only if \(C\) is bounded above by some \(\alpha < \Omega\).
2.1. (\(\impliedby\)) If \(C\) is closed and bounded by \(\alpha < \Omega\), then \(C\) is a closed subset of the compact interval \([0, \alpha]\), hence \(C\) is compact.
2.2. (\(\implies\)) If \(C\) is unbounded in \(S_\Omega\), the collection of open rays \(\{[0, \beta) : \beta \in S_\Omega\}\) forms an open cover of \(C\) with no finite subcover (since the supremum of any finite set of elements in \(S_\Omega\) is strictly less than \(\Omega\)). Thus \(C\) cannot be compact.
3. Construction of the bijection \(f: S_\Omega^* \to \overline{S}_\Omega\): Define \(f: S_\Omega^* \to \overline{S}_\Omega\) by: \[f(x) = x \quad \text{for } x \in S_\Omega, \qquad f(\infty) = \Omega.\] This is a bijection: it is injective because the restriction to \(S_\Omega\) is the identity (so distinct points of \(S_\Omega\) have distinct images) and \(\Omega\) is not in \(S_\Omega\) (so \(\infty\) is the only point mapping to \(\Omega\)); it is surjective because every point of \(\overline{S}_\Omega = S_\Omega \cup \{\Omega\}\) is hit, the points of \(S_\Omega\) by themselves and \(\Omega\) by \(\infty\).
4. \(f\) is a homeomorphism: Proof:
4.1. For \(x \in S_\Omega\), a local neighborhood basis at \(x\) in \(S_\Omega^*\) consists of open intervals \((a, b) \subseteq S_\Omega\) containing \(x\).In \(\overline{S}_\Omega\), open intervals \((a, b)\) with \(b < \Omega\) form the exact same local neighborhood basis at \(x\).
4.2. A neighborhood basis of \(\infty\) in \(S_\Omega^*\) consists of sets of the form \((S_\Omega \setminus [0, \alpha]) \cup \{\infty\} = (\alpha, \Omega) \cup \{\infty\}\) for \(\alpha < \Omega\) (by <1>2).
4.3. Under \(f\), the image of \((\alpha, \Omega) \cup \{\infty\}\) is \((\alpha, \Omega) \cup \{\Omega\} = (\alpha, \Omega]\), which is precisely the standard basic open neighborhood of \(\Omega\) in \(\overline{S}_\Omega\).
4.4. Therefore \(f\) is an open continuous bijection, so \(f\) is a homeomorphism.Q.E.D.