Show that for \(X\) an arbitrary topological space, the one-point (Alexandroff) compactification \(X^* = X \cup \{\infty\}\) is compact.
Prove that every path-connected topological space is connected.
Solution
1. Let \(X^*=X\cup\{\infty\}\) carry the Alexandroff topology: the open sets not containing \(\infty\) are the open subsets of \(X\), while a neighborhood of \(\infty\) has the form\[ \{\infty\}\cup(X\setminus K) \]with \(K\subseteq X\) compact and closed.
2. The space \(X^*\) is compact.
Let \(\mathcal U\) be an open cover of \(X^*\). Choose \(U_\infty\in\mathcal U\) containing \(\infty\). Then for some compact closed \(K\subseteq X\), \[ \{\infty\}\cup(X\setminus K)\subseteq U_\infty. \] The remaining points lie in \(K\). The sets \(U\cap X\), for \(U\in\mathcal U\), cover \(K\), so compactness of \(K\) gives finitely many of them covering \(K\). Together with \(U_\infty\) they form a finite subcover of \(X^*\).
3. Every path-connected space is connected.
Fix \(x_0\in X\). For each \(x\in X\), choose a path from \(x_0\) to \(x\). Each path image is connected, all such images contain \(x_0\), and their union is \(X\). A union of connected subsets with a common point is connected.