Let \(Y=(0,1)\times(0,1)\) and let \(X=Y\cup\{*\}\). Let \(\mathcal T_1:=\{U\mid U\text{ is a Euclidean open subset of }Y\}\), \(\mathcal T_2:=\{X-C\mid C\text{ is a compact subset of the Euclidean space }Y\}\), and let \(\mathcal T:=\mathcal T_1\cup\mathcal T_2\). Then \((X,\mathcal T)\) is a topological space called the one-point compactification of \((0,1)\times(0,1)\) (you do not need to prove this). Show that the space \((X,\mathcal T)\) is compact.
Solution
Let \(\mathcal U\) be an arbitrary open cover of \(X\).
1. Some member \(U_0\in\mathcal U\) has the form\[ U_0=X\setminus C \]for a compact subset \(C\subseteq Y\).
The cover \(\mathcal U\) contains a member \(U_0\) with \(*\in U_0\). No member of \(\mathcal T_1\) contains \(*\), so \(U_0\in\mathcal T_2\). By definition of \(\mathcal T_2\), there is a compact set \(C\subseteq Y\) such that \[ U_0=X\setminus C. \]
2. For every \(U\in\mathcal U\), the intersection \(U\cap Y\) is open in the Euclidean topology on \(Y\).
If \(U\in\mathcal T_1\), then \(U\subseteq Y\) is Euclidean open, so \(U\cap Y=U\) is open in \(Y\). If \(U\in\mathcal T_2\), write \[ U=X\setminus K \] with \(K\subseteq Y\) compact. The Euclidean space \(Y\) is Hausdorff, so its compact subset \(K\) is closed in \(Y\). Hence \[ U\cap Y=Y\setminus K \] is open in \(Y\).
3. Finitely many members of \(\mathcal U\) cover the compact set \(C\).
2. Compactness of \(C\) gives members\[ U_1,\ldots,U_m\in\mathcal U \]such that\[ C\subseteq (U_1\cap Y)\cup\cdots\cup(U_m\cap Y) \subseteq U_1\cup\cdots\cup U_m. \]
4. The finite family\[ U_0,U_1,\ldots,U_m \]covers \(X\).
By 1, \[ X=U_0\cup C. \] By 3, \(C\subseteq U_1\cup\cdots\cup U_m\). Therefore \[ X\subseteq U_0\cup U_1\cup\cdots\cup U_m. \] Thus every open cover of \(X\) has a finite subcover, so \(X\) is compact.