Let \(X\) be a metric space and let \((x_n)\) converge to \(x_0\in X\). Prove that \[ C=\{x_0,x_1,x_2,\ldots\} \] is compact.
Solution
1. Let \(\mathcal U\) be an arbitrary open cover of \(C\). There are\(U_0\in\mathcal U\) and an integer \(N\geq2\) such that\[ x_0\in U_0 \qquad\text{and}\qquad x_n\in U_0\quad\text{for every }n\geq N. \]
Because \(\mathcal U\) covers \(C\), there is \(U_0\in\mathcal U\) with \(x_0\in U_0\). Since \(U_0\) is open, there is \(\varepsilon>0\) such that \[ B(x_0,\varepsilon)\subseteq U_0. \] The convergence \(x_n\to x_0\) gives an integer \(N\geq2\) such that \[ n\geq N \quad\Longrightarrow\quad d(x_n,x_0)<\varepsilon. \] Hence every \(x_n\) with \(n\geq N\) lies in \(U_0\).
2. The finite set\[ \{x_1,\ldots,x_{N-1}\} \]is covered by finitely many members of \(\mathcal U\).
For each integer \(n\) with \(1\leq n<N\), choose \(U_n\in\mathcal U\) such that \(x_n\in U_n\). There are only finitely many such indices.
3. The family\[ \{U_0,U_1,\ldots,U_{N-1}\} \]is a finite subcover of \(C\).
1.Every \(x_n\) with \(1\leq n<N\) lies in \(U_n\) by step
2. Thus the displayedfinite family covers every point of \(C\).
4. Q.E.D.
The open cover \(\mathcal U\) was arbitrary, and step 3 gives it a finite subcover. Therefore \(C\) is compact.