Let \(f:\mathbb R\to\mathbb R\) be bounded. Suppose the graph of \(f\) is a closed subset of \(\mathbb R^2\). Prove that \(f\) is continuous.
Solution
1. If \(x_n\to x\) in \(\RR\), then \(f(x_n)\to f(x)\).
Suppose instead that \(f(x_n)\) does not converge to \(f(x)\). Then there are \(\varepsilon>0\) and a subsequence \((x_{n_k})\) such that \[ \abs{f(x_{n_k})-f(x)}\geq\varepsilon \] for every \(k\). Since \(f\) is bounded, the real sequence \((f(x_{n_k}))\) is bounded. By the Bolzano–Weierstrass theorem, it has a convergent subsequence; write \[ f(x_{n_{k_j}})\longrightarrow y. \] Also \(x_{n_{k_j}}\to x\), so \[ \bigl(x_{n_{k_j}},f(x_{n_{k_j}})\bigr)\longrightarrow(x,y). \] Every point on the left lies in the graph of \(f\). Since the graph is closed, \((x,y)\) also lies in it, and therefore \(y=f(x)\). But passing to the limit in \[ \abs{f(x_{n_{k_j}})-f(x)}\geq\varepsilon \] gives \(\abs{y-f(x)}\geq\varepsilon\), a contradiction.
2. The function \(f\) is continuous on \(\RR\).
Step 1 shows that for every \(x\in\RR\) and every sequence \(x_n\to x\), one has \(f(x_n)\to f(x)\). By the sequential criterion for continuity on \(\RR\), \(f\) is continuous at every \(x\).
3. Q.E.D.
Step 2 is the required conclusion.