Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 4, Problem 6

Problem

If $f$ is defined on $E$, the graph of $f$ is the set of points $(x,f(x))$, for $x \in E$. In particular, if $E$ is a set of real numbers, and $f$ is real-valued, the graph of $f$ is a subset of the plane.

Suppose $E$ is compact, and prove that $f$ is continuous on $E$ if and only if its graph is compact.

Answer

$\Rightarrow$: Consider a map $\phi(x):=(x,f(x))$. Since $\phi$ is continuous and $E$ is compact, the graph $\phi(E)$ is compact.

$\Leftarrow$: Suppose $f$ is not continuous on $E$. Then for any sequence $x_n \rightarrow x$, $f(x_n) \not\rightarrow f(x)$. Since $\phi(E)$ is compact, the sequence $(x_n,f(x_n)) \in \phi(E)$ has a convergent subsequence $(x_{n_k},f(x_{n_k})) \rightarrow (p,q) \in \phi(E)$. Since $x_n \rightarrow x$, $x_{n_k} \rightarrow x = p$. Furthermore, $q=f(p)$ because $(p,q)$ is on the graph. Hence, $f(x_{n_k}) \rightarrow f(x)$, which is a contradiction.

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