Exercise 6, chapter 4 Rudin's "Principles of Mathematical Analysis":
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 rea-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.
The following is my solution.
Let $g(x)={(x,f(x)):x\in E}$. For every $\epsilon >0$ there exists $\delta > 0$ such that $d(g(x), g(p))=\sqrt{(x-p)^2+(f(x)-f(p))^2} < \epsilon$ for all points $x \in E$ for which $|x-p|<min(\frac{\epsilon}{\sqrt{2}}, \delta)$. For all points $x \in E$ for which $|x-p|<\delta$, we have $|f(x)-f(p)|<\frac{\epsilon}{\sqrt{2}}$. We can surely find such $\delta$ since $f$ is continuous. We can conclude that $g$ is continuous at $p$, and therefore $g(E)$, the graph is compact.
I would like to ask two questions:
- Is my solution to the forward part correct?
- How to prove the inverse part?
Thank you in advance.
Best Answer
Your proof in the forward direction looks fine. For the reverse, suppose that $E$ is compact but $f$ is not continuous. Then we can find a sequence $\{x_n,f(x_n)\}$ so that $\{x_n\}$ converges to $x$ but $\{f(x_n)\}$ does not converge to $f(x)$. By compactness of $E$ we extract a convergent subsequence $\{x_{n_k}, f(x_{n_k})\}$. We know that $f(x_{n_k})$ does not converge to $f(x)$, say it converges to some other value $y$. Then the point $(x,y)$ is the limit point of $\{x_{n_k}, f(x_{n_k})\}$ but is also not contained in $E$ a contradiction.