Vtyjkyuk

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 $g(x)=(x,f(x)):xin E$. 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 $|x-p|<min(fracepsilonsqrt2, delta)$. For all points $x in E$ for which $|x-p|<delta$, we have $|f(x)-f(p)|<fracepsilonsqrt2$. 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:

1. Is my solution to the forward part correct?


2. How to prove the inverse part?

Thank you in advance.

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.

Try to add details to the answers of @Sheel Stueber.

For the reverse, assume that $E$ is compact but $f$ is not continuous at $x_0 in E$. Therefore, $x_0$ is a limit point of $E$ (every function is continuous at isolated point). We can find a sequence $p_n$ in $E$ such that $lim_n to infty p_n = x_0$ (Theorem 3.2(d)).

Construct a sequence $p_n, f(p_n) in g(E)$, (refer to the question). Because $g(x)=(x, f(x))$ is compact, by Theorem 3.6(a), we can find a subsequence $p_n_k, f(p_n_k)$ converges to $(x_1, y_1) in g(E)$. By Theorem 3.2(a), $(x_1, y_1)$ is a limit point of $g(E)$.

Obviously, $p_n_k$ is a subsequence of $p_n$, and $lim_n to infty p_n = x_0$, so $lim_n_k to infty p_n_k = x_0$. In other words, $x_0=x_1$ and $(x_0, y_1)$ is a limit point of $g(E)$.

Furthermore, because $g(E) in R^2$ is compact, by Theorem 2.41, $g(E)$ is closed. Then the limit point $(x_0, y_1) in g(E)$. By definition of function, $y_1=f(x_0)$.

So far, we know $(x_0, f(x_0))$ is a limit point of $g(E)$, i.e., for any $epsilon > 0$, there is a neighborhood $N_epsilon (x_0, f(x_0))$ such that $N_epsilon^* (x_0, f(x_0)) cap g(E) neq emptyset$. In other words, for any $epsilon > 0$, for all $x in g^-1(N_epsilon (x_0, f(x_0)) cap g(E))$, we have $|f(x)-f(x_0)|<epsilon$ (since $(x-x_0)^2+(f(x)-f(x_0))^2< epsilon ^2$). We can conclude $f$ is continuous at $x_0$ which contradicts our assumption.