# $f$ is continuous on $E$ if and only if its graph is compact.

This question may be asked before under different formulation, the original problem is Chapter 4, Exercise 7 of Rudin's text: The Principles of Mathematical Analysis:

Problem: If $f$ is defined on $E$, the graph is the set of points $(x, f(x))$, for $x \in E$.

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

My Attempt: Let $\Gamma(f) = \{(x, f(x)): x \in E\} \subset E \times \mathbb{R}^1$. I was trying to define a function $F: E \rightarrow E \times \mathbb{R}^1$ by $F(x) = (x, f(x))$ and show $F$ is continuous on $E$. Since $E$ is compact, it follows that $F(E) = \Gamma(f)$ is compact.

For the converse, the earlier thread A real function on a compact set is continuous if and only if its graph is compact stated that the projection function $\pi$ is continuous on $E \times \mathbb{R}^1$.

My question is: without knowing any specific metric on $E$ or on $E \times \mathbb{R}^1$, it looks hard to me to show that $F$ and $\pi$ aforementioned are continuous. Since the condition doesn't give any information about what I concern, how should I proceed?

• We can get a metric on this product by defining an induced metric on the product of metric spaces. For example, given $(X, d_X)$ and $(Y,d_Y)$, it is common to define the induced metric on $X \times Y$ by $$d_{X \times Y}((x_1,y_1),(x_2,y_2)) = d_X(x_1,x_2) + d_Y(y_1,y_2)$$ Jul 6, 2015 at 22:47
• @Omnomnomnom Thank you for your reply. I knew we might can do so, but I am actually trying to avoid such "subjective" proof. I am wondering is there any "objective" proof that without imposing any metric by ourselves? I think the existence of $d_X$ is fine, the annoying part is can we proceed without giving any specific metric form on $d_{X \times \mathbb{R}^1}$? Jul 6, 2015 at 23:02
• the problem is that you have to say what it means for $\Gamma(f)$ to be compact, for which we need some kind of topology on $E \times \Bbb R$. If you're not going to get this topology using a "subjective" choice of metric on the product, then perhaps you should do so using the definition of the product topology, ignoring the underlying metrics altogether. Jul 6, 2015 at 23:06
• @Zhanxiong the metric .@Omnomnomnom defined induces the product topology, as there are only finitely many factors. So it isn't "subjective". Whatever that in this context might mean. Jul 6, 2015 at 23:13
• So literally, we can use any metric on $E \times \mathbb{R}^1$, such as $d_{E \times \mathbb{R}^1}((x_1, y_1), (x_2, y_2)) = \max(d_E(x_1, x_2), |y_1 - y_2|)$, right? What I mean by "subjective" is like this --- the proof details depend on the specification of the metric on $E \times \mathbb{R}^1$, which I understand is acceptable but not that perfect... Jul 7, 2015 at 1:20

Suppose that $f$ is continuous, and define $F(x):=(x,f(x))\subset E\times\mathbb R$. Fix $x\in E$ and consider a sequence $(x_n)\subset E$ converging to $x$. Since $f$ is continuous, $f(x_n)\to f(x)$ and hence $F(x_n)=(x_n,f(x_n))\to(x,f(x))=F(x)$, which shows that $F$ is continuous. Now your argument works fine, since now $\Gamma(f)=F(E)$ is compact.
For the converse assume that $\Gamma(f)=F(E)$ is compact. Fix $x\in E$ and a sequence $(x_n)\subset E$ converging to $x$. Assume that $(x_n,f(x_n))$ does not converge to $(x,f(x))$. Since $\Gamma(f)$ is compact, there is a subsequence $(x_{n_k},f(x_{n_k})$ that converges in $\Gamma(f)$ to some $(x',y)\neq(x,f(x))$. But since $(x_n)$ converges to $x$, then so does $(x_{n_k})$ (because it's a subsequence), which implies $x'=x$. But since $(x',y)\in\Gamma(f)$, we must have $y=f(x')=f(x)$ and consequently $(x',y)=(x,f(x))$, a contradiction. This shows that $(x_n,f(x_n))\to(x,f(x))$; in particular, $f(x_n)\to f(x)$, which proves that $f$ is continuous at $x$ (and that $F$ is continuous at $x$ as well).
• Does this generalize ? If $E$ is any compact space and $D$ is any space, is it true that $f:E\to D$ is continuous iff the graph of $f$ is compact?......+1 Jan 9, 2019 at 20:55