8
$\begingroup$

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?

$\endgroup$
9
  • 1
    $\begingroup$ 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 $$ d_{X \times Y}((x_1,y_1),(x_2,y_2)) = d_X(x_1,x_2) + d_Y(y_1,y_2) $$ $\endgroup$ Jul 6, 2015 at 22:47
  • $\begingroup$ @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 $d_{X \times \mathbb{R}^1}$? $\endgroup$
    – Zhanxiong
    Jul 6, 2015 at 23:02
  • $\begingroup$ 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. $\endgroup$ Jul 6, 2015 at 23:06
  • $\begingroup$ @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. $\endgroup$
    – user251257
    Jul 6, 2015 at 23:13
  • $\begingroup$ 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... $\endgroup$
    – Zhanxiong
    Jul 7, 2015 at 1:20

1 Answer 1

11
$\begingroup$

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).

$\endgroup$
1
  • $\begingroup$ 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 $\endgroup$ Jan 9, 2019 at 20:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.