I have seen answers to this question, which go beyond my understanding of compactness and continuity. I was wondering whether we can cook up a proof using sequential compactness and certain equivalent definitions of continuity such as the inverse image of any closed set is closed.
Here is what I have been able to conjure up so far.
Assume that the graph of $f$ is compact. This means that it is also closed and bounded. The graph is a closed and bounded subset of $A \times f(A)$. All we need to show is that $f(A)$ is compact, and we are are home free, right? (since continuous functions take compact sets to compact sets).
Question is: how do we show that $f(A)$ using the fact that the graph is compact. Can we claim that $f(A)$ is closed and bounded (since by Heine-Borel, any closed and bounded subset of $\mathbb R$ is compact)?
I feel like I am really close. Can anyone help me out?
Best Answer
Corrected 2 December 2021.
Suppose that $f$ is not continuous. Then there are a point $x\in A$, an $\epsilon>0$, and a sequence $\langle x_n:n\in\Bbb N\rangle$ in $A$ converging to $x$ such that $\left|f(x_n)-f(x)\right|\ge\epsilon$. (Why?) Let $G$ be the graph of $f$. Then $\big\langle\langle x_n,f(x_n)\rangle:n\in\Bbb N\big\rangle$ is a sequence in the compact metric space $G$, so it has a convergent subsequence $\big\langle\langle x_{n_k},f(x_{n_k})\rangle:k\in\Bbb N\big\rangle$.
Show that the limit of this subsequence must be of the form $\langle x,\alpha\rangle$ for some $\alpha\in\Bbb R$. (Recall that $x$ is the limit of $\langle x_n:n\in\Bbb N\rangle$.)
Show that $\alpha\ne f(x)$. Conclude that $\langle x,\alpha\rangle\notin G$.
This contradicts the compactness of $G$; how?