# Baby Rudin ch4 ex6: $f$ is continuous on $E$, which is compact, if and only if its graph is compact. I will explain how I solved the problem and will appreciate it if you check if my solution is right.

(if $$f$$ is continuous on $$E$$, its graph is compact)

Let $$\{A_\alpha\}$$ be an open cover of the graph. For each $$x \in E$$, $$\exists \alpha_x$$ such that $$(x, f(x)) \in A_{\alpha_x}$$.

Since $$A_\alpha$$ is open, $$\exists \epsilon_x$$ such that an open ball $$B((x, f(x)), \epsilon_x) \subset A_{\alpha_x}$$

Because $$f$$ is continuous at x, for the given $$\epsilon_x$$, $$\exists \delta_x>0$$ such that if $$d(y,x)<\delta_x$$, $$d(f(x), f(y))<\epsilon_x$$. Let $$r_x = min(\delta_x, \epsilon_x)$$. Then, $$\forall y \in B(x, r_x)$$, $$f(y)\in B(f(x),r_x)$$; therefore, $$(y, f(y)) \in B((x, f(x)), r_x) \subset B((x, f(x)), \epsilon_x)$$. Thus, $$\{B(x, r_x)\}$$ is an open cover of $$E$$.

Because $$E$$ is compact, we can pick $$x_1, ... x_n$$ such that $$U^n_{k=1}B(x_k, r_k)=E$$. Therefore, the graph $$\subset U^n_{k=1}B((x_k, f(x_k)), \epsilon_x) \subset U^n_{k=1} A_{\alpha_{x_k}}$$.

(if the graph of $$f$$ is compact, $$f$$ is continuous on $$E$$)

Pick $$x \in E$$ and an open cover of the graph $$\{B_\alpha((x, f(x)), \epsilon_x) \}$$. Because the graph is compact, the graph $$\subset U^n_{\alpha = 1} B_\alpha((x, f(x)), \epsilon_x)$$. If $$(y, f(y)) \in B((x, f(x)), \epsilon_x)$$, $$y \in B(x, \epsilon_x)$$ and $$f(y) \in B(f(x), \epsilon_x)$$.

Choose an open cover of $$E$$, and because $$E$$ is compact, $$U^m_{\beta=1} B_\beta (x, \delta_x)$$. Pick $$r_x = min(\epsilon_x, \delta_x)$$. If $$y \in B(x, r_x), f(y) \in B(f(x), \epsilon_x)$$. Thus, $$f$$ is continuous.

Thank you in advance!

The proof of "$$f$$ continuous then $$\Gamma(f)$$ (the graph of $$f$$) compact" has a right idea, but could be much more easily proved, if you use the fact that the continuous image of a compact space is compact, a thing you're reproving in part for this special case, it seems to me.

The reverse direction is not correctly shown: for your approach you should start with a given $$\epsilon>0$$ for $$x$$ and $$f(x)$$, and find a correct $$\delta$$ for that $$\epsilon$$. But easier is to note that $$f^{-1}[C] = \pi_1[\Gamma(f) \cap (X \times C)]$$ for any closed $$C$$ in the codomain (and where $$\pi_1$$ is the first projection onto $$E$$).

A different approach. Tools: (1): A continuous image of a compact space (set) is compact. (2). A subset of $$\Bbb R$$ or of $$\Bbb R^2$$ is compact iff it is closed and bounded.

(I). If $$f$$ is continuous: Then $$f''E=\{f(x):x\in E\}$$ is compact. So $$\Gamma(f)=\{(x,f(x):x\in E\}$$ is bounded because it is a subset of the bounded set $$E\times f''E.$$

Now if $$(x_n,f(x_n))$$ is a sequence in $$\Gamma(f)$$ converging to $$(x,y)\in \Bbb R^2$$ then $$x=\lim_{n\to \infty} x_n\in E$$ because $$E$$ is closed. By the continuity of $$f$$, if $$(x_n)_n$$ is a sequence in $$E$$ converging to $$x\in E$$ then $$(f(x_n))_n$$ converges to $$f(x)$$. So $$(x,y)=(x,f(x))\in \Gamma(f).$$ So $$\Gamma(f)$$ is closed.

So if $$f$$ is continuous then $$\Gamma(f)$$ is bounded and closed, hence compact.

(II). If $$f$$ is not continuous: Then there exists a sequence $$(x_n)_n$$ in $$E$$ converging to some $$x\in E,$$ such that for some $$r>0$$ we have $$|f(x_n)-f(x)|>r$$ for all $$n.$$ Now we have

(i) If $$(f(x_n)_n$$ is unbounded then $$\Gamma(f)$$ is unbounded, so $$\Gamma(f)$$ is not compact.

(ii). If $$(f(x_n)_n$$ is bounded then there is a subsequence $$(f(x_{j_n}))_n$$ converging to some $$y\in \Bbb R.$$ Now $$(x_{j_n})_n$$ converges to $$x,$$ so $$((x_{j_n},f(x_{j_n}))_n$$ converges to $$(x,y).$$

But $$y=\lim_{n\to \infty}f(x_{j_n})\ne f(x)$$ because $$|f(x_{j_n})-f(x)|\geq r$$ for all $$n.$$ So $$(x,y)\in \overline {\Gamma(f)}\setminus \Gamma(f),$$ so $$\Gamma(f)$$ is not closed, hence not compact.

Remark: For a function $$f$$ and a set $$S\subset dom (f)$$ the notation $$f''S$$ (read "$$f$$-double-prime-$$S$$'') denotes $$\{f(x):x\in S\}.$$