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 $\{f(x):x\in S\}.$


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.