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Rudin_ex6

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!

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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 $E$).

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

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