Exercise 4.6

Question

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Exercise 6:
    If $f$ is defined on $E$, the {\it graph} of $f$ is the set of points $\bigl(x,f(x)\bigr)$, for $x\in E$. Suppose $E$ is compact, and prove that $f$ is continuous on $E$
    if and only if its graph is compact.

ghostofgarborg · Accepted Answer

Assume $f : X 	o Y$ is continuous. Then the map $\Gamma(x) = (x, f(x))$ 
    is also continuous by thm. 4.10a. Since $E$ is compact, so is $\Gamma(E)$, which is the graph of $F$.

If $\Gamma(E)$ is compact, let $V$ be a closed subset of $Y$. The set 
    $V^\prime = (X 	imes V) \cap \Gamma(E)$ is closed in $\Gamma(E)$, hence compact. 
    The projection $ \pi : X 	imes Y 	o X$ is continuous, so 
    $ f^{-1}(V) = \pi(V^\prime) $ is compact, hence closed 
    (since $X$ is a metric space and therefore Hausdorff). This makes $f$ continuous.

Amit Awasthi · Answer

\begin{proof} First prove that the continuity of $f$ implies the graph of $f$ is compact. Let $g(x) = (x,f(x))$. $g(x)$ is continuous since given $\epsilon > 0$ and $x \in E$, there exists $\delta > 0$ such that $d(x,y) < \delta$ implies $d(f(x),f(y)) < \epsilon / 2$ since $f(x)$ is continuous. In particular let $d(x,y) < \mathrm{min}(\delta,\epsilon/2)$. Then, \begin{align*} &d(g(x),g(y))\ &\quad=((x,f(x),(y,f(y)))\ &\quad\quad\le d((x,f(x)),(x,f(y)) + d((x,f(x)),(y,f(x))\ &\quad\quad\quad\quad= d(f(x),f(y)) + d(x,y)\ &\quad\quad\quad\quad\quad< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \end{align*} Thus, $g(x)$ is continuous. Since $E$ is compact and $g(x)$ is continuous, $g(E)$, the graph of $f$, is compact. \par Now prove that the compactness of the graph of $f$ implies the continuity of $f$ on $E$. Let $g(x) = (x,f(x))$. We can construct a sequence $\{x_n\}$ where $x_n \in E$ for all $n \in \mathbb{N}$. We can also construct a sequence $\{(x_n,f(x_n))\}$ where $(x_n,f(x_n)) \in g(E)$ for all $n \in \mathbb{N}$. Now, without a loss generality, we can take subsequences of both $\{x_n\}$ and $\{x_n,f(x_n)\}$ such that both subsequences converge because $E$ and $g(E)$ are both compact. These subsequences converge to some point $x \in E$ and $(x,f(x)) \in g(E)$ respectively. Now we can consider only the second coordinate of $\{x_n,f(x_n)\}$, $\{f(x_n)\}$, which converges since this coordinate converges to point $f(x) \in f(E)$. Then, the continuity of $f(x)$ follows since for $\epsilon > 0$, we can find a $\delta > 0$ such that $|f(x_n) - f(x_m)| < \epsilon$ since $0 < |x_n - x_m| < \delta$ where $N=\mathrm{max}(N(\epsilon),N(\delta))$ and $n,m \ge N$. Thus, $f(x)$ is continuous. \end{proof}

Exercise 4.6

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Exercise 4.6

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