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The problem: If $f$ is defined on $E$, the graph of $f$ is the set of points $(x,f(x))$, for $x \in E$. In particular, if $E$ is a set of real numbers, and $f$ is real-valued, the graph of $f$ is a subset of the plane. Suppose $E$ is compact, and prove that $f$ is continuous on $E$ if and only if its graph is compact.

I'm struggling with a couple things with this problem:

  1. I'm having trouble understanding the graph definition. As I understand it, the graph is a subset of the set $E \times f(E)$?

Below is my proof for the if direction.

(if direction) Let the graph of $f$ be notated by $G(f)$.

Let $\phi_1: G(f) \rightarrow E$ be defined as $\phi_1((x,f(x))) = x$ for $(x,f(x)) \in G(f).$ Also let $\phi_2: G(f) \rightarrow f(E)$ be defined as $\phi_2((x,f(x))) = f(x)$ for $(x,f(x)) \in G(f)$.

Let $\{G_{\alpha}\}_{\alpha \in A}$ be an open cover of $G(f)$. Because $f$ is continuous and $E$ is compact, $f(E)$ is also compact.

Hence there exists $M \in \mathbb{N}$ such that $E \subset \bigcup\limits_{i = 1}^{M} \phi_1(G_{\alpha_i}),$ where $G_{\alpha_i} \in \{G_{\alpha}\}_{\alpha \in A}$ for $i = 1,2,..., m$. There also exists $N \in \mathbb{N}$ such that $E \subset \bigcup\limits_{i = 1}^{N} \phi_2(G_{\beta_i}),$ where $G_{\beta_i} \in \{G_{\alpha}\}_{\alpha \in A}$ for $i = 1,2,..., n$.

From this it follows that $G(f) \subset \left(\bigcup\limits_{i = 1}^{M} G_{\alpha_i}\right)\cup \left(\bigcup\limits_{i = 1}^{N} G_{\beta_i}\right),$ so we have a finite subcover of the open cover above that contains $G(f)$.

I'm completely sure that my proof is non-rigorous, as I need to prove that the $\phi$ functions allow for the open cover defined to be open covers for $E$ and $f(E)$. Also the leap from the finite subcovers of $E$ and $f(E)$ to a finite subcover of $G(f)$ seems unclear. I would really appreciate if someone could look over my misunderstandings and proofs! Thank you.

alexion
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1 Answers1

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The question about producing open covers of $E$ and $f(E)$ using $\phi_1$ and $\phi_2$ can be taken care of; $\phi_1$ and $\phi_2$ are projection maps, and projection maps are always open. This means that $\phi_1(G_{\alpha})$ and $\phi_2(G_{\alpha})$ are open for each open set $G_{\alpha}$. You can also check that the sets $\{\phi_1(G_{\alpha})\}$ cover $E$ (and similarly for $f(E)$).

However, going from the finite subcovers of $E$ and $f(E)$ to a finite subcover of $G(f)$ doesn't work. Here's an example of what goes wrong. Let $E = [0, 3] \subseteq \mathbb{R}$ and $f(x) = x$. Suppose that $$ G_1 = (-1, 4) \times (-1, 1) \qquad\text{and}\qquad G_2 = (2, 4) \times (-1, 4) $$ are two of the subsets in your open cover of $G(f)$. Now $\phi_1(G_1)$ covers $E$ and $\phi_2(G_2)$ covers $f(E)$, but $G_1 \cup G_2$ doesn't cover $G(f)$; the point (1.5, 1.5) isn't in either $G_1$ or $G_2$.

The good news is that there's a quick way to prove that $G(f)$ is compact; it's the image of the compact set $E$ under the continuous map $x \mapsto (x, f(x))$.

Justin Barhite
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  • Is my proof attempt a deadend or is there a way to continue? Thank you for your helpful response, by the way! – alexion Dec 02 '19 at 12:19
  • You're welcome! I can't definitively promise there isn't a way to modify your idea to make it work, but it feels like a dead end to me, because of the example I described; open sets in $E \times \mathbb{R}$ whose projections cover $E$ and $f(E)$ might not cover the graph of $f$. – Justin Barhite Dec 02 '19 at 12:31