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The following is question 6 from page 99 of Walter Rudin's Principles Of Mathematical Analysis. I'm having trouble understanding what the metric of the graph might be (which, as far as I can tell, is not defined in the text or 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 think I've been able to prove the forward result. Suppose that E is compact. Rudin proves a theorem in the text that states the image of a compact metric space under a continuous function is also compact. Therefore, we know that $f(E)$ is compact. Now suppose that $\lbrace G_\alpha \rbrace, \alpha \in A$ is an open cover of the graph, where $G_i = B_i \times C_i$ and $B_i \in E, C_i \in f(E)$. Then $\lbrace A_\alpha \rbrace$ and $\lbrace B_\alpha \rbrace$ are open covers for E and $f(E)$, respectively. Because these sets are compact, their open covers contain finite subcovers, $\lbrace A^\prime_\beta \rbrace$ and $\lbrace B^\prime_\gamma \rbrace$, respectively. Thus, the set of all combinations of $(A^\prime_\beta, B^\prime_\gamma)$ forms a finite open subcover of the graph, proving that the graph is compact.

Actually, I'm really confused at this point, because it's just occured to me while typing the above that I cannot assume that each set $G_i$ can be represented as a set $\lbrace (x,y) \mid x \in A_i, y\in B_i \rbrace$ for open sets $A_i \subseteq E$ and $B_i \subseteq f(E)$.

So at this point, I'm not sure what to do, since I am unable to figure out what the distance metric might be in the metric space containing the graph. Is there a convention for this sort of problem? Did Rudin want the reader to only consider real-valued functions for f?

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Based on the words "the graph of $f$ is a subset of a plane," I would interpret this as asking about functions $\mathbb{R} \rightarrow \mathbb{R}$. The point in the proof you're struggling with is a general issue in showing that a product of compact spaces is compact. Since you get to be in a metric space, I'd recommend using the convergent subsequence definition of compactness instead. –  Kevin Carlson Aug 12 '12 at 2:23
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@KevinCarlson Thank you Kevin, that makes sense! If you wouldn't mind retyping your comment as an answer, I can mark it correct. –  Andrew Aug 12 '12 at 2:37
    
Andrew-that's done, glad it helped. –  Kevin Carlson Aug 12 '12 at 6:54

2 Answers 2

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Based on the words "the graph of f is a subset of a plane," I would interpret this as asking about functions $\mathbb{R}\rightarrow \mathbb{R}.$ The point in the proof you're struggling with is a general issue in showing that a product of compact spaces is compact. Since you get to be in a metric space, I'd recommend using the convergent subsequence definition of compactness instead.

I'll note, however, that you don't need the full strength of this result here, as Brian's answer shows.

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Here, $E$ could be any compact metric space.

Define the function $f_1: E \to E \times E$ by $f_1(x) = (x, f(x))$. Clearly this function is continuous, so $f_1(E)$ which is the graph is compact. Now assume that the graph is compact. We show that the inverse of image of every closed subset of of $E$ is closed in $E$. Define the maps $q_x, q_y$ by $(x, y) \to x$ and $(x, y) \to y$ respectively. If $U \subset E$ is closed, then $f^{-1}(U) = q_x(q_y^{-1}(U) \cap f_1(E))$. Both $q_x$ and $q_y$ are continuous, so $q_y^{-1}(U)$ is closed, so $q_y^{-1}(U) \cap f_1(E)$ is a closed subset of the compact set $f_1(E)$ and is therefore compact itself. By continuity of $q_x$, $f^{-1}(U)$ is compact in $E$, and since $E$ is Hausdorff (all metric spaces are), $f^{-1}(U)$ is closed.

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You do a lot of really cool stuff in this answer. However, I disagree with the line "Clearly this function is continuous ..." Let $d_{E \times E}:E \times E \to \mathbb{R}$ be the metric $d_{E \times E}(x,y) = 1$ if $x \neq y$ and $d_{E \times E}(x,y) = 0$ otherwise. Continuity of $f_1$ then fails to hold. –  Andrew Aug 12 '12 at 12:41
    
It will always be the case that $f_1$ is continuous. To see this, take $U \subset E \times E$ to be open. Then $U = \bigcup_{i \in A} V_i \times K_i$, where $V_i, K_i$ are open in $E$, so $f_1^{-1}(U) = \bigcup_{i \in A} f_1^{-1}(V_i \times K_i)$ = $\bigcup_{i \in A} V_i \cap f^{-1}(K_i)$, which is open. The metric on $E \times E$ depends on $E$, so if $E \times E$ is a discrete space, then $E$ will be discrete as well. In that case, the continuity of $f_1$ is even more obvious. –  Vectk Aug 12 '12 at 15:41
    
What do you mean when you say "The metric of $E \times E$ depends on E"? Can you explain the dependence? Also, why should it be possible to write that "$U = \bigcup_{i \in A} V_i \times K_i$ where $V_{i},K_{i}$ are open in $E$"? Does this follow from the axioms of a metric space? It's possible that there are some added axioms imposed for Cartesian product of generalized metric spaces. But at this point, I'm inclined to believe that the original intent of the question was to assume that $E \subseteq \mathbb{R}$. –  Andrew Aug 12 '12 at 17:04
    
I think the original question from the text was somewhat misworded. –  Andrew Aug 12 '12 at 17:11
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