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Exercise 6, chapter 4 Rudin's "Principles of Mathematical Analysis":

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 that the question is flawed. Show that the graph is compact as a subset of which metric space? Under what metric? The compactness of a set can depend on a metric as explained here: Does compactness depend on the metric?

For the "only" if direction, I defined a new metric space $X \times Y$ where $Y$ is the codomain of $f$ with metric $d_{X \times Y}=d_X+d_Y$. I was able to show that in this new metric space, the graph is indeed compact. But,I feel as if this answers a different question.

Moreover, what would I even do for the "if" direction? I start by supposing that the graph is compact, but what does this even mean? There is no defined metric and no given underlying metric space for which the graph is a compact subset.

Arctic Char
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user1923
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  • The graph in this setting is compact in the product space. Topologically, there's only one product space (for finitely many components). Metric-wise there are multiple, but they are all topologically equivalent. The metric that you described induces the product topology. Because compactness is purely topological, which metric you have on the product space does not matter, as long as it induces the product topology. – Ian Feb 26 '23 at 22:41
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    Compactness a purely topological notion. It requires no metric. – Anne Bauval Feb 26 '23 at 22:42
  • @Ian I appreciate the response. But this is above the level of the text. The terms "topologically equivalent" and "product space" are absent from the classic baby rudin text. Although I am sure you answer my question with this comment, it does not help me in my current situation: working through baby rudin. – user1923 Feb 26 '23 at 22:44
  • @AnneBauval So is the answer to the question linked in my post wrong? Moreover, I don't understand: whether a set is open or not certainly depends on the metric and so what constitutes an open cover must also depend on the metric. – user1923 Feb 26 '23 at 22:45
  • How far into topology did they get at this stage? Not everything I said is really necessary to do this one problem. – Ian Feb 26 '23 at 22:46
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    @peanut The linked answer is correct, but the point is that "compactness depends on the metric" is correct but can be misleading, because you can change the metric without changing which sets are compact. It is more correct to say "compactness depends on the topology". Thinking in terms of "compactness depends on the metric" makes you wonder about whether, say, $d_{X \times Y}((x_1,y_1),(x_2,y_2))=\sqrt{ d_X(x_1,x_2)^2+d_Y(y_1,y_2)^2}$ has the same compact sets as the $d_{X \times Y}$ you wrote in your question. – Ian Feb 26 '23 at 22:49
  • @Ian Yes, compactness depends on the topology. But Rudin does not introduce the concept of a topological space, he only works with metric spaces. In particular, he only defines compactness for subsets of metric spaces. – Paul Frost Feb 26 '23 at 22:59

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Rudin writes

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.

This makes clear that Rudin considers a special situation: One has $E \subset \mathbb R$ and $graph(f) \subset \mathbb R^2$. It is now clear that $graph(f)$ is a subset of $\mathbb R^2$ with its standard metric.

Paul Frost
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  • Rudin does not make it clear that we are considering a special situation. His comment is nothing more than an example elucidating the concept of a graph in $\mathbb R^2$. – user1923 Feb 26 '23 at 22:53
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    I think you're right, but I would still say that needing to assume that $E$ is a subset of $\mathbb{R}$ and $f$ is real-valued based on that second sentence is...not exactly ideal. But I don't have the context of what came before this in the book. – Ian Feb 26 '23 at 22:53
  • @Ian 's comment is what I was trying to say. – user1923 Feb 26 '23 at 22:55
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    In his first sentence Rudin does not even say what the codomain of $f$ is. Is it a metric space? Or $\mathbb R$? This suggests to me that he considers the special situation described in the second sentence. But you are right, we have scope for interpretation. – Paul Frost Feb 26 '23 at 23:15
  • suggesting soemthing and makeing something clear are two completely different things – user1923 Feb 26 '23 at 23:17
  • What is this discussion? Clearly the second sentence is meant to be part of the question. – Funktorality Feb 26 '23 at 23:35