If $f$ is continuous then $G_f \subset M \times \mathbb{R}$ is closed.

Question

Consider a function $f: M \rightarrow \mathbb{R}.$ The graph of $f$ is the set $$G_f := \{(x,y) \in M \times \mathbb{R} : y = fx\}.$$

Prove that if $f$ is continuous then $G_f \subset M \times \mathbb{R}$ is closed.

It follows from the compactness of $M$ and the continuity of $f$ that $f(M)$ is compact. Notice that $G_f = M \times f(M)$ is the cartesian product of two compact sets and is therefore compact. All compact sets are bounded and closed.

Short and sweet, or incorrect?

$G_f$ is not a Cartesian product of two compact sets but a (closed) subset of such a product. — Henno Brandsma, Oct 16 '18 at 04:40
Is it not true that a closed subset of a compact set is also compact? — Rafael Vergnaud, Oct 16 '18 at 04:41
You need the compactness to see that a closed graph implies continuity. Not for this direction. — Henno Brandsma, Oct 16 '18 at 04:44
@RafaelVergnaud Is $M$ a metric space? A topological space? A subspace of $\mathbb{R}$? — qualcuno, Oct 16 '18 at 04:44
The book doesn't specify. However, the book is an introductory book to analysis. I think it discusses topology only in the context of metric spaces. Moreover, the author does always refer to a metric space as $M$ or $N.$ However, again, it doesn't specify. The problem is problem 44 in Pugh's Real Mathematical Analysis. — Rafael Vergnaud, Oct 16 '18 at 04:45
https://math.stackexchange.com/questions/45227/what-conditions-are-required-on-the-domain-for-closed-graph-theorem-of-topologic?noredirect=1&lq=1 — Clemens Bartholdy, Sep 02 '24 at 08:42

qualcuno · Accepted Answer · 2018-10-16T04:56:24.260

It is not true that $G_f = M \times f(M)$. Take for example the identity on $\mathbb{R}$, in which case

$$ G_{id} = \Delta = \{(x,x) : x \in \mathbb{R}\} \neq \mathbb{R} \times \mathbb{R} = M \times id(M). $$

Hint: a set is closed if limits of nets (or sequences if your space is metric) on the space that converge, are elements of the space itself. Take an arbitrary convergent net (or sequence) on $G_f$. Use that projections are continuous (and the equivalence of continuity that is related to sequences/nets). Conclude that the limit point are elements of $G_f$.

Since you mention that there is no treatment of topological spaces which aren't metric, here's a solution for the case of metric spaces:

Take $\{(x_n,f(x_n))\}_n$ a sequence on $G_f$ that converges to some element $(s,t)$. By continuity of $\pi_1$, we have that $x_n \to s$. By continuity of $f$, we have $f(x_n) \to f(s)$. Since we also have $f(x_n) \to t$, necessarily $t = f(s)$ and so $(s,t) = (s,f(s)) \in G_f$.

@RafaelVergnaud no problem! I've added a solution for metric spaces, just in case that you want to check your approach. — qualcuno, Oct 16 '18 at 04:51

If $f$ is continuous then $G_f \subset M \times \mathbb{R}$ is closed.

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