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Proposition: let $f:X\to Y$ be a continuous function and $Y$ be Hausdorff. Show that the graph $$G_f= \{(x,f(x))\mid x \in X\}$$ is closed in $X\times Y$.

Proof:

I will show that the closure of the graph equals itself. To do this, take $(x,y)\in \text{Cl(}{G_f})$ and suppose $y \not =f(x)$. From this we will derive a contradiction, which implies that all limit points of $G_f$ are inside itself (since $y=f(x)$), hence implying it is closed.

Since $y$ and $f(x)$ are distinct in $Y$, there are neighborhoods $V_1$ and $V_2$ of respectively $y$ and $f(x)$ that are disjoint. And, since $f$ is continuous, there is a neighborhood $U$ of $x$ such that $f(U)\subseteq V_2$.

Now, the set $U\times V_1$ is an open set containing $(x,y)$ and since this is a limit point of $G_f$, every open set containing this point must intersect $G_f$. Suppose $(z,f(z)$ is such a point in the intersection. Since this means $z\in U$, then $f(z)\in f(U)\subseteq V_2$. But then $$(z,f(z))\in(U\times V_1)\cap(U\times V_2),$$ which is a contradiction with $V_1$ and $V_2$ being disjoint.

Clemens Bartholdy
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Stijn D'hondt
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  • Yes, the proof is fine. – Najib Idrissi Sep 01 '20 at 15:03
  • @CatharticEncephalopathy ...but that question is about the special case $\mathbb{R} \to \mathbb{R}$, which is less general? – Ben Steffan Sep 01 '24 at 12:52
  • https://math.stackexchange.com/questions/2488309/f-is-continuous-if-and-only-if-its-graph-is-closed-in-x-times-y?rq=1 There is also this dupe @BenSteffan – Clemens Bartholdy Sep 01 '24 at 16:09
  • @CatharticEncephalopathy Then why tag this (and the other dozen or so questions you did this to) as a duplicate of the wrong question? Now everything's a mess. – Ben Steffan Sep 01 '24 at 18:19
  • Well, the question would, either way not be upto the sites current standards. There are so many dupes even other than the one that I mentioned, that it doesn't matter that much. – Clemens Bartholdy Sep 01 '24 at 18:36

1 Answers1

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You have a typo in the second paragraph: you meant that $U$ is a nbhd of $x$, not of $f(x)$. Apart from that it’s fine. You don’t actually need to use contradiction, however: if $\langle x,y\rangle\in(X\times Y)\setminus G_f$, let $V_1$ and $V_2$ be disjoint open nbhds of $y$ and $f(x)$, respectively, and let $U$ be an open nbhd of $x$ such that $f[U]\subseteq V_2$. Then $(U\times V_1)\cap G_f=\varnothing$: if $\langle u,v\rangle\in U\times V_1$, then $\langle u,f(u)\rangle\in U\times V_2$, so $v\ne f(u)$, and $\langle u,v\rangle\notin G_f$.

Brian M. Scott
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