$f:X\rightarrow Y$, with $Y$ compact, is continuous $\Rightarrow$ its graph is closed

Asked May 02 '19 at 18:15

Active May 02 '19 at 18:15

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Let $X,Y$ topological spaces, $f:X\rightarrow Y$ a function, with $Y$ compact. I need to show that, if $f$ is continuous, so $$G_{f}:=\{(x,f(x)):x\in X\}$$ is closed in $X\times Y$.

I saw that question with $Y$ Hausdorff, so is easy to prove that for every $(x,y)$ with $y\neq f(x)$, there is an open set $U\times V$ such that $U\times V\subset (G_{f})^{c}$ by taking a neighborhood $V$ of $y$ in $Y$ with $f(x)\notin V$.

Can I do that for $Y$ compact? If not, how can I solve this question?

asked May 02 '19 at 18:15

Mateus Rocha

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You need to assume at least $Y$ is $T_1$. For example, take $Y$ with the chaotic topology $\left{\varnothing,Y\right}$ and any function $f\colon X\to Y$. Then the graph of $f$ is dense in $X\times Y$. – Luiz Cordeiro May 02 '19 at 18:16
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In some texts compact is defined as Hausdorff + compact. It is possible the problem came from one that does so. – logarithm May 02 '19 at 19:01
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Indeed there can be counterexamples if $Y$ is not Hausdorff, see my post here for a simple example. If $Y$ is Hausdorff, no compactness at all is needed, just continuity of $f$. – Henno Brandsma May 02 '19 at 21:33
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Possible duplicate of Show that if $f:X\to Y$ is a continuous function if and only if the graph of $f$ is a closed subset of $X\times Y$ (As pointed out by Henno Brandsma) – YuiTo Cheng May 02 '19 at 23:25

$f:X\rightarrow Y$, with $Y$ compact, is continuous $\Rightarrow$ its graph is closed

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