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Given a function $f:X\to Y$, we define the graph of $f$ as the set $$G(f)=\{(x,f(x)),x\in X\}$$ Show that if $X$ is compact then $f$ is a continous function if and only if $G(f)$ is a closed subset of $X\times Y$.

As i found the answer here...Show that if $f:X\to Y$ is a continuous function if and only if the graph of $f$ is a closed subset of $X\times Y$ but im not getting in my head as im not able to understand this paragraph......

If $Y$ is not Hausdorff, then the first assertion may fail: Let $X=\{0,1\}$ with discret topology, $Y=\{0,1\}$ with trivial topology, and $f=\operatorname{id}$. It's easy to find the contradition...

PLiz elaborate this this answer with proper proof,,,,,

jasmine
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1 Answers1

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If $Y=\{0,1\}$ with trivial topology, then any map to $Y$ is continuous. So in particular, this is the case if we take $X=\{0,1\}$ with discrete topology (a compact space) and $f\colon x\mapsto x$. So the claim tells us that $G(f)$ should be a closed subset of $X\times Y$, or equivalently, its complement open. However, any open subset of $X\times Y$ that contains $(1,0)$ must also contain $(0,0)$. We conclude that $G(f)$ is not closed.

Hagen von Eitzen
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