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Let $f: X\to Y$ be a function between topological spaces, where $Y$ is Hausdorff and $G_f:=\{(x,f(x)):x\in X\}\subseteq X\times Y$

Show that:

a) If $f$ is continuous, then $G_f$ is closed.

b) If $X$ and $Y$ are compact and $G_f$ closed, then $f$ is continuous.

Hello,

I have a problem with this task and might need some help.

To show that $G_f$ is closed, I tried to show that $G_f$ only contains boundary points. So $\partial G_f=G_f$.

$G_f: X\to X\times Y, x\mapsto (x,f(x))$

I have a general question. Do you always use the product topology when you work with sets which are products of sets? I might use the projection $pr_X: X\times Y\to X, (x,y)\mapsto x$ and $pr_Y:X\times Y\to Y, (x,y)\mapsto y$ then.

To show that $\partial G_f=G_f$ I have to show that for every element $(x,y)\in G_f$ and every neighbourhood of this point, the neighbourhood contains elements which are not included in $G_f$.

Could this be a possible way to solve this? I would be thankful for a hint.

Thanks in advance.

Will R
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Mr.Topology
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2 Answers2

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First observe that if $Y$ is Hausdorff, then the diagonal $D=\{(y,y):y\in Y\}\subset Y\times Y$ is closed. You can see this in the following way:

If $(x_1,x_2)$ is a point in $Y\times Y$ that is not in $D$ (i.e. $x\neq y$) then there are open neighbouhoods $x_1\in U_1$ and $x_2\in U_2$ with $U_1\cap U_2=\emptyset$. $U_1\times U_2$ is an open neighbourhood of $(x_1,x_2)$ that does not intersect $D$.

If $f$ is continuous, so is the map $f\times 1 \colon X\times Y \to Y\times Y$ that maps $(x,y)$ to $(f(x),y)$. $G_f$ is the preimage of $D$ under this map, so it is closed.

For the second statement one needs to show that preimages of closed sets in $Y$ are closed in $X$. Let $A\subset Y$ be closed. Then $X\times A\subset X\times Y$ is closed and the same is true for $G_f \cap (X\times A)$. Because Y is compact, the projection $\pi\colon X\times Y\to X$ is a closed map (see here). This implies that $\pi(G_f \cap (X\times A))=f^{-1}(A)$ is a closed subset of $X$.

Community
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user337041
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  • Thanks for your answer. I have some question. Why does $U_1\times U_2$ not intersect $D$? What do you mean with the notation $f\times 1: X\times Y\to Y\times Y$? What does the $1$ stand for? – Mr.Topology May 04 '16 at 21:00
  • If $U_1\times U_2$ intersects $D$, there is a point of the form $(x,x)\in U_1\times U_2$ but this means that $x\in U_1\cap U_2=\emptyset$. By $1$ I mean the identity map from $Y$ to $Y$. The map $f\times 1$ is the product of $f$ with the identity on $Y$, i.e. $(x,y)\mapsto (f(x),y)$. Maybe $f\times id_Y$ would be a bit more intuitive. – user337041 May 04 '16 at 21:05
  • Ah, I see. Thank you. Can you explain your notation $f\times 1$ too? – Mr.Topology May 04 '16 at 21:06
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For (a) : For $(x,y)\in X \times Y$ with $y\ne f(x),$ let $B,B'$ be disjoint open subsets of $Y$ with $y\in B$ and $f(x)\in B'.$

Let $A$ be an open subset of $X$ with x$\in A,$ and such that $\forall x'\in A\;(f(x')\in B').$ Then $C=A\times B$ is open in $X\times Y;$ and $(x,y)\in C$ while $C\cap G_f=\phi.$

So any $(x,y)\in (X\times Y)$ \ $G_f$ belongs to an open subset ($C$) of $X\times Y$ which is disjoint from $G_f.$ So $(X\times Y)$ \ $G_f$ is open in $X\times Y.$

DanielWainfleet
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  • Where do you use, that $f$ is continous? – Mr.Topology May 05 '16 at 16:39
  • The existence of the set $A$ with its stated properties comes from the continuity of $f$. – DanielWainfleet May 05 '16 at 16:49
  • Where exactly is the continutiy of $f$ needed for the set A? I do not see it... – Mr.Topology May 05 '16 at 17:32
  • $f:X\to Y$ is continuous iff : Whenever $x\in X$ and $B'$ is open in $Y$ with$ f(x)\in B',$ there exists open $A$ in $X$ with $x\in A$ and $\forall x'\in A;(f(x')\in B').$ This is the topological generalization ot the "$\delta , \epsilon$" version of continuity of real functions. It is one of several equivalent definitions of continuity. – DanielWainfleet May 05 '16 at 18:09