Showing a graph is closed


by radou
Tags: graph, showing
radou
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Jul16-10, 01:58 PM
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Me again.

Problem. Let X be a topological space, and Y a T2-space (i.e. a Haussdorf topological space). Let f : X --> Y be a continuous function. One needs to show that the graph of , i.e. the set G = {(x, f(x)) : x is in X} is closed in X x Y.

Attempt of proof. To show what we need to show, we have to prove that the complement of G is open. Now, since Y is a Haussdorf space, one can show that it is a T1-space too, i.e. that every set containing a single point is closed. So, for every x, {f(x)} is closed in Y. Further on, since f is continuous, the preimage of every closed set is a closed set too, so, for any f(x), {x} is closed in X. I feel I'm getting very warm, but I got stuck anyway. The complement of G in X x Y is X x Y \ G = X x Y \ U (x, f(x)) (where the union goes through all x in X) = [itex]\cap[/itex] [(X x Y) \ (x, f(x))]. Now, if X and Y are T1-spaces, can one show that X x Y is a T1 space, too? But then we'd in general have an infinite intersection of open sets, which doesn't need to be open.

Thanks in advance for a push.
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eok20
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Jul16-10, 02:34 PM
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Some comments: the preimage of {f(x)} is not necessarily x.

One way that I think will work is that G is the inverse image of {(y,y)} < Y x Y under a nice map. Using the fact that Y is Hausdorff I think you can show that the diagonal set ({(y,y)}) is closed.
radou
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Jul17-10, 06:01 AM
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Quote Quote by eok20 View Post
Some comments: the preimage of {f(x)} is not necessarily x.
Oh, of course! For some weird reason, I assumed f to be injective.

Quote Quote by eok20 View Post
One way that I think will work is that G is the inverse image of {(y,y)} < Y x Y under a nice map. Using the fact that Y is Hausdorff I think you can show that the diagonal set ({(y,y)}) is closed.
OK, I'll try to look into it.

radou
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Jul21-10, 04:18 AM
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Showing a graph is closed


Another attempt of a proof, a bit different than suggested.

So, to review, given f : X --> Y, with f continuous and Y Haussdorf (or T2), we have to show that the graph G of f is closed in X x Y (the closed graph theorem).

If we show that Cl(G) is a subset of G, then Cl(G) = G, and G is closed.

Let x = (a, b) be from Cl(G). Then there exists a net n : D --> G such that n --> (a, b). This implies the convergence of its components, i.e. n = (n1, n2) -- > (a, b). Since f is continuous, and n1 is a net in X which converges to a, the composition net f o n1 must converge to f(a). So, the net n' = (n1, f o n1) converges to (a, f(a)).

Let p1 and p2 be the projections from X x Y to X and Y respectively, defined with p1(x, y) = x and p2(x, y) = y. Since they are continuous, for every net N : D --> X x Y which converges to some (x0, y0), we have p1 o N --> x0 and p2 o N --> y0.

If we apply the projections p1 and p2 with the nets n and n', we obtain p1 o n --> a, p2 o n = b, p1 o n' --> a and p2 o n' --> f(a).

Now, the part I'm not exactly sure about (regardless of the same limits, of course) - does p1 o n equal p1 o n', since the first components od n and n' are equal? If so, can we conclude that n and n' are equal? If so, it follows easily from the other two relations that b = f(a), since Y is Haussdorf, so x = (a, b) = (a, f(a)) and hence lies in G.

Thanks in advance for anyone with enough patience for this.
radou
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Jul21-10, 04:22 AM
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Actually, after thinking a bit more, it seems pretty obvious that p1 o n = p1 o n' doesn't imply n = n', because it's only about the first components.

Still, I feel I could be on the right track with the work up to this absurd conclusion.


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