Continuity and compactness

[math8]

X, Y metric spaces. f:X-->Y and X is compact.
How do I prove that f is continuous if and only if G(f)={(x,f(x)):x in X} C X x Y is compact.

I think for the forward direction, since f is continuous and X is compact, then f(X) is compact. Hence, G(f)=X x f(X) is compact as a cross product of compact sets.

But for the backward direction, I am totally lost.

 

[math8]

compactness and continuity

X, Y metric spaces. f:X-->Y and X is compact.
How do I prove that f is continuous if and only if G(f)={(x,f(x)):x in X} C X x Y is compact.

I think for the forward direction, since f is continuous and X is compact, then f(X) is compact. Hence, G(f)=X x f(X) is compact as a cross product of compact sets.

But for the backward direction, I am totally lost.

 

[tim_lou]

compactness in metric space is equivalent to sequential compactness, namely that every infinite sequence has a subsequence that converges. Further, continuity in a metric space is equivalent to limit point continuity, namely that for any x_n -> x, f(x_n)-> x (this is true for any first countable space).

relating these two qualities should give you what you want. (also note that convergences in the product topology means each coordinate converges separately).

 

[tim_lou]

see my comments in the other thread. It's not a good idea to post in two separate forums in general.

 

[math8]

So since X is compact, we have that X is seq. compact. So every seq in X has a subseq. which converges in X.
Now we let (xn) be a sequence in X. Hence (xn, f(xn)) is a seq. in G(f). Let xn converges to x. so any subsequence (xnk) of (xn) converges to x.
Now, since G(f) is sequentially compact, there is a subsequence (xnk, f(xnk)) which converges to (x, y) in G(f). So since (x, y) is in G(f), y=f(x). So (xnk, f(xnk)) --> (x, f(x)).
So f(xnk) --> f(x).
Now we show f(xn)-->f(x). That's where I am not sure how to argue this, should I say by way of a contradiction, assuming that f(xn) does not converge to f(x), then there exists an epsilon positive such that d(f(xn),f(x)) greater than epsilon for infinitely many n.
But since f(xnk) --> f(x), there exists an N s.t. for all k>N, d(f(xnk),f(x))<epsilon. (i.e. d(f(xnk),f(x))<epsilon for all but finitely many k) But this is a specific subsequence, so it does not work for all subsequences :((. Would that give me the contradiction I need? What am I missing?

 

[math8]

I know, it was mistake, I realized that I posted the other one in the algebra section, that's why I decided to repost this one in the analysis section.

 

[HallsofIvy]

I've merged the other thread into this one.

 

[tim_lou]

You pretty much have most of it down. However, you didn't exploit the property of a graph enough to complete the proof. Namely that for every x, there can only be one point, (x,f(x)).

Now, fix another subsequence (x_nj, f(x_nj)) that doesn't converge (edit: it should be epsilon away from) to (x, f(x)). If it is infinite, then it's got to have a subsequence that converges to something else. However, this convergence violates the fact I mentioned above.

 

[mathwonk]

well basically, that isn the graph of f, and the graph of f is bijectively equivalent to th domain of f. so presumab;y, the map X-->XxY taking x to (x,f(x)) is continuousn iff f is, and it takes X then homeomorphically to the graph of f, which should imply graph f is compact also.

so the harder part is to show if graphf is compact then that map is continuous.

lets see, i guess the map f is essentially the same as the projection from garaphf onto the target space Y, so if graph f is compact then that projection takes, hmmm,i don't see it immediately,...

 

[mathwonk]

oh i se it. it boils down to showing the map taking x to (x,f(x)) is continuous, i.e. the pullback of ma clsoed set is clsoed. but a closed set in a compact set (the graph) is compact, so its projection down to X, which is also its pullback, is also compact, hence closed. so we are done.