Function continuous, then a subset is closed

[missavvy]

Homework Statement

Let M, N be two metric spaces. For f: M --> N, define the function on M,

graph(f) = {(x,f(x)) \inMxN: x\inM}

show f continuous => graph(f) is closed in MxN

Homework Equations

The Attempt at a Solution

I can't figure out what method to use.

I have written out many equivalent defintions of continuity and closed sets, so far I know I want to show that.. basically since f is continuous, we have convergent sequences mapped to convergent sequences, ie xn -> x, fn -> f

And I need to show that (x,f(x)) \in graph(f)

It's either extremely trivial, because from the definition of continuity we see that all the limits belong to that subset..

But anyways, I don't think I can say that or make that a proof by just stating the obvious..

So I tried creating a neighbourhood and showing that M\graph(f) is open.

Assume f is continuous. Then for all x in M, for all ε, there exists δ=θ/2 s/t dM(x,y)<δ => dN(f(x),f(y))<ε

So consider B(y,ε'), choose ε' = δ = θ/2.
For any z in B(y,ε')

d(z,x) ≤ d(z,y) + d(x,y) < ε' + δ = θ

Then d(z,x) < θ and z is in M\graph(f)...

This is all probably very wrong.. I didn't use anywhere the definition of graph(f) which worries me.

eek.. Help?

 

[Bacle]

Have you consider seeing what happens to points that _are not_ in the graph, i.e., to see what happens to the complement of the graph in MxN? Seeing closed sets as complements of open ones. of course.

 

[missavvy]

hm, I am not sure I understand.. so if I take points in MxN, that are not in the graph, then we have (x,f(x)) but x is not in M, which wouldn't make sense, would it?

 

[Bacle]

No, I mean, MxN, and then take (m,n), with n not in f(M), i.e., n is in N-f(M); since f is a function, every argument m in M can have--has--exactly one output, so that if (m,f(m)) is in the graph, then (m,n') , with n'≠ f(m) _cannot_ be in the graph of f.

 

[missavvy]

Ok.. So to clarify, we have points in MxN,

(m,n) , n not in f(M)
(m,n') not in graph(f)

Then do I use the defn of continuity on these two sets of points?

 

[HallsofIvy]

Which definition of "continuous" are you given? One is "f is continuous if and only if, for every open set U, f^{-1}(U) is open". That may be what Bacle was thinking when he suggested that you look at complements which will be open if the given sets are closed.

 

[missavvy]

I'm just give that f is continuous... I'm not told that any set U is open though, so I don't know if I can use that definition.

Is there a way to do this using a delta epsilon definition? Or neighbourhoods?

 

[Bacle]

What I was suggesting was using continuity to construct an open set UxV; U open in M, V open in N (so that UxV is open in the product MxN) , as an open set containing a point (m,n), with n≠ f(m), i.e., (m,n) is not in the graph of f , so that UxV misses the point (m,f(m)), then UxV is an open set containing a generic point in the complement, so that the complement of the graph is open, and so the graph is closed. Specifically, the continuity of f allows you to contain the image of the graph of f within an ε-ball of f(x)--where you can , again by continuity, make this ε as small as you want. Consider, e.g., the graph of f(x)=x², and a point not in the graph, e.g., a point p=(x,x²+a) ; a≠ 0. Can you see how to construct an open set containing p, that misses the graph? Here M=N=the reals.