Is a Closed Graph of a Function on a Closed Interval Indicative of Continuity?

[Mystic998]

Homework Statement

All right, so this appeared on my final. The intervals are in the reals:

If f : [a, b] -> [c, d] , and the graph of f is closed, is f continuous?

Homework Equations

The Attempt at a Solution

Well, my gut reaction is no, just because it seems like a fairly strong claim, but I couldn't really come up with a decent counterexample. On the other hand, I can see that the graph has to be compact. And I think that would imply that f([a, b]) has to be compact (if it wasn't you could just take any open cover of [a, b] and an open cover of f([a, b]) that doesn't have a finite subcover, then the Cartesian product would be an open cover without finite subcover, I think). But I'm not sure where I can go from there or if that's even useful information.

 

[Dick]

If the sequence x_i->x in [a,b] then the sequence f(x_i) has cluster points in [c,d] because [c,d] is compact. Suppose it has two unequal ones y1 and y2. Then since the graph is closed y1 and y2 are both on the graph. Is this compatible with f being a single valued function?

 

[quasar987]

But what guarentees that the limit of f(x_i) is f(x)?

 

[zhentil]

Bolzano-Weierstrass

 

[Dick]

But what guarentees that the limit of f(x_i) is f(x)?

If f(x_i) has a limit y, then (x,y) is on the graph since the graph is closed. The graph is the set of all points (x,f(x)).

 

[quasar987]

That's true. That was a good little problem!

 

[morphism]

What about the converse: if f:[a,b]->[c,d] is continuous, then is its graph closed?

 

[quasar987]

That one's easy. If (x_i, f(x_i)) is a converging sequence in the graph, then it means x_i -->x, for some x in [a,b], and by continuity, f(x_i)-->f(x). Since (x,f(x)) is in the graph, it's closed.