# Homework Help: Quick topology question

1. Dec 18, 2007

### Mystic998

1. The problem statement, all variables and given/known data
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?

2. Relevant equations

3. 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.

2. Dec 19, 2007

### 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?

3. Dec 19, 2007

### quasar987

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

4. Dec 19, 2007

### zhentil

Bolzano-Weierstrass

5. Dec 19, 2007

### Dick

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)).

6. Dec 19, 2007

### quasar987

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

7. Dec 19, 2007

### morphism

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

8. Dec 19, 2007

### 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.