# Monotonic and Continuous function is homeomorphism

1. Jun 29, 2010

### ForMyThunder

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

If H:I$$\rightarrow$$I is a monotone and continuous function, prove that H is a homeomorphism if either
a) H(0) = 0 and H(1) = 1
or b) H(0) = 1 and H(1) = 0.

2. Relevant equations

3. The attempt at a solution

So if I can prove H is a homeomorphism for a), b) follows from the fact that the map defined by t$$\rightarrow$$H(1-t) is also a homeomorphism because it is the composite of two homeomorphisms.

H is obviously one-to-one, but I don't know how to "show" this.

At first I figured that I should assume |H(t1)-H(t2)| < $$\delta$$ for some $$\delta$$ > 0 and try to show that there exists a $$\epsilon$$ > 0 such that |t1 - t2| < $$\epsilon$$. I didn't know where to go after this, so I tried the sequence definition for continuity and I got nowhere.

Any suggestions?

2. Jun 29, 2010

### Dick

Use monotone to prove it's one-to-one. Use the Intermediate Value Theorem to prove it's onto.

3. Jun 29, 2010

### some_dude

By "monotone" do you mean strictly monotone? (x < y implies f(x) < f(y))? If so it's trivial, but if not then I don't think you'll be able to prove this. E.g.

f(x) = 0 for x in [0, 1/2) and = 2(x - 1/2) otherwise.

4. Jun 29, 2010

### some_dude

Perhaps you need to add the hypothesis f is dffble?

5. Jun 29, 2010

### ForMyThunder

Thanks. But I also need a little help with showing that the inverse function is also continuous.

6. Jun 29, 2010

### tmccullough

What class is this for? The fact that the continuous image of a compact set is compact coupled with the Heine-Borel theorem is the super easy way to conclude that $$H$$ is closed/open, so $$H^{-1}$$ is continuous.

That seems like it might be too fancy for your class, though. Disregard if you don't know what I'm talking about.

You can show directly that the image of an open set is open, just using the definition of monotone and continuity. More explicitly, consider the image of an open interval, given that the map is continuous - it's the same as showing that it's onto, except replace 0 and 1 with different values.