Monotonic and Continuous function is homeomorphism

[ForMyThunder]

Homework Statement

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.

Homework Equations

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?

 

[Dick]

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

 

[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.

 

[some_dude]

Perhaps you need to add the hypothesis f is dffble?

 

[ForMyThunder]

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

 

[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.