# Continuity of an inverse function

ohreally1234

## Homework Statement

Prove that the a continuous function with compact domain has a continuous inverse. Also prove that the result does not hold if the domain is not compact.

## The Attempt at a Solution

I tried using the epsilon delta definition of continuity but didn't get anywhere.
This is supposed to be a pretty standard proof in any real analysis book but unfortunately the one I'm using has it listed as an exercise...

Homework Helper
First, you need to add that the function is 1-1 to your premises. I'll give you a hint as to why you need compactness by giving you the usual counterexample f:[0,2*pi)->R^2 defined by f(x)=(cos(x),sin(x)). Why does that show the result doesn't hold without compactness?

ohreally1234
Hmm, how do I prove the first part (that it is continuous)

Homework Helper
Think about it. Suppose f:X->Y. A function is continuous if for every convergent sequence in X, x_n->x that f(x_n)->f(x). f^(-1) is continuous if for every convergent sequence y_n->y in Y that f^(-1)(y_n)->f^(-1)(y). You want to show f^(-1) is continuous given f is continuous, 1-1, and X is compact. Hint: define x_n such that f(x_n)=y_n. Does the sequence x_n have a limit point? Why?

ohreally1234
First, you need to add that the function is 1-1 to your premises. I'll give you a hint as to why you need compactness by giving you the usual counterexample f:[0,2*pi)->R^2 defined by f(x)=(cos(x),sin(x)). Why does that show the result doesn't hold without compactness?

But isn't f(x) = (cos(x), sin(x)) not a 1-1 function (because it's a circle)?

Homework Helper
But isn't f(x) = (cos(x), sin(x)) not a 1-1 function (because it's a circle)?

No. Why? Show me two values of x that map to the same value of (cos(x),sin(x)) in the domain [0,2*pi).

ohreally1234
oh sorry, i also forgot to add that
f:[a,b] => R
I got the continuity part down (your hint really helped!), but I'm having trouble with the compact part.

Homework Helper
oh sorry, i also forgot to add that
f:[a,b] => R
I got the continuity part down (your hint really helped!), but I'm having trouble with the compact part.

What kind of 'trouble with the compact part'? You are given that the domain is compact, you don't have to prove it.

ohreally1234
I'm having trouble with giving a counter example of a continuous 1-1 function f:[a,b] => R whose inverse is not continuous (does it even exist)?.

Homework Helper
I'm having trouble with giving a counter example of a continuous 1-1 function f:[a,b] => R whose inverse is not continuous (does it even exist)?.

You won't find one for f:[a,b]->R. [a,b] is compact. You are going to have to be more creative and pick a non-compact domain.

ohreally1234
If the domain is non-compact, does such a function exist?

Homework Helper
If the domain is non-compact, does such a function exist?

Yes. Pick the domain X to be [0,1) union [2,3]. Notice the open end on the first interval. Can you define a SIMPLE function f:X->R whose inverse isn't continuous? Concentrate on using the open endpoint.

ohreally1234
sorry I am stumped... i tried playing with functions such as f(x)=1/(x-1) but I am not sure what to do