Continuity of an inverse function

  • #1
ohreally1234
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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.


Homework Equations





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...
 

Answers and Replies

  • #2
Dick
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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?
 
  • #3
ohreally1234
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Hmm, how do I prove the first part (that it is continuous)
 
  • #4
Dick
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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?
 
  • #5
ohreally1234
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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)?
 
  • #6
Dick
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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).
 
  • #7
ohreally1234
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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.
 
  • #8
Dick
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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.
 
  • #9
ohreally1234
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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)?.
 
  • #10
Dick
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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.
 
  • #11
ohreally1234
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If the domain is non-compact, does such a function exist?
 
  • #12
Dick
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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.
 
  • #13
ohreally1234
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sorry I am stumped... i tried playing with functions such as f(x)=1/(x-1) but I am not sure what to do
 
  • #14
Dick
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Try f(x)=x on [0,1) and f(x)=x-1 on [2,3]. Since I just gave that to you, for free, you have to explain to me why it works. Is it continuous? What's the range? Is it 1-1? Finally give me a formula for the inverse (like the one for f) and tell me why it's discontinuous.
 
  • #15
ohreally1234
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ooo that makes so much more sense:
range of f is [0,2] but the inverse f^-1: [0,2] => [0,1) U [2,3] is clearly discontinuous at x=1.

thanks so much
 
  • #16
Dick
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Sure, but can you help a little more next time, and show some more of approaches you have tried?
 

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