Inverse of a continuous function.

[bloodhound]

In my revision on linear algebra i came across this statement.

 

If a continuous function is invertible then its inverse need not be continuous!

 

except i am having a hard time finding examples.

 

Can someone give me such a function. cheers.

[matt grime]

How abstract do you wish to be?

 

It may help if you recall that a function is continuous iff the inverse image of an open set is open. Thus all you need to do is find an invertible continuous function that sends an open set to a closed set. Of course if you're allowed to play around with the topology then it's very easy.

 

Let S be any set with the discrete topology, and let S' be a copy of S with any other topology, then the identity map from S to S' is continuous (the inverse image of an open set is open) but its inverse is not continous, otherwise S' would have the discrete topology.

[bloodhound]

hmm.. i havent come across topology yet. what is a discrete topology?

[matt grime]

the discrete topology is one in which all sets are declared to be "open", though if you don't know what a topology on a space is this is meaningless. i don't propose explaining it here since there are lots of references out there that can explain it better than an off the cuff one here and now.

 

Anyway, here is a continuous bijection that is not invertible:

 

let S be the subset of R given by [-1,0]u(1,2]

 

let f(x)=x if x is in [-1,0] and x-1 if x is in (1,2] then this map is continuous but not with continuous inverse onto the interval [-1,1]

[bloodhound]

thanks for that.

 

is topology on a space something to do with metric spaces?

[matt grime]

sort of, every metric gives a topology on the space. a topological space is a generalization of the ideas of continuity to spaces where distance might not make sense.

[jcarlson]

[math]f(x) = tan^{-1}(x)[/math]

 

The inverse of this function is:

 

[math]x = tan^{-1}(f(x))[/math]

 

[math]f(x) = tan(x); -\frac{\pi}{2} < x < \frac{\pi}{2} [/math]

 

The domain is restricted on the inverse because [math] f(x) = tan^{-1}(x) [/math] is restricted to the range [math]-\frac{\pi}{2} < f(x) < \frac{\pi}{2}[/math]. Thus the function [math]f(x) = tan^{-1}(x)[/math] is continuous on the domain of all real numbers, but its inverse is not.

 

*edit*

 

This is the case for every function whose domain is all the real numbers, but whose range is restricted due to a horizontal asymptote or an absolute minimum. The inverse will not be continuous on the domain of all real numbers. Perhaps an even better example would be the function [math]f(x) = x^{2}[/math], whose inverse, [math]f(x) = \sqrt{x}[/math] is not only not defined for x < 0, but isnt a function at all, by the vertical line test.

[matt grime]

It's inverse isn't defined on R, arctan isn't an invertible function from R to R so this is a non-example. You don't seem to understand that the domain and range are part of the definition are functions - it makes little sense to say tan is invertible with continuous and inverse and isn't.