Is there a Simple Proof for the Continuity of the Inverse Function?

[CarmineCortez]

Homework Statement

Let I be an interval in R, and let f: I-->R be one-to-one, continuous function. Then prove that f^(-1):f(I)-->R is also continuous.

Homework Equations

The Attempt at a Solution

I started a thread yesterday and had some responses but the proofs became quite complicated and my proof ended up wrong, so I tried more basic approach.

So if f is continuous then
lim_x->af(x) = f(a) now if f^-1 is continuous then
lim_f(x)->f(a)f^-1(f(x)) = a
which is equivalent to lim_f(x)->f(a)x=a
which is true since f is continuous.

Does this make any sense, I'm not really familiar with taking the limit of the domain as the function is changing. Is there any hope for this proof?

Thanks

 

[CarmineCortez]

any ideas

 

[morphism]

It's not a good idea to start a duplicate thread. You should've just used your previous thread. Anyway, I'll give you more detailed hints this time.

The first order of business is to prove that f is either decreasing or increasing on I. The idea is to use the intermediate value theorem. However, since the IVT only applies to closed and bounded intervals, we need to be a little careful here. Take any two points a<b in I. Since f is one-to-one, then either f(a)<f(b) or f(a)>f(b). Suppose wlog that f(a)<f(b). Now our goal is to prove that f is increasing on I. If c<d are two other points in I, define g:[0,1]->R by

g(t) = f((1-t)b + td) - f((1-t)a + tc).

Why is it natural to look at this? (Draw a picture.) Use the IVT to prove that g>0 for all t. Conclude that f is increasing on I.

Having done this, the continuity of f^(-1) follows easily from an epsilon-delta argument.

 

[CarmineCortez]

It's not a good idea to start a duplicate thread. You should've just used your previous thread. Anyway, I'll give you more detailed hints this time.

The first order of business is to prove that f is either decreasing or increasing on I. The idea is to use the intermediate value theorem. However, since the IVT only applies to closed and bounded intervals, we need to be a little careful here. Take any two points a<b in I. Since f is one-to-one, then either f(a)<f(b) or f(a)>f(b). Suppose wlog that f(a)<f(b). Now our goal is to prove that f is increasing on I. If c<d are two other points in I, define g:[0,1]->R by

g(t) = f((1-t)b + td) - f((1-t)a + tc).

Why is it natural to look at this? (Draw a picture.) Use the IVT to prove that g>0 for all t. Conclude that f is increasing on I.

Having done this, the continuity of f^(-1) follows easily from an epsilon-delta argument.

Does the epsilon delta proof use the property f^-1(f(x)) = x so

d(x,x_o)<delta and f(x)-f(x_0)<epsilon then for the inverse

f(x)-f(x_0)<epsilon d(f^-1(f(x)) -f^-1(f(x₀)) )=(x-x₀)<delta
but what if delta does not equal epsilon...

 

[morphism]

Why are you starting with "d(x,x_o)<delta and f(x)-f(x_0)<epsilon"? We're given an epsilon and we want to find a delta>0 such that |f^-1(y) - f^-1(y_0)|<epsilon when |y-y_0|<delta.

Of course everything in the domain of f^-1 looks like f(something). So we may suppose that y=f(x) and y_0=f(x_0).

Drawing a picture will make the task of finding a suitable delta easier.