If f is a one-to-one continuous function defined on an interval (a,b)...

If f is a one-to-one continuous function defined on an interval (a,b), then its inverse function f^-1 is also continuous.

>brainlets can't prove this

We will not do your homework

This isn't hw, brainlet, I already know the answer.

Do your own homework you idiot.

roofFirst we show that if f is both one-to-one and continuous on sa, bd, then it must
be either increasing or decreasing on sa, bd. If it were neither increasing nor decreasing,
then there would exist numbers x1, x2, and x3 in sa, bd with x1 , x2 , x3 such that
f sx2 d does not lie between f sx1d and f sx3 d. There are two possibilities: either (1) f sx3 d
lies between f sx1d and f sx2 d or (2) f sx1d lies between f sx2 d and f sx3 d. (Draw a picture.)
In case (1) we apply the Intermediate Value Theorem to the continuous function
f to get a number c between x1 and x2 such that f scd − f sx3 d. In case (2) the Intermediate
Value Theorem gives a number c between x2 and x3 such that f scd − f sx1d. In either
case we have contradicted the fact that f is one-to-one.
Let us assume, for the sake of definiteness, that f is increasing on sa, bd. We take
any number y0 in the domain of f 21 and we let f 21sy0 d − x0; that is, x0 is the number
in sa, bd such that f sx0d − y0. To show that f 21 is continuous at y0 we take any « . 0
such that the interval sx0 2 «, x0 1 «d is contained in the interval sa, bd. Since f is
increasing, it maps the numbers in the interval sx0 2 «, x0 1 «d onto the numbers in the
interval s f sx0 2 «d, f sx0 1 «dd and f 21 reverses the correspondence. If we let denote
the smaller of the numbers 1 − y0 2 f sx0 2 «d and 2 − f sx0 1 «d 2 y0, then the
interval sy0 2 , y0 1 d is contained in the interval s f sx0 2 «d, f sx0 1 «dd and so is
mapped into the interval sx0 2 «, x0 1 «d by f 21. (See the arrow diagram in Figure 3.)
We have therefore found a number . 0 such that
if |y 2 y0 | , then | f 21syd 2 f 21sy0 d | , «

OP what's the point? Why are you posting this thread?

desu im high af and just wanted to call people >brainlets

It's not true in general

it's pretty simple to prove when you consider that f^-1 is just a reflection in y=x

no its not

f isnt necessarily a homeomorphism

>defined on an interval

this can literally be done in 3 steps. do you really need help on this OP?

>then it must
>be either increasing or decreasing
wew lad want to try again?

>babbys first analysis course

...

Stewart's Calculus. You're the brainlet. Consider the trigonometric functions that have inverses. Moreover those functions are neither strictly increasing or decreasing unless you only consider support on (0, 2pi).

>muh invariance of domain

Doesn't this immediately follow from continuity preserving open sets, and injectivity inducing a bijection on the open sets of the domain and range of f?

continuity doesn't preserve open sets. well it does, but in a very different way.

Well obviously you're only going to consider 0-2pi as those functions have circular motion as opposed to linear motion

Yeah I said it the wrong way (preimage of every open set is open). But wouldn't they both be open mappings too?

>linear motion
Yeah I should have restricted the support for each trigonometric function, so my bad there, but you cannot deny that the functions must not be monotone as stated in the original post that I replied to. As for linearity, WGAF, we're talking about all C(a,b), eg universal quantification.

How we speak/write is VERY important. We must be precise about our intentions or trivial counterexamples are easy to find.

This is a terribly over-complicated proof for this. I'm pretty sure I proved this in my class without even referencing the IVT. Just use the definition of continuity and an injection.

well it IS true that an injective continuous map from R^n to R^n is an open map, but it is a non-trivial fact. or at least for me

even if you are sending (a,b) to S^1?