I need a counterexample for the following:
If f:[a,b] -> R is ONE-TO-ONE and satisfies the intermediate value property, then f is continuous on [a,b].
I know that this is a false statement if you exclude the one-to-one property. The example I received before was f(x) = sin(1/x), but this function is not one-to-one. I am having a really hard time coming up with an example of a ONE-TO-ONE function that satisfies the IVP and is discontinuous on an interval. Can you please help me?
Also, is there anyway of correcting this statement to make it true, and if so, can you prove it?
Consider the following function on the interval
0 <= x <= 2
(where "<=" means "less or equal", that is this interval is [0,2])
f = x for 0<=x<1
f = x-2 for 1<=x < 2
f = 1 for x = 2
This function has the following properties:
(1) It maps one-to-one the interval [0,2] onto interval [-1,1]
(2) It satisfies the intermediate value property
(3) It is not continuous
Therefore it is a counterexample of the kind you request.
A possible way to correct ...
Discontinuous counter example statements are provided. A ONE-TO-ONE function that satisfies the IVP and discontinuous intervals are determined.