Thread: Give example and prove it

1. Give example and prove it

An open subinterval of the closed interval [a, b] is an open interval (c, d)
such that (c, d) subset of [a, b] (i.e. a <= c < d <= b). Give an example of numbers a < b and of a function f : [a, b] --> R which is increasing on
every open subinterval of [a, b] and is not increasing on [a, b]. Be sure
to justify your example (Hint: Choose f to be discontinuous.)

I am really bad at something that contains with intervals

2. The first thing I would notice is that (a, b) is an open subinterval of [a,b]. So your example function must be increasing on (a, b) (and there is a very easy example of such a function). And then you would need to pick values for f(a) and f(b) so that the function on [a,b] is NOT increasing. Again, fairly easy.

3. Originally Posted by HallsofIvy
The first thing I would notice is that (a, b) is an open subinterval of [a,b]. So your example function must be increasing on (a, b) (and there is a very easy example of such a function). And then you would need to pick values for f(a) and f(b) so that the function on [a,b] is NOT increasing. Again, fairly easy.

I have got an example
f(x) = 1 if x = 0,
f(x) = x if 0<x<2,
f(x) = 1.5 if x = 2

This function is continuous and increasing in every open subinterval of [0,2]
But it is not increasing in [0,2].

but I dont know how to prove it, can you show me the proof?