# Thread: Analysis - Continuous Functions

1. ## Analysis - Continuous Functions

Right, i'm doing semester 2 analysis and i'm completely stumped, have no idea where to start on this question:
Let f be a map from an interval I to the reals (not necessarily continuous). Prove the equivalence of the two statements:
(i) if x,y are contained within I with f(x)<f(y), then for all c in (f(x),f(y)) there exists z in (x,y) union (y,x) : f(z) =c
(ii) if J, a subset of I, is any interval, then f(J) is an interval

any help would be appreciated, thanks

2. Originally Posted by danilo
Right, i'm doing semester 2 analysis and i'm completely stumped, have no idea where to start on this question:
Let f be a map from an interval I to the reals (not necessarily continuous). Prove the equivalence of the two statements:
(i) if x,y are contained within I with f(x)<f(y), then for all c in (f(x),f(y)) there exists z in (x,y) union (y,x) : f(z) =c
(ii) if J, a subset of I, is any interval, then f(J) is an interval

any help would be appreciated, thanks
What have you tried? Is there a particular point you are getting stuck at? Do you know the concept of connectedness? These are all relevant questions.

3. I'm not really sure about the last part you wrote in part (i) but lets see if I understand this:

I'll start you off with the easier direction:

Proof (ii) implies (i)

If J is an interval then f(J) is an interval,

so chose x, y in I, f(x) < f(y) and look at the interval [x,y] (or [y,x])

By assumption, f([x,y]) is an interval that obviously contains (if not is equal to) the interval [f(x), f(y)]. Thus, if c is in (f(x), f(y)), then c is in f([x,y]), and so there must exists a z in (x,y) such that f(z) = c.