# Analysis - Continuous Functions

• Mar 6th 2010, 12:14 PM
danilo
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
• Mar 6th 2010, 09:50 PM
Drexel28
Quote:

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.
• Mar 7th 2010, 09:26 AM
southprkfan1
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.