# preimage of a continous function is an interval

• Oct 3rd 2013, 08:34 PM
director
preimage of a continous function is an interval
To prove: a continuous real-valued function (f) defined on set S has an image (S') that is an interval then S must be an interval.

Tools: definition of continuity, definition of a connected set, connected set in R is an interval,

If S' is not a singleton; if $f(a), f(b) \in S'$ such that $f(a) < f(b)$ and $f(c) \in S'$ such that $f(a) < f(c) < f(b)$, can it be shown that $a < c < b$?
I'm looking for something to get me started. Thanks.
• Oct 4th 2013, 06:45 AM
johng
Re: preimage of a continous function is an interval
Hi,
I don't think you've stated your question correctly. Given what you said:
Let f be a continuous function with domain S (subset of R) and image S' = f(S). If S' is an interval, then S is an interval.

This is clearly false:

Let $S=[-2,-1]\cup[1,2]$ and f(x)=x2. Then f(S)=[1,4], but S is not an interval.

Did you omit the hypothesis that f is one to one? But if so, the problem is trivial as you point out.
• Oct 4th 2013, 02:38 PM
director
Re: preimage of a continous function is an interval
Thanks John, it's possible that I made a mistake. Here's the original problem statement:

Quote:

A non-empty set S of real numbers is an interval <=> every continuous real-valued function on S has an interval as its image
• Oct 4th 2013, 03:00 PM
Plato
Re: preimage of a continous function is an interval
Quote:

Originally Posted by director
Thanks John, it's possible that I made a mistake. Here's the original problem statement:
A non-empty set S of real numbers is an interval <=> every continuous real-valued function on S has an interval as its image

It is certainly in one direction.
Theorem: The continuous image of a connected set is a connected set.
Now in the real numbers intervals are connected.
Thus if $\mathcal{I}$ is an interval and $f$ is continuous then $f(\mathcal{I})$ is an interval.

Now you have been given a counterexample for the other direction.
• Oct 4th 2013, 06:24 PM
director
Re: preimage of a continous function is an interval
I understand. I didn't give much thought to the possibility of the second part being false. The counter example is clear. Thanks.
• Oct 4th 2013, 06:42 PM
johng
Re: preimage of a continous function is an interval
Hi again,
As Plato points out, the main implication is: if S is an interval and f is continuous on S, then f(S) is an interval.
However, the converse as written is trivially true: if every continuous function f on S has f(S) an interval, then S is an interval -- take f(x) = x.

Kind of a funky statement, but it's true.