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 $\displaystyle f(a), f(b) \in S'$ such that $\displaystyle f(a) < f(b)$ and $\displaystyle f(c) \in S'$ such that $\displaystyle f(a) < f(c) < f(b)$, can it be shown that $\displaystyle a < c < b$?

I'm looking for something to get me started. Thanks.

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 $\displaystyle S=[-2,-1]\cup[1,2]$ and f(x)=x^{2}. 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.

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

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 $\displaystyle \mathcal{I}$ is an interval and $\displaystyle f$ is continuous then $\displaystyle f(\mathcal{I})$ is an interval.

Now you have been given a **counterexample** for the other direction.

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.

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.