1. ## Intermediate Value Theorem

Hi,

I would like to know how to proof the following:
Let f : (a,b) --> Z be continuous on (a,b) where Z is the integers.
Show that f must be a constant function.

I know we have to suppose f is nonconstant but next, how do we use the Intermediate Value Theorem to show the complete proof?

2. Originally Posted by zxcv
I would like to know how to proof the following:
Let f : (a,b) --> Z be continuous on (a,b) where Z is the integers.
Show that f must be a constant function.
We use proof by contradiction. Suppose that $\displaystyle f\colon(a,\,b)\to\mathbb{Z}$ is a nonconstant function that is continuous on its domain. Then there must exist $\displaystyle c$ and $\displaystyle d$ in $\displaystyle (a,\,b)$ such that $\displaystyle f(c)\neq f(d)$ (for, otherwise, $\displaystyle f$ would be constant). Choose any noninteger $\displaystyle k$ between $\displaystyle f(c)$ and $\displaystyle f(d).$ Then, since $\displaystyle f$ is a function of a real variable that is continuous on the interval $\displaystyle [c,\,d],$ we may apply the intermediate value theorem to conclude that there is some $\displaystyle e\in[c,\,d]$ such that $\displaystyle f(e)=k.$ But this would mean that the noninteger $\displaystyle k$ is in the codomain of $\displaystyle f,$ which is a contradiction. $\displaystyle \square$

3. Hi,

I was just wondering if we can choose a non integer k between f(c) and f(d) since f is (a,b)-->Z and Z is the integers.

4. Originally Posted by zxcv
Hi,

I was just wondering if we can choose a non integer k between f(c) and f(d) since f is (a,b)-->Z and Z is the integers.
I am not sure I understand what you are asking. $\displaystyle f(c)$ and $\displaystyle f(d)$ must be distinct integers, and there are infinitely many real numbers between any two given integers. Pick one of them, and the intermediate value theorem says that $\displaystyle f$ must take on that value somewhere (which is impossible, since $\displaystyle f$ only gives integral values).