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 $f\colon(a,\,b)\to\mathbb{Z}$ is a nonconstant function that is continuous on its domain. Then there must exist $c$ and $d$ in $(a,\,b)$ such that $f(c)\neq f(d)$ (for, otherwise, $f$ would be constant). Choose any noninteger $k$ between $f(c)$ and $f(d).$ Then, since $f$ is a function of a real variable that is continuous on the interval $[c,\,d],$ we may apply the intermediate value theorem to conclude that there is some $e\in[c,\,d]$ such that $f(e)=k.$ But this would mean that the noninteger $k$ is in the codomain of $f,$ which is a contradiction. $\square$

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. $f(c)$ and $f(d)$ 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 $f$ must take on that value somewhere (which is impossible, since $f$ only gives integral values).