Multivariate Calculus: continuous functions I was reading about continuity and i saw this problem as a exercise but I cannot find a way to prove it. 
Suppose that $f$ is continuous on a region and $f$ is different of zero, show that $f$ has only one sign. 
Thanks 
 A: A region in this context is a connected open set $\Omega\subset{\mathbb R}^n$. It is well known that a region is automatically path connected: For any two points  $a$, $b\in\Omega$ there is a continuous curve
$$\gamma:\quad [0,1]\to\Omega$$
with $\gamma(0)=a$, $\gamma(1)=b$. If $f(a)\,f(b)<0$ then the continuous auxiliary function $\phi(t):=f\bigl(\gamma(t)\bigr)$ would have a zero at some $\tau\in\>]0,1[\>$, hence  for $\xi:=\gamma(\tau)\in\Omega$ we would have $f(\xi)=0$, contrary to assumption.
A: If your definition of “region” is such that each region is connected and if your function is real-valued, you can prove it as follows: since your region $R$ is connected, then $f(R)$ is connected too, and the connected non-empty subsets of $\mathbb R$ are the intervals. So, $f(R)$ is an interval to which $0$ does not belong and therefore either $f(R)\subset(0,+\infty)$ or $f(R)\subset(-\infty,0)$.
Otherwise, your statment is false. Just let $R^\star$ be a connected component of $R$ and consider the continuous map$$\begin{array}{rccc}f\colon&R&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}1&\text{ if }x\in R^\star\\-1&\text{ otherwise.}\end{cases}\end{array}$$
