Theorem   Let $f:X\rightarrow Y$ be a continuous function with $X$ a connected space and $Y$ a totally ordered set in the order topology. If $x_1,x_2\in X$ and $y\in Y$ lies between $f(x_1)$ and $f(x_2)$ , then there exists $x\in X$ such that $f(x)=y$ .

Proof. The sets $U=f(X)\cap(-\infty,y)$ and $V=f(X)\cap(y,\infty)$ are disjoint open subsets of $f(X)$ in the subspace topology, and they are both non-empty, as $f(x_1)$ is contained in one and $f(x_2)$ is contained in the other. If $y\notin f(X)$ , then $U\cup V$ constitutes a separation of the space $f(X)$ , contradicting the hypothesis that $f(X)$ is the continuous image of the connected space $X$ . Thus there must exist $x\in X$ such that $f(x)=y$ .