Why do we need surjectivity in this theorem?

In class we proved the following theorem:

Given $X_1,X_2$ ordered sets. Then any surjective increasing $\phi: X_1 \to X_2$ is continuous wrt the interval topology on $X_1$ and $X_2$.

I was asked to find an example to prove that the surjectivity condition above is essential: What does this mean? Do I need to find an increasing $f$ which is not surjective but continuous? How can I construct one?

Solutions Collecting From Web of "Why do we need surjectivity in this theorem?"

The problem is asking you to find a strictly increasing map $\phi : X_1 \rightarrow X_2$ that isn’t continuous, and to prove that $\phi$ isn’t continuous. The phrasing of the question offers an important hint, namely that you should be thinking about non-surjective choices of $\phi$, since no surjective choice is going to work.

Here’s a pretty substantial hint. Take $X_1$ equal to $\mathbb{N}$ with an element $a_0$ adjoined “at the end”, and $X_2$ equal to $\mathbb{N}$ with two elements $b_0$ and $b_1$ adjoined “at the end,” such that $b_0 \leq b_1$. Let $\phi : X_1 \rightarrow X_2$ be the map that takes $a_0$ to $b_1$, while leaving all the natural numbers unchanged. Now prove that $\phi$ isn’t continuous.

Edit: I now see that I misunderstood what you mean by increasing; you mean what I would call “strictly increasing.” I’ll leave this answer here anyway, since it shows that we cannot weaken this condition in the theorem.

Original answer. This isn’t true even if we assume surjectivity. Let $P = \{-1,1\}$ denote the ordered set such that $-1 \leq 1$. Then every subset of $P$ is open with respect to the interval topology. Now define $f : \mathbb{R} \rightarrow P$ by asserting that $f(x) = 1$ iff $x$ is positive. Then despite that $\{-1\} \subseteq P$ is open, nonetheless $f^{-1}(\{-1\})$ is not.