This is a follow up question to the question I asked here: The range of a continuous function on the order topology is convex
Let $(X,\mathcal{T})$ denote the unit square with the lexicographical order. Suppose that $f:[0,1] \rightarrow X$ is continuous, where $[0,1]$ has the usual topology and $f:(0) = (0,0)$ and $f(1) = (1,1)$
a. Show that $f$ is surjective.
b. Show that there are no functions with properties described above.
Attempt at a solution: a.Since the range of a continuous function from the usual topology to a linear order topology is convex as proven in the linked question and since $f:(0) = (0,0)$ and $f(1) = (1,1)$ then $(f(0), f(1)) \subseteq f([0,1])$ for the interval defined over $X$, so for every $p \in X \ \exists \ x \in [0,1]$ s.t $f(x) = p$ and hence, $f$ is surjective.
b. I think this function won't be continuous, for given $\left|x - y\right| < \delta$, $f(x) < f(y)$ where $f(y)$ is of the form $(a, 0)$; $\left|f(y) - f(x)\right| < \varepsilon$ since $f(y)$ does not have an immediate predecessor.
Your argument for surjectivity is fine, but your argument for (b) doesn’t really make sense: no point in either ordered set has an immediate predecessor. Here’s a pretty big HINT:
For each $x\in[0,1]$ let $U_x=\{x\}\times(0,1)$; then $U_x$ is open in $W$, and the sets $U_x$ are pairwise disjoint. Now suppose that $f$ is continuous, and consider the sets $f^{-1}[U_x]$ for $x\in[0,1]$.