Discontinuity points in linearly ordered space

Question

It is a simple well known result that given a monotonic function $f:\mathbb{R}\to \mathbb{R}$, its discontinuity points can only be jump discontinuities. My question is probably a simple one. Given an order preserving map $f:(X,\mathcal{T}_X)\to (Y,\mathcal{T}_Y)$, where these space are linearly topological ordered spaces, can we say that a discontinuity point $x_0\in X$ of $f$ is a jump discontinuity in general? i.e., can we say that $$ \underset{x<x_0}{\sup} f(x)< \underset{x>x_0}{\inf} f(x)? $$

Do we need to assume further that $Y$ is a complete lattice, or is this true in the general case?

Answer 1

Hint/idea : for any order preserving function $f$ we already know, for any $x_0$ (which is not the minimal or maximal element of $X$) that

$$\sup_{x < x_0} f(x) \le f(x_0) \le \inf_{x > x_0} f(x)\tag{1}$$

So it suffices to show that if equality holds between the leftmost and rightmost term, $f$ is in fact continuous at $x_0$. I think this is not too hard using the order topology definitions and order preservingness.

Answer 2

I'll try to write an answer following the hint of Henno Brandsma, hoping that if it is wrong someone will point to a mistake.

Attempt:

We assume that

$$ \sup_{x < x_0} f(x) = f(x_0) = \inf_{x > x_0} f(x) $$

Basic open sets in a linearly ordered topological space $Z$ are of the form

$$ L_Z(z):=\{ z'\in X: z'<z \} \quad \text{and} \quad U_Z(z):=\{ z'\in Z: z'>z \}. $$

It is enough to check continuity against basic open sets in $Y$. Hence, it will suffice to show that:

• $f^{-1}\big[ L_Y(u) \big]$ contains an open set of the form $L_X(x_u)$ for $u>f(x_0)$ when $\inf_{x>x_0}f(x)=f(x_0)$ and
• $f^{-1}\big[ U_Y(\ell) \big]$ contains an open set of the form $U_X(x_\ell)$ for $\ell<f(x_0)$ when $\inf_{x>x_0}f(x)=f(x_0)$.

Since $u>f(x_0)$, there exists $x_u>x_0$ such that $u>f(x_u)\geq f(x_0)$. Since $f$ is order preserving, we conclude that

$$ x_0\in L_X(x_u) \subseteq f^{-1}\big[ L_Y(u) \big] . $$

By similar arguments, for all $\ell<f(x_0)$ there exists $x_\ell<x_0$ satisfying $\ell< f(x_\ell)\leq f(x_0)$. Using again the fact that $f$ is order preserving yields

$$ x_0\in U_X(x_\ell) \subseteq f^{-1}\big[ U_Y(\ell) \big]. $$

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Does this argument seem flawed to anyone? Also I use the fact that sets bounded from above have a supremum and sets bounded from below have an infimum. Does anyone think I can omit this condition?