# Does lub-preservation imply monotonicity?

A function $f$ is continuous exactly when it is

• monotonic, i.e. $d\sqsubseteq d' \Rightarrow f(d)\sqsubseteq f(d')$, and
• lub-preserving, i.e. for any chain $d_0\sqsubseteq d_1 \sqsubseteq \dots$ we have $f(\bigsqcup_n d_n) = \bigsqcup_n f(d_n)$ (call this (1))

We can demonstrate that the first property is implied by the second (and is thus unnecessary) by reasoning as follows. Suppose $f$ is lub-preserving.

Lub-preservation is a property of all chains, so let’s consider the particular chain $d \sqsubseteq d'$. The lub of this chain is $d'$, so the LHS of (1) becomes $f(d')$. The RHS is $f(d) \sqcup f(d')$, and since this is required to be equal to $f(d')$ we can deduce that $f(d) \sqsubseteq f(d')$. We have deduced monotonicity.

Sadly, the reasoning is invalid. Since $\sqcup$ is a binary operation on chains rather than just sets, it might not be defined for $f(d) \sqcup f(d')$. We could respond by saying that it must be defined because it’s equal to $f(d')$! This, however, is dependent on your definition of equality. A perfectly sensible definition of equality between $x$ and $y$ could be: if $x$ and $y$ are defined, then the values at which they are defined are the same. This definition makes $x=y$ vacuously true should either be undefined.

In conclusion: the answer is no; to show continuity we need to show both monotonicity and lub-preservation.