Let $X$ be a finite set. $\mathcal{P}(X)$ denotes the set of all subsets of $X$. Let $\Gamma$ be a sub-lattice of $\mathcal{P}(X)$, i.e. $\Gamma$ is a collection of subsets of $X$ closed under union and intersection.

Now let $f:\Gamma\to\mathbb{R}$ be an increasing function, i.e. for all $A,B\in\Gamma,\ A\subseteq B\Rightarrow f(A)\leq f(B)$.

Is it always possible to extend $f$ to an increasing function $\widetilde f\!:\mathcal{P}(X)\to\mathbb{R}$?

I found this statement in a paper, but I'm not aible to prove it (any attempt I did was wrong) and on the other hand I'm not aible to find a counter-example.


1 Answer 1


The answer should be yes. It suffices to define

$$\widetilde f:\mathcal{P}(X)\to\mathbb{R}\,,\quad \widetilde f(A)=f(\tilde A)\quad\text{with }\tilde A:=\bigcap\{B\in\Gamma\,|\,B\supseteq A\} \;.$$

$\tilde A\in\Gamma$ thanks to the lattice property. If $A\in\Gamma$ then $\tilde A= A$. It is easy to check that $\tilde f$ is increasing if $f$ is increasing.


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