Is it possible to extend any poset to have least upper bound property?

Question

Is it possible to extend any poset to have least upper bound property?

For instance, we could extend natural number to real number which has lub by Dedekind Cut.

Like this, the extension step works for any poset with countable many elements?

Answer 1

The answer is yes: let $X$ be any partially ordered set. A subset $L \subseteq X$ is called a lower set of $X$, if

$$(\forall x \in X)(\forall y \in L) \big( x < y \Rightarrow x \in L \big).$$

Let $\overline{X}$ denote the set of all lower sets of $X$ and consider it ordered by inclusion. Then:

• $i : X \to \overline{X}$ given by $i(x) = \{ a \in X : a \leqslant x \}$ is injective and order preserving, i.e. for each $x, y \in X$ we have that $x < y \iff i(x) \subsetneq i(y)$.

• Each subset $S \subseteq \overline{X}$ has a least upper bound, namely: $\displaystyle \sup S = \bigcup S$.

The above construction shows that every partially ordered set can be embedded into a partially ordered set with least upper bound property. There is, however, one little problem: this embedding is not lub-preserving, i.e. suppose a subset $A \subseteq X$ already has a least upper bound in $X$. Then it may happen that

$$i \left( \sup_X A \right) \neq \sup_{\overline{X}} i[A].$$

We can slightly complicate the construction above to fix this problem: let $j : X \to \overline{X}$ be given by

$$j(x) = \{ a \in X : x \not \leqslant a \}.$$

Then again $j$ is injective and order preserving, but this time whenever $\sup A$ exists in $X$,

$$j \left( \sup_X A \right) = \sup_{\overline{X}} j[A],$$

which is proved by the following property: $\sup A \leqslant b \iff (\forall a \in A) \, a \leqslant b$.