< [[nlab:Sandbox]] Does LPO imply the following statement: > Let $\Sigma$ be the initial $\sigma$-frame. For any set $A$, any entire $\Sigma$-open relation $R:A \times \mathbb{2} \to \Sigma$ from $A$ to the boolean domain contains a functional entire relation, i.e. there exists a function $f:A \to \mathbb{2}$ such that for all $x$ in $A$ $R(x, f(x)) = \top$. \begin{theorem} Any coequalizer of two non-intersecting $\Sigma$-open injections has a section if and only if every $\Sigma$-open subset is a decidable subset. \end{theorem} Suppose LPO holds. This means that every $\Sigma$-open subset $B$ of a set $A$ is a [[decidable subset]] with complement $\overline{B}$. Suppose we have two $\Sigma$-open subsets $B \subseteq A$ and $C \subseteq A$ which are isomorphic $B \simeq C$ and disjoint $B \cap C = \emptyset$. This implies that $A$ is the disjoint union of $B$ and $C$ and $\overline{B} \cap \overline{C}$, which is isomorphic to the disjoint union $B + B + (\overline{B} + \overline{C})$ by the isomorphism $B \simeq C$. [[!redirects Sandbox &gt; history]]