Proposition 7

Let $(x_\iota)_{\iota \in I}$ be a family of elements of an ordered set $\mathrm{E}$, and let $(\mathrm{J}_\lambda)_{\lambda \in L}$ be a covering of the index set $\mathrm{I}$. Suppose that each of the subfamilies $(x_\iota)_{\iota \in J_\lambda}$ has a least upper bound in $\mathrm{E}$. For the family $(x_\iota)_{\iota \in I}$ to have a least upper bound in $\mathrm{E}$, it is necessary and sufficient that the family $\left( \sup\limits_{\iota \in J_\lambda} x_\iota \right)_{\lambda \in L}$ should have a least upper bound in $\mathrm{E}$, and then we have

Let $b_\lambda = \sup\limits_{\iota \in J_\lambda} x_\iota$. Suppose that $(x_\iota)_{\iota \in I}$ has a least upper bound $a$. Then $a \geqslant b_\lambda$ for each $\lambda \in \mathrm{L}$ (Corollary to Proposition 5). On the other hand, if $c \geqslant b_\lambda$ for each $\lambda \in \mathrm{L}$, then we have $c \geqslant x_\iota$ for each $\iota \in \mathrm{I}$, because $(\mathrm{J}_\lambda)_{\lambda \in L}$ is a covering of $\mathrm{I}$; hence $c \geqslant a$, which proves that

Conversely, suppose that the family $(b_\lambda)_{\lambda \in L}$ has a least upper bound $a'$. Then $a' \geqslant x_\iota$ for all $\iota \in \mathrm{I}$. On the other hand, if $c' \geqslant x_\iota$ for all $\iota \in \mathrm{I}$, then we have in particular

for each $\lambda \in \mathrm{L}$, so that $c' \geqslant a'$ and therefore