Question. Let $P$ denote a poset with all binary joins, and all chain-shaped joins (of arbitrary cardinality).
Is $P$ necessarily a complete lattice?
Motivation. Let $f$ and $g$ denote closure operators on $P$. Then we get a new closure operator $f + g$ on $P$ as follows: given $x \in P$, we define $$(f + g)(x) = \bigvee_{\alpha : \mathbf{On}}(f \vee g)^\alpha(x)$$
It follows that $+$ is the binary join in the poset of closure operators. But this only works if $P$ has all binary joins, and also all chain-shaped joins. Hence the question.
Let $P$ be such a poset, and let $S\subseteq P$. Enumerate $S=\{x_\xi:\xi<\kappa\}$, where $\kappa=|S|$. For $\eta<\kappa$ let $S_\eta=\{x_\xi:\xi<\eta\}$. If $y_\eta=\bigvee S_\eta$, let $y_{\eta+1}=y_\eta\lor x_\eta$; then $y_{\eta+1}=\bigvee S_{\eta+1}$. Now suppose that $\eta\le\kappa$ is a limit ordinal, and we’ve defined $y_\xi$ for each $\xi<\eta$. Let $C_\eta=\{y_\xi:\xi<\eta\}$; $C_\eta$ is a chain, so we can let $y_\eta=\bigvee C_\eta$. The construction goes through to $\kappa$, and $y_\kappa=\bigvee S$.