Let $K$ denote the class of all cardinal numbers. Then a poset admits all joins with cardinality in $K$ iff it admits all meets with cardinality in $K$.
Question. For which other classes of cardinal numbers $K$ does this equivalence hold?
I know, for example, that if $K$ is replaced with the finite non-zero cardinals, the equivalence fails. Consider, for example, the poset $\{0,a,b\}$ with $0 < a$ and $0 < b.$
But, what happens if $K$ is something else? Like, all finite cardinals, including zero? Or what if $K = \{0,1,2,\cdots\} \cup \{\aleph_0\}$?
What's the general principle?
I'm going to be making the standard assumption that if a poset has joins (resp. meets) of sets of cardinality $\kappa$, then it also has joins (resp. meets) of any non-zero cardinality smaller than $\kappa$. Notice that $1$ can be added or removed from any such class without altering the property, so we'll assume $1\notin K$ by default. It follows thus that a set of cardinalities $K$ as above is completely determined by the least cardinal in it and by either being unbounded, or by the least upper bound of it. We will show that if $K\ne \emptyset$ (which is a trivial possibility), then $0\in K$ and $K$ is unbounded, from which will follow that $K$ is the class of all cardinalities.
Let $K\ne \emptyset$ be such class of cardinalities. Assume $0\notin K$ and let $\kappa\ne 1$ be a cardinality in $K$. Let $S$ be a set of cardinality $\kappa$ and let $V=\mathcal P_*(S)$ be the set of all non-empty subsets of $S$. Clearly, $V$ has all joins except the empty join, so it has all joins of cardinalities from $K$. But, $V$ does not have meets of cardinality $\kappa$, since for instance, the set of all singletons from $S$ has cardinality $\kappa$, but no meet in $V$. So, we conclude that $0\in K$.
Assume that $K$ is bounded by some infinite cardinal $\kappa$. Let $S$ be a set of cardinality $2^\kappa$ and consider $V=\mathcal P_\sharp(S)$, the set of all subsets of $S$ of cardinality $\le \kappa$. Clearly, $V$ has all unions of cardinality $\le \kappa$, and thus all joins of cardinalities from $K$. But, $V$ does not have empty meets since $V$ clearly does not contain a top element. This contradicts $0\in K$ and thus we conclude that $K$ is not bounded.
(not really necessary, but just to take care of the different flavour of finite cardinals, in case you was another argument why $K$ must contain all finite cardinals: Let $\kappa$ be the least cardinal in $K-\{0,1\}$. Assume $\kappa>2$. But then let $S$ be a doubleton and $V=\mathcal P(S)-\{S\}$. Then $V$ has empty joins (the bottom element) and (vacuously) all joins of other cardinalities from $K$. But, $V$ does not have empty meets, again a contradiction. So $2\in K$. However, since the claim "a poset has binary joins/meets" is equivalent to "a poset has all finite joins/meets", it follows that all finite non-empty cardinals are in $K$.)
To conclude, the only possible classes of such cardinalities are $\emptyset$, $\{1\}$, and the class of all cardinalities.
I hope my answer does not include terribly redundant things because I really love the question.