Does closed under unions of chains imply closed under unions of upward directed families of sets?

Question

In the book "A Course in Universal Algebra" from Burris and Sankappanavar, in the section 1.5, during the exercises, there is something like that:

"Given a set $A$ and a family $K$ of subsets if $A$, $K$ is said to be closed under unions of chains if whenever $C\subseteq K$ and $C$ is a chain (under $\subseteq$) then $\bigcup C\in K$; and $K$ is said to be closed under unions of upward directed families of sets if whenever $D\subseteq K$ is such that $A_1,A_2\in D$ implies $A_1\cup A_2\subseteq A_3$ for some $A_3\in D$, then $\bigcup D\in K$. A result of set theory says that $K$ is closed under unions of chains iff $K$ is closed under unions of upward directed families of sets."

I tried to prove that every upward directed family $D$ has a cofinal chain $C$ by taking for example a maximal chain, but this is not true, since the family of finite subsets of an uncountable set does not have a cofinal chain.

I do not even know if the fact asserted in the book is correct.

Answer 1

That result was published by J. Mayer-Kalkschmidt and E. Steiner, Some theorems in set theory and applications in the ideal theory of partially ordered sets, Duke Math. J. 2 (1964), 287–289.
Inasmuch as I don't have access to the paper, which is hidden behind a paywall (the result is stated on the first page, which is freely available), the proof below may not be exactly the same as the one in the paper. I write "directed" for "directed upwards".

Lemma. If $D$ is a directed set of cardinality $\kappa$, an infinite cardinal, then $D$ is the union of a chain of directed subsets, each of cardinality less than $\kappa$.

Proof. We consider three cases.

Case 1. $\kappa=\aleph_0$.

Let $D=\{d_n:n\lt\omega\}$ be an enumeration of $D$. We will define a sequence of finite directed sets $D_n$. Let $D_0=\{d_0\}$. For $n\gt0$, having defined $D_{n-1}$, let $u_n$ be an upper bound for $D_{n-1}\cup\{d_n\}$, and let $D_n=D_{n-1}\cup\{d_n,u_n\}$. Then $\{D_n:n\lt\omega\}$ is a chain of finite directed sets whose union is $D$.

Case 2. $\kappa$ is an uncountable regular cardinal.

Let $D=\{d_\alpha:\alpha\in\kappa\}$. For $\beta\in\kappa$, let $D_\beta=\{d_\alpha:\alpha\lt\beta\}$. Then $B=\{\beta\in\kappa:D_\beta\text{ is directed}\}$ is unbounded in $\kappa$, so that $\{D_\beta:\beta\in B\}$ is a chain of directed sets whose union is $D$, and of course $|D_\beta|\lt\kappa$ for each $\beta$. (To see that $B$ is unbounded, given an ordinal $\beta_o\lt\kappa$, consider the limit of a sequence $\beta_0\lt\beta_1\lt\beta_2\lt\cdots\lt\beta_n\lt\cdots\lt\kappa$ where every finite subset of $D_{\beta_n}$ has an upper bound in $D_{\beta_{n+1}}$.)

Case 3. $\kappa$ is a singular cardinal.

Let $D=\bigcup_{\alpha\in\lambda}E_\alpha$ where $|E_\alpha|\lt\kappa$ and $\lambda=\operatorname{cf}\kappa$. Recursively define directed sets $D_\alpha\subseteq D\ (\alpha\in\lambda)$ so that $|D_\alpha|\lt\kappa$ and $E_\alpha\cup\bigcup_{\xi\lt\alpha}D_\xi\subseteq D_\alpha$. Then $\{D_\alpha:\alpha\in\lambda\}$ is a chain of directed sets whose union is $D$.

Theorem. If a family $K$ of sets is closed under unions of chains, then $K$ is closed under directed unions.

Proof. Assuming the contrary, let $D$ be a directed subfamily of $K$ of minimum cardinality whose union does not belong to $K$. Of course $D$ must be infinite. By the lemma, we can write $D=\bigcup_{\alpha\in\lambda}D_\alpha$, where $\{D_\alpha:\alpha\in\lambda\}$ is a chain of directed families, and $|D_\alpha|\lt|D|$ for each $\alpha$. By the minimality of $|D|$, we have $d_\alpha=\bigcup D_\alpha\in K$ for each $\alpha$. But then, since $\{d_\alpha:\alpha\in\lambda\}$ is a chain in $K$, we have $\bigcup D=\bigcup_{\alpha\in\lambda}d_\alpha\in K$, contradicting our assumption that $\bigcup D\notin K$.