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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.

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  • $\begingroup$ Hint: consider a maximal subset of $\bigcup D$ which is in $K$ (why does it exist? why is it the whole thing?) $\endgroup$ – Wojowu Jan 22 at 20:39
  • $\begingroup$ You threw out the hypothesis of closed for upper directed families. $\endgroup$ – William Elliot Jan 23 at 4:00
  • $\begingroup$ Burris and Sankappanavar give no reference for that "result of set theory"? Do they cite Mayer-Kalkschmidt & Steiner at all? $\endgroup$ – bof Feb 19 at 5:57
  • $\begingroup$ They do not cite it at all. $\endgroup$ – Daniel Kawai Feb 27 at 23:50
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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 $\bigcup D\notin K$.

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