The following is taken from a set of lecture notes found here Lecture Notes 1: The Language of Sentential Logic. Bold face emphasis added by me.

Let $X$ be a set, and let $\mathscr{P}X=\left\{ A\vert A\subseteq X\right\} $ be the powerset of $X$. Let $\Phi:\mathscr{P}X\to\mathscr{P}X$ be a monotone operation, i.e., for all $A,B\in\mathscr{P}X$, we have $A\subseteq B$ implies $\Phi\left(A\right)\subseteq\Phi\left(B\right)$.

Definition. We say that a set $A\subseteq X$ is closed under $\Phi$ if $\Phi\left(A\right)\subseteq A$.

Proposition 1. (a) Let $\left(A_{i}\right)_{i\in I}$ be a family of sets such that each $A_{i}$ is closed under $\Phi$. Then the intersection $A=\bigcap_{i\in I}A_{i}$ is also closed under $\Phi$.

(b) If $B\subseteq X$ is any set, then there exists a smallest subset $\bar{B}\subseteq X$ such that $B\subseteq\bar{B}$ and $\bar{B}$ is closed under $\Phi$.

Proof. (a) Let $i\in I$ be arbitrary. By hypothesis, we have that $A_{i}$ is closed under $\Phi$, so $\Phi\left(A_{i}\right)\subseteq A_{i}$. Since $A\subseteq A_{i}$ for all $i$, we have $\Phi\left(A\right)\subseteq\Phi\left(A_{i}\right)$ by monotonicity of $\Phi$. Hence $\Phi\left(A\right)\subseteq A_{i}$. Since $i$ was arbitrary, it follows that $\Phi\left(A\right)=\bigcap_{i\in I}A_{i},$ therefore $\Phi\left(A\right)\subseteq A$ and $A$ is closed under $\Phi$.

(b) This is a trivial consequence of (a). Let $\bar{B}$ be the intersection of all sets $A\in\mathscr{P}X$ such that $B\subseteq A$ and $A$ is closed under $\Phi$. By (a), $\bar{B}$ is itself closed under $\Phi$, and it also contains $A$. Since it is the intersection of all such sets, it is therefore the smallest with those properties.

There are two conclusions which I am not understanding. The first is in the proof of (a) where it is stated that $\Phi\left(A\right)=\bigcap_{i\in I}A_{i}.$

Is equality justified here, or should it be $\Phi\left(A\right)\subseteq\bigcap_{i\in I}A_{i}?$

The second problem is the claim in the proof of part (b) that $\bar{B}$ contains $A$. Should $A$ be replaced with $B$ in that statement?

  • $\begingroup$ Looking at the link, I don't see the equality statement. $\endgroup$ Commented Feb 7, 2019 at 15:31
  • $\begingroup$ Indeed. My mistake. I had originally wanted to ask about part (b) and transcribed the text for that purpose. I obviously made a mistake in doing so. $\endgroup$ Commented Feb 7, 2019 at 15:45

1 Answer 1


Equality is not justified here; once you know this counterexamples are easy to construct by considering the simplest case, where $I=\{i\}$ is a singleton and $\Phi(A_i)$ is a proper subset of $A_i$. Or another simplest case, where $\Phi(A)=\varnothing$ for all $A\subset X$, and $X\neq\varnothing$.

And indeed $A$ should be replaced with $B$ in that part of the statement. As it stands, the statement makes no sense as no particular $A$ has been specified.


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