# Is this a correct proof of the closure proposition for subsets of a powerset?

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?

• Looking at the link, I don't see the equality statement. Commented Feb 7, 2019 at 15:31
• 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. Commented Feb 7, 2019 at 15:45

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.