Let $X$ be a set and let $\mathcal{U}\in \mathcal{P}(\mathcal{P}(X))$.
Since $\mathcal{P}(X)$ is posetal, it suffices to prove the condition $(\star )$. So let $\mathcal{U},\mathcal{V}\in \mathcal{P}(\mathcal{P}(X))$ with $\mathcal{U}\subset \mathcal{V}$. We claim that
\[ \bigcap _{V\in \mathcal{V}}V\subset \bigcap _{U\in \mathcal{U}}U. \]
Indeed, if $x\in \bigcap _{V\in \mathcal{V}}V$, then $x\in V$ for all $V\in \mathcal{V}$. But since $\mathcal{U}\subset \mathcal{V}$, it follows that $x\in U$ for all $U\in \mathcal{U}$ as well. Thus $x\in \bigcap _{U\in \mathcal{U}}U$, which gives our desired inclusion.
We have
\begin{align*} \bigcap _{A\in \mathcal{A}}\left(\bigcap _{U\in A}U\right) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ x\in X\ \middle |\ \begin{aligned} & \text{for each $A\in \mathcal{A}$,}\\ & \text{we have $x\in \bigcap _{U\in A}U$} \end{aligned} \right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ x\in X\ \middle |\ \begin{aligned} & \text{for each $A\in \mathcal{A}$ and each}\\ & \text{$U\in A$, we have $x\in U$} \end{aligned} \right\} \\ & = \left\{ x\in X\ \middle |\ \begin{aligned} & \text{for each $U\in \displaystyle \bigcup _{A\in \mathcal{A}}A$,}\\ & \text{we have $x\in U$} \end{aligned} \right\} \\ & \subset \left\{ x\in X\ \middle |\ \begin{aligned} & \text{for each $U\in \displaystyle \bigcap _{A\in \mathcal{A}}A$,}\\ & \text{we have $x\in U$} \end{aligned} \right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\bigcap _{U\in {\scriptsize \displaystyle \bigcap _{A\in \mathcal{A}}A}}U. \end{align*}
Since $\mathcal{P}(X)$ is posetal, naturality is automatic (Chapter 11: Categories, Item 4 of Proposition 11.2.7.1.2). This finishes the proof.
We have
\begin{align*} \bigcap _{V\in \left\{ U\right\} }V & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ x\in X\ \middle |\ \begin{aligned} & \text{for each $V\in \left\{ U\right\} $,}\\ & \text{we have $x\in U$} \end{aligned} \right\} \\ & = \left\{ x\in X\ \middle |\ x\in U \right\} \\ & = U.\end{align*}
This finishes the proof.
If $U=\text{Ø}$, then we have
\begin{align*} \bigcap _{\left\{ u\right\} \in \chi _{X}(U)}\left\{ u\right\} & = \bigcap _{\left\{ u\right\} \in \text{Ø}}\left\{ u\right\} \\ & = X,\end{align*}
so $\bigcap _{\left\{ u\right\} \in \chi _{X}(U)}\left\{ u\right\} =X\neq \text{Ø}=U$. When $U$ is nonempty, we have two cases:
-
1.
If $U$ is a singleton, say $U=\left\{ u\right\} $, we have
\begin{align*} \bigcap _{\left\{ u\right\} \in \chi _{X}(U)}\left\{ u\right\} & = \left\{ u\right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}U.\end{align*}
-
2.
If $U$ contains at least two elements, we have
\begin{align*} \bigcap _{\left\{ u\right\} \in \chi _{X}(U)}\left\{ u\right\} & = \text{Ø}\\ & \subset U.\end{align*}
This finishes the proof.
Item 5: Interaction With Unions I
We have
\begin{align*} \bigcap _{W\in \mathcal{U}\cup \mathcal{V}}W & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ x\in X\ \middle |\ \begin{aligned} & \text{for each $W\in \mathcal{U}\cup \mathcal{V}$,}\\ & \text{we have $x\in W$} \end{aligned} \right\} \\ & = \left\{ x\in X\ \middle |\ \begin{aligned} & \text{for each $W\in \mathcal{U}$ and each}\\ & \text{$W\in \mathcal{V}$, we have $x\in W$} \end{aligned} \right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ x\in X\ \middle |\ \begin{aligned} & \text{for each $W\in \mathcal{U}$,}\\ & \text{we have $x\in W$} \end{aligned} \right\} \\ & \phantom{={}}\mkern 4mu\cap \left\{ x\in X\ \middle |\ \begin{aligned} & \text{for each $W\in \mathcal{V}$,}\\ & \text{we have $x\in W$} \end{aligned} \right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left(\bigcap _{W\in \mathcal{U}}W\right)\cap \left(\bigcap _{W\in \mathcal{V}}W\right)\\ & = \left(\bigcap _{U\in \mathcal{U}}U\right)\cap \left(\bigcap _{V\in \mathcal{V}}V\right). \end{align*}
This finishes the proof.
Item 6: Interaction With Unions II
Omitted.
Item 7: Interaction With Intersections I
We have
\begin{align*} \left(\bigcap _{U\in \mathcal{U}}U\right)\cap \left(\bigcap _{V\in \mathcal{V}}V\right) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ x\in X\ \middle |\ \begin{aligned} & \text{for each $U\in \mathcal{U}$,}\\ & \text{we have $x\in U$} \end{aligned} \right\} \\ & \phantom{={}}\mkern 4mu\cup \left\{ x\in X\ \middle |\ \begin{aligned} & \text{for each $V\in \mathcal{V}$,}\\ & \text{we have $x\in V$} \end{aligned} \right\} \\ & = \left\{ x\in X\ \middle |\ \begin{aligned} & \text{for each $W\in \mathcal{U}\cap \mathcal{V}$,}\\ & \text{we have $x\in W$} \end{aligned} \right\} \\ & \subset \left\{ x\in X\ \middle |\ \begin{aligned} & \text{for each $W\in \mathcal{U}\cup \mathcal{V}$,}\\ & \text{we have $x\in W$} \end{aligned} \right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\bigcap _{W\in \mathcal{U}\cap \mathcal{V}}W.\end{align*}
Since $\mathcal{P}(X)$ is posetal, naturality is automatic (Chapter 11: Categories, Item 4 of Proposition 11.2.7.1.2). This finishes the proof.
Item 8: Interaction With Intersections II
Omitted.
Item 9: Interaction With Differences
Let $X=\left\{ 0,1\right\} $, let $\mathcal{U}=\left\{ \left\{ 0\right\} ,\left\{ 0,1\right\} \right\} $, and let $\mathcal{V}=\left\{ \left\{ 0\right\} \right\} $. We have
\begin{align*} \bigcap _{W\in \mathcal{U}\setminus \mathcal{V}}U & = \bigcap _{W\in \left\{ \left\{ 0,1\right\} \right\} }W\\ & = \left\{ 0,1\right\} , \end{align*}
whereas
\begin{align*} \left(\bigcap _{U\in \mathcal{U}}U\right)\setminus \left(\bigcap _{V\in \mathcal{V}}V\right) & = \left\{ 0\right\} \setminus \left\{ 0\right\} \\ & = \text{Ø}. \end{align*}
Thus we have
\[ \bigcap _{W\in \mathcal{U}\setminus \mathcal{V}}W=\left\{ 0,1\right\} \neq \text{Ø}=\left(\bigcap _{U\in \mathcal{U}}U\right)\setminus \left(\bigcap _{V\in \mathcal{V}}V\right). \]
This finishes the proof.
Item 10: Interaction With Complements I
Let $X=\left\{ 0,1\right\} $ and let $\mathcal{U}=\left\{ \left\{ 0\right\} \right\} $. We have
\begin{align*} \bigcap _{W\in \mathcal{U}^{\textsf{c}}}U & = \bigcap _{W\in \left\{ \text{Ø},\left\{ 1\right\} ,\left\{ 0,1\right\} \right\} }W\\ & = \text{Ø}, \end{align*}
whereas
\begin{align*} \bigcap _{U\in \mathcal{U}}U^{\textsf{c}} & = \left\{ 0\right\} ^{\textsf{c}}\\ & = \left\{ 1\right\} . \end{align*}
Thus we have
\[ \bigcap _{W\in \mathcal{U}^{\textsf{c}}}U=\text{Ø}\neq \left\{ 1\right\} =\bigcap _{U\in \mathcal{U}}U^{\textsf{c}}. \]
This finishes the proof.
Item 11: Interaction With Complements II
This is a repetition of Item 12 of Proposition 4.3.6.1.2 and is proved there.
Item 12: Interaction With Complements III
This is a repetition of Item 11 of Proposition 4.3.6.1.2 and is proved there.
Item 13: Interaction With Symmetric Differences
Let $X=\left\{ 0,1\right\} $, let $\mathcal{U}=\left\{ \left\{ 0,1\right\} \right\} $, and let $\mathcal{V}=\left\{ \left\{ 0\right\} ,\left\{ 0,1\right\} \right\} $. We have
\begin{align*} \bigcap _{W\in \mathcal{U}\mathbin {\triangle }\mathcal{V}}W & = \bigcap _{W\in \left\{ \left\{ 0\right\} \right\} }W\\ & = \left\{ 0\right\} , \end{align*}
whereas
\begin{align*} \left(\bigcap _{U\in \mathcal{U}}U\right)\mathbin {\triangle }\left(\bigcap _{V\in \mathcal{V}}V\right) & = \left\{ 0,1\right\} \mathbin {\triangle }\left\{ 0\right\} \\ & = \text{Ø}, \end{align*}
Thus we have
\[ \bigcap _{W\in \mathcal{U}\mathbin {\triangle }\mathcal{V}}W=\left\{ 0\right\} \neq \text{Ø}=\left(\bigcap _{U\in \mathcal{U}}U\right)\mathbin {\triangle }\left(\bigcap _{V\in \mathcal{V}}V\right). \]
This finishes the proof.
Item 14: Interaction With Internal Homs I
This is a repetition of Item 10 of Proposition 4.4.7.1.3 and is proved there.
Item 15: Interaction With Internal Homs II
This is a repetition of Item 11 of Proposition 4.4.7.1.3 and is proved there.
Item 16: Interaction With Internal Homs III
This is a repetition of Item 12 of Proposition 4.4.7.1.3 and is proved there.
Item 17: Interaction With Direct Images
This is a repetition of Item 4 of Proposition 4.6.1.1.5 and is proved there.
Item 18: Interaction With Inverse Images
This is a repetition of Item 4 of Proposition 4.6.2.1.3 and is proved there.
Item 19: Interaction With Codirect Images
This is a repetition of Item 4 of Proposition 4.6.3.1.7 and is proved there.
Item 20: Interaction With Unions of Families I
This is a repetition of Item 20 of Proposition 4.3.6.1.2 and is proved there.
Item 21: Interaction With Unions of Families II
This is a repetition of Item 21 of Proposition 4.3.6.1.2 and is proved there.