If F is a monotonic function then $F(Z) \subseteq Z$ where Z is the intersection of power sets. I have been presented the following problem:
Let $A$ be any set and $p(A)$ be the power set of A. Show that for any function $p(A)$ to $p(A)$, the set of all subsets X of A such that $F(X) \subseteq X$ is nonempty, and therefore the intersection Z of all such subsets exists. A function F from $p(A)$ to $p(A)$ is called monotone if for any subsets $X$ and $Y$of $A$, if $X \subseteq Y$, then $F(X) \subseteq F(Y)$.
Show that if F is monotone then $F(Z) \subseteq Z$.
Where I'm at so far:
In the question, X is defined as $p(A)$, thus the cardinal of X would be the cardinal of $p(A)$. Therefore, for any $x \in X$, $x \in p(A)$.
Monotonic functions are functions wherein ordered sets either have their order preserved or reversed. Thus, I understand the statement "A function F from $p(A)$ to $p(A)$ is called monotone if for any subsets $X$ and $Y$of $A$, if $X \subseteq Y$, then $F(X) \subseteq F(Y)$. "
What I don't understand is this portion of the question:
Show that for any function $p(A)$ to $p(A)$, the set of all subsets X of A such that $F(X) \subseteq X$ is nonempty, and therefore the intersection Z of all such subsets exists.
Can you help me visualise/understand the function $p(A)$ to $p(A)$ and the statements "$F(X) \subseteq X$ is non-empty therefore there exists an intersection Z of all such subsets"?
A simple example or something would help me a lot.
Thank you so much!
 A: $X$ is not defined at all in the question: there is no specific set called $X$. Rather, the dummy variable $X$ is used to define a certain family of subsets of $X$. This might be clearer if we give that family a name:
$$z(A)=\{X\in p(A):F(X)\subseteq X\}\,.$$
Then $Z$ is defined as
$$Z=\bigcap z(A)=\bigcap\{X\in p(A):F(X)\subseteq X\}\,.$$
This makes $Z$ an intersection of subsets of $A$ and therefore itself just a subset of $A$.
For example, if $A=\{0,1,2\}$, and $F(X)=A\setminus(X\cup\{0\})$ for each $X\in p(A)$, we have:
$$\begin{align*}
&F(\varnothing)=\{1,2\},\\
&F(\{0\})=\{1,2\},F(\{1\})=\{2\},F(\{2\})=\{1\},\\
&F(\{0,1\})=\{2\},F(\{0,2\})=\{1\},F(\{1,2\})=\varnothing,\\
&F(A)=\varnothing
\end{align*}$$
For this function $F$ we can see that $F(X)\subseteq X$ if and only if $X=A$ (since $F(A)=\varnothing\subseteq A$) or $X=\{1,2\}$ (since $F(\{1,2\})=\varnothing\subseteq\{1,2\}$). Thus, $z(A)=\big\{A,\{1,2\}\big\}$, and
$$Z=\bigcap z(A)=A\cap\{1,2\}=\{1,2\}\,.$$
The first paragraph that you wrote describing where you are now is far off the mark. If $X\in z(A)$, then by definition $X$ is a subset of $A$, so $|X|\le|A|$; in particular, $|X|<|p(A)|$, since $|A|<|p(A)|$ no matter what $A$ is. However, cardinality is simply not relevant to this problem. Finally, if $X\in z(A)$ and $x\in X$, then $x\in A$; in general $x\notin p(A)$.
Your first task is to show that $z(A)\ne\varnothing$, i.e., that $A$ has at least one subset $X$ such that $F(X)\subseteq X$. By definition $F(X)\subseteq A$ for all $X\in p(A)$ (why?), and $A\in p(A)$, so $F(A)\subseteq A$. Thus, $A\in z(A)$ no matter what function $F$ is mapping subsets of $A$ to subsets of $A$. This means that we can form $Z=\bigcap z(A)$; it will be some subset of $A$.
Your second task is to show that if $F$ is monotone — i.e., if $F(X)\subseteq F(Y)$ whenever $X\subseteq Y\subseteq A$ — then $F(Z)\subseteq Z$. In other words, you’re to show that if $F$ is monotone, then $Z\in z(A)$: the intersection of all of the sets in $z(A)$ is itself in $z(A)$.
HINT: We know that $Z\subseteq X$ for each $X\in z(A)$. (Why?) Thus, if $F$ is monotone we know that $F(Z)\subseteq F(X)\subseteq X$ for each $X\in z(A)$. In particular, $F(Z)\subseteq X$ for each $X\in z(A)$. What does this tell you about the relationship between $F(Z)$ and $\bigcap z(A)$?
A: To visualize a function $F$ just take a couple of examples.
First with $A = \{1,2\}$ you have $$\mathcal P(A)= \{\emptyset, \{1\},\{2\}, \{1,2\}\}.$$ And you can take $F_1(\emptyset) = \{1,2\}$, $F_1(\{1\}) = \{2\}$, $F_1(\{2\}) = \{1\}$ and $F_1(\{1,2\}) = \emptyset$. However $F_1$ is not increasing, I let you tell why.
Another example of increasing $F_2$ also defined on $\mathcal P(A)$ is $F_2(x) = \{1,2\}$ for all $x \in \mathcal P(A)$ which is increasing.
If you want an example of an increasing map with $A$ infinite, take $A = \mathbb N$ and $F (x) = x \cup \{1, 2\}$ for all $x \in \mathcal P(A) = \mathcal P(\mathbb N)$.
Now coming to the question you are asked. Proving that the set
$$S = \{X \in \mathcal P(A) \mid F(X) \subseteq X\}$$ is not empty is pretty easy... Just notice that $A$ belongs to $S$! And as $S \neq \emptyset$
$$\bigcap S$$ exists.
