Monotone Induction System counter example. On page 31 of Fundamentals of Mathematical Logic, there is the following exercise.

Exercise 1.2.22  A pair $(X, \mathcal{I})$ such that $\mathcal{I}: \mathscr{P}(X) \rightarrow \mathscr{P}(X)$ and for $Y \subseteq Z \subseteq X$, $\mathcal{I}(Y) \subseteq \mathcal{I}(Z)$ is called a $\textbf{monotone induction system}$.
For $Y \subseteq X$, say that $Y$ is $\mathcal{I}$-closed if and only if $\mathcal{I}(Y) \subseteq Y$.

*

*Show that there exists a unique smallest $\mathcal{I}$-closed set $\overline{X}$.

*Let $X_0 = \mathcal{I}(\emptyset)$ and for all $n$, $X_{n + 1} = \mathcal{I}(X_n)$. Prove that for all $n \in \omega$, $X_{n} \subseteq X_{n + 1} \subseteq \overline{X}$.

*Given an example to show that $\bigcup_{n \in \omega} X_n$ may $\textit{not}$ be $\mathcal{I}$-closed.

*Find a further hypothesis on $(X, \mathcal{I})$ which implies that $\bigcup_{n \in \omega} X_n$ $\textit{is}$ $\mathcal{I}$-closed and hence equal to $\overline{X}$ (other than the trivial one $\mathcal{I}(\emptyset) = \emptyset$).


For 1., it is easy to see that if $C$ is the set of all closed subsets of $X$, then $\overline X = \bigcap C$ is the desired set, and 2. is straightforward.
However, I have been unable to find a counterexample for part 3.
I moved on to 4. and I think the condition they are looking for is that the sets $\{X_n\}_{n \in \omega}$ (or for any increasing sequence of sets), will stabilize. This guarantees that the union will indeed be closed. I'm not sure if this is the one that they are looking for since this does not give an if and only if condition (though they don't explicitly ask for one). For example, if we have $X = \omega$, and $\mathcal{I}(Y) = \{2 \cdot n: n \in Y\} \cup \{1\}$, then it can be verified that $X_n = \{2^i: 0 \leq i \leq n\}$.
Then $I \left( \bigcup_{n \in \omega} X_n\right) = \bigcup_{n \in \omega} X_n$ so the union is indeed closed, yet clearly this sequence doesn't stabilize.
Hence, if I want to find a counter example, I recognize that the set $X$ need be infinite, and that the sequence must not stabilize, but I haven't been able to construct one that doesn't close. Any hints would be appreciated.
 A: In 4, they're actually looking for a much weaker condition, and this is related to 3.
Let's start with 3. We'll build a counterexample in a very silly way, by taking a well-behaved operator and tweaking it to make it ruder. Specifically, working on $\mathbb{N}$ for simplicity, let $$\mathcal{I}_0(A)=\{2\}\cup\{a+1: a\in A\}.$$ Iterating $\mathcal{I}_0$ builds the sequence of finite sets $$\{2\}, \{2,3\}, \{2,3,4\}, ...$$ Now we define $\mathcal{I}$ by modifying $\mathcal{I}_0$ to be "unsatisfied at the limit:"

 Let $\mathcal{I}(A)=\mathcal{I}_0(A)$ if $A$ is finite, and let $\mathcal{I}(A)=\mathcal{I}_0(A)\cup\{1\}$ if $A$ is infinite.

This is perfectly valid, and gives a counterexample to 3 as desired. On the other hand, it's also rather annoying; in a precise sense, it's "discontinuous," and the relevant continuity property is exactly what 4 is looking for:

 Say $\mathcal{I}$ is continuous iff for every $A$ we have $$\mathcal{I}(A)=\bigcup_{B\subseteq_{fin}A}\mathcal{I}(B),$$ where "$\subseteq_{fin}$" means "is a finite subset of."

