Given a subset $A \subseteq B$, the closure of $A$ under $f$, denoted by $cl_f(A)$, is the smallest subset of $B$ which contains $A$ and is closed under $f$.
Define a sequence $\{A_n\}_{n \in \mathbb{N}}$ of subsets of B recursively as follows:
$$A_0 = A$$ $$A_{n+1}=f[A_n]\cup A_n \text{ for } n\in \mathbb{N}.$$
Prove $cl_f(A)=\bigcup_{n \in \mathbb{N}}A_n$.
How do I go about this proof? I am having trouble starting it.
On
You need to show two things:
$\bigcup_{n\in\mathbb{N}}A_n$ is closed under $f$.
Every set containing $A$ closed under $f$ contains $\bigcup_{n\in\mathbb{N}}$.
For the second one, try proof by induction. Do you see why every set containing $A$ which is closed under $f$ must contain $A_0$? OK, now supposing every set containing $A$ which is closed under $f$ must contain $A_n$, do you see how to show that every set containing $A$ which is closed under $f$ must contain $A_{n+1}$?
For the first, suppose $a\in\bigcup_{n\in\mathbb{N}}A_n$. You want to show that $f(a)$ is also in $\bigcup_{n\in\mathbb{N}}A_n$. Well, if something is in a union of sets, then it's in one of the sets in the union - so $a\in A_n$ for some $n$ ...
On
Writing $C = \bigcup_{n \in \mathbb{N}} A_n$, it suffices to prove:
Clearly $C$ contains $A$. To prove that $C$ is closed under $f$ note that for all $n \in \mathbb{N}$ and all $x \in A_n$, you have $f(x) \in A_{n+1}$.
To prove that $C$ is a subset of all other $f$-closed subsets containing $A$, let $C'$ be such a subset and prove (e.g. by induction) that $A_n \subseteq C'$ for all $n \in \mathbb{N}$.
$cl_f(A)\subset\cup_nA_n$ since $\cup_nA_n$ it contains $A$ and is stable by $f$.
On the other hand, suppose that $A_n\subset cl_f(A)$, $f(A_n)=A_{n+1}\subset cl_f(A)$, since $A_0=A\subset cl_f(A)$, this implies recursively that $\cup_nA_n\subset cl_f(A)$.