(This is not my question yet.)

Suppose $\displaystyle f:A \longrightarrow A$. A set C is said to be closed under $\displaystyle f$ if $\displaystyle \forall x\in C (f(x) \in C)$ Now suppose $\displaystyle B \subseteq A$. The closure of $\displaystyle B$ under $\displaystyle f$ is the smallest set $\displaystyle C$ such that $\displaystyle B \subseteq C \subseteq A$ and $\displaystyle C$ is closed under $\displaystyle f$, if there is such a smallest set.

Let

$\displaystyle B_1 = B$

$\displaystyle \forall n \geq 1, B_{n+1} = f(B_n) = {f(x)| x \in B_n}$

Then the closure of B under f is given by the set

$\displaystyle C = \cup_{n \in Z^+}B_n$.

Here is my question.

Suppose $\displaystyle \Im$ is a family of functions from A to A, and $\displaystyle B \subseteq A$. The closure of Bunder $\displaystyle \Im$is the smallest set C such that $\displaystyle B \subseteq C \subseteq A$ and $\displaystyle \forall f \in \Im, $ C is closed under $\displaystyle \Im$, if there is such a smallest set.

What should be the closure of B under $\displaystyle \Im$ in this case?

I try to prove, but am unsure.

Should it be $\displaystyle C = \cap_{f \in \Im} f\left(\cup_{n \in Z^+}B_n\right)$ or $\displaystyle \cup_{f \in \Im} f\left(\cup_{n \in Z^+}B_n\right)$ ??

Or could it be $\displaystyle \cup_{n \in Z^+} \cap_{f \in \Im} f \left(B_n\right)$ ?

I tried to prove $\displaystyle \cap_{f \in \Im} f\left(\cup_{n \in Z^+}B_n\right)$ is the closure of B under $\displaystyle \Im$. Every property, closedness under f, and $\displaystyle \cap_{f \in \Im} f\left(\cup_{n \in Z^+}B_n\right)$ is the smallest set has been proved, except $\displaystyle B \subseteq \cap_{f \in \Im} f\left(\cup_{n \in Z^+}B_n\right) \subseteq A$. (I'm kinda stucked.)

I tried to use info from the first part. Let $\displaystyle C_f$ be the closure of B under $\displaystyle f$ Then I claimed that $\displaystyle \cup_{f \in \Im}C_f$ is the closure of B under $\displaystyle \Im$, but my friend said it is not closed. For example,

$\displaystyle B={0}$

$\displaystyle f(x)=x+2$,

$\displaystyle g(x)=x+3$

$\displaystyle C_f=${0,2,4,6,8,10,...}

$\displaystyle C_g=${0,3,6,9,12,15,...}

$\displaystyle C_f \cup C_g $ does not contain 5, but $\displaystyle f\left(g(0)\right) = 5$

Thanks a lot!