# Thread: What is the closure under F in this question?

1. ## What is the closure under F in this question?

(This is not my question yet.)

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

Let
$B_1 = B$
$\forall n \geq 1, B_{n+1} = f(B_n) = {f(x)| x \in B_n}$

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

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

Here is my question.

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

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

I try to prove, but am unsure.

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

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

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

I tried to use info from the first part. Let $C_f$ be the closure of B under $f$ Then I claimed that $\cup_{f \in \Im}C_f$ is the closure of B under $\Im$, but my friend said it is not closed. For example,
$B={0}$
$f(x)=x+2$,
$g(x)=x+3$

$C_f=${0,2,4,6,8,10,...}
$C_g=${0,3,6,9,12,15,...}

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

Thanks a lot!