Remark 11.1.3 (Set Theory Background).label Let $X\neq \emp$ be a set we denote the powerset by $\cali{P}(X)=2^{X}=\curl{2:2\suf X}$. Given $E,F\suf X$ denote the difference by $E\del F=E-F=\curl{x\in X:x\in E,x\nin F}$, symmetric difference by $E\triangle F=(E\del F)\cup (F\del E)$, $E^{c}=X\del E$ for $E\suf X$. Given $f:X\to Y$ and $E\suf Y$ then $f\inv(E)=\curl{x\in X:f(x)\in E}$, $f(D)=\curl{f(x):x\in D\suf X}$. We have some properties for $f:X\to Y$ with $E_{\al}\suf X$ for $\al\in A$

1. (1)

   $f\inv(\cups{}{\al\in A }E_{\al} )=\cups{}{\al\in A }f\inv(E_{\al})$

2. (2)

   $f\inv(\caps{}{\al\in A }E_{\al} )=\caps{}{\al\in A }f\inv(E_{\al})$

3. (3)

   $f\inv(E^{c})=\parens{f\inv(E)}^{c}$

An equivalence relation is a relation $R\suf X\ti X$ satisfying

1. (1)

   $\fall x\in X,xRx$

2. (2)

   $\fall x,y\in X, xRy\iff yRx$

3. (3)

   $\fall x,y,z\in X$ such that $xRy$ and $yRz$ then $xRz$

A partial ordering on $X\neq\emp$ is a relation $R\suf X\ti X$ ($R=\leq$) such that

1. (1)

   $x\leq y,y\leq z\implies x\leq z$

2. (2)

   $x\leq y,y\leq x\implies x=y$

3. (3)

   $x\leq x$

The pair $(X,\leq)$ is a poset. A maximal (resp. minimal) element $x\in X$ such that $y\geq x$ (resp. $y\leq x$) implies $y=x$. If $\leq$ is a total ordering if $\fall x,y\in X$ either $x\leq y$ or $y\leq x$ and $(X,\leq)$ is totally (linearly) ordered.

Let $(X,\leq)$ be a poset such that every linearly-ordered subset has a maximal element. Zorn’s Lemma states that $X$ has a maximal element which is equivalent to the Axiom of Choice. Let $A$ be an index set and $\curl{X_\al}_{\al\in A}$ is a nonempty collection of sets. Define the product set $X\define\prods{}{\al\in A}X_{\al}=\curl{f:A\to \cups{}{\al\in A}X_\al:\fall \al\in A,f(\al)\in X_\al}$ which comes with natural projections $\pi_{\al}:X\to X_{\al}$ defined by $\pi_{\al}(f)=f(\al)$. When $A$ is countable $\prods{}{\al\in A}X_{\al}$ is naturally isomorphic to the Cartesian product. AC states that under the above setting $X\neq\emp$.

Furthermore for $X\neq\emp,\curl{E_n}_{1}^{\infty}\suf \cali{P}(X)$ then

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