Explanation:
In mathematics a \( \mathcal{B} = \{ A_{\alpha} : \alpha \in \Lambda \} \subseteq \mathcal{P}(X) \) subset family is called a \( \textbf{Borel Family} \) defined on \( X \) if and only if \( \forall n \in \mathbb{N} \) for \( A_{n} \in \mathcal{B} \) which holds,
$$\bigcup_{n=1}^{\infty} A_{n} \in \mathcal{B} ~~~~~ \textit{and} ~~~~~ \bigcap_{n=1}^{\infty} A_{n} \in \mathcal{B}$$
these conditions. The theorem that we are about to prove is a more simple answer for a theorem that we're going to prove in the future. It is a continuation from this theorem. Currently it might not be clear but as we progress and finish the remaining theorems it will be much more easy to understand.
Theorem:
Let \( \emptyset \neq X \). Any \( A \subseteq \mathcal{P}(X) \) family will cause a \( \mathcal{B}_{A} \) Borel family that holds \( A \subseteq \mathcal{B} \). Besides \( |\mathcal{B}_{A}| \leq \mathbf{c} \cdot (|A|)^{\aleph_{0}} \) holds. If \( |A| = \mathbf{c} \) then \( |\mathcal{B}_{A}| = \mathbf{c} \).
Proof:
\( b,k: (\mathcal{P}(X))^{\omega} \longrightarrow \mathcal{P}(X) \) functions are for all \( \lbrace A_{n} \rbrace^{\infty}_{n=1} \in (\mathcal{P}(X))^{\omega} \) defined as,
$$b(\lbrace A_{n} \rbrace^{\infty}_{n=1}) = \bigcup^{\infty}_{n=1} A_{n} ~~~~~ \text{and} ~~~~~ k(\lbrace A_{n} \rbrace^{\infty}_{n=1}) = \bigcap^{\infty}_{n=1} A_{n}$$
And let \( \mathcal{F} = \lbrace b,k \rbrace\). In that case \( \mathcal{F} \subseteq (\mathcal{P}(X))^{(\mathcal{P}(X))^{\omega}} \) holds. Besides, if this special serie of subsets \( \lbrace A_{n} \rbrace^{\infty}_{n=1} \in (\mathcal{P}(X)) \), \( \forall n \in \mathbb{N} \) defined as \( A_{2n} = X \) and \( A_{2n - 1} = \emptyset \), \( b,k \) functions that are defined on \( (\mathcal{P}(X))^{\omega} \) are different because of,
$$b(\lbrace A_{n} \rbrace^{\infty}_{n=1}) = X \neq \emptyset = k(\lbrace A_{n} \rbrace^{\infty}_{n=1})$$
And \( |\mathcal{F}| = 2 \) and moreover \( \mathcal{A} \subseteq \mathcal{F}(\mathcal{A}) = b(\mathcal{A}^{\omega}) \cup k(\mathcal{A}^{\omega}) \) holds since any given \( A \in \mathcal{A} \), \( \forall n \in \mathbb{N} \) that \( A_{n} = A \). For this case to hold there should be an equality such that \( b(\lbrace A_{n} \rbrace^{\infty}_{n=1}) = A = k(\lbrace A_{n} \rbrace^{\infty}_{n=1}) \) or to make it more clear \( A \in b(\mathcal{A}^{\omega}) \cap k(\mathcal{A}^{\omega}) \subseteq b(\mathcal{A}^{\omega}) \cup k(\mathcal{A}^{\omega}) = \mathcal{F}(\mathcal{A}) \) for the special serie \( \lbrace A_{n} \rbrace^{\infty}_{n=1} \in \mathcal{A} \)
In that case there exists \( \mathcal{B}_{\mathcal{A}} \subseteq \mathcal{P}(X) \) that holds \( \mathcal{A} \subseteq \mathcal{B}_{\mathcal{A}} = \mathcal{F}(\mathcal{B}_{\mathcal{A}}) \). This family is called the \( \textbf{Borel family of $\mathcal{A}$ on $X$} \). Moreover since \( 2 = |\mathcal{F}| \leq |\mathcal{F}| \lor |\mathcal{A}| \) and \( (|\mathcal{F}|)^{\aleph_{0}} = \mathbf{c} \) holds, it is relatively easy to see that \( |\mathcal{B}_{\mathcal{A}}| \leq \mathbf{c} \cdot (|\mathcal{A}|)^{\aleph_{0}} \) holds. If \( |\mathcal{A}| = \mathbf{c} = 2^{\aleph_{0}} \) then \( (|\mathcal{A}|)^{\aleph_{0}} = (2^{\aleph_{0}})^{\aleph_{0}} = 2^{\aleph_{0} \cdot \aleph_{0}} = 2^{\aleph_{0}} = |\mathcal{A}| \) holds and from here we can say \( |\mathcal{A}| \leq |\mathcal{B}_{\mathcal{A}}| \leq \mathbf{c} \cdot (|\mathcal{A}|)^{\aleph_{0}} = \mathbf{c} \cdot \mathbf{c} = |\mathcal{A}| \),
\( \therefore |\mathcal{B}_{\mathcal{A}}| = |\mathcal{A}| \)