Theorem 1
Let \( \emptyset \neq X \) and \( \emptyset \neq \mathcal{F} \subseteq X^{X^{\omega}} \). \( \forall A \in \mathcal{P}(X) \) let us write,
$$ \mathcal{F}(A) = \bigcup_{f \in \mathcal{F}} f(A^{\omega}) \subseteq X ~~~~~ (*)$$
There exists \( A_{M} \subseteq X \) for the condition \( A_{0} \subseteq A_{M} = \mathcal{F}(A_{M}) \) in response to any subset \( A_{0} \) for the condition \( A_{0} \subseteq \mathcal{F}(A_{0}) \)
Proof:
First thing first let us define a function that is an element of \( \mathcal{F} \), \( f : X^{\omega} \longrightarrow X \) that maps an element of \( X \) to each serie that is defined on the set \( X \). Since this function for the set of functions \( \mathcal{F} \) holds for \( A^{\omega} \subseteq B^{\omega} \) in a case where \( A \subseteq B \subseteq X \) and \( f(A^{\omega}) \subseteq f(B^{\omega}) \) for each \( f \in \mathcal{F} \). Let us observe a case where the following holds,
$$\mathcal{F}(A) = \bigcup_{f \in \mathcal{F}} f(A^{\omega}) \subseteq \bigcup_{f \in \mathcal{F}} f(B^{\omega}) = \mathcal{F}(B)$$
\( \forall A \subseteq X \) do not forget that the definition of \( \mathcal{F}(A) \) is (*). From here let us define a function \( \varphi : W(\omega_{1}) \longrightarrow \mathcal{P}(X) \) in a way like the following,
$$\varphi(A)=A_{0}, ~~~~~ \varphi(\alpha) = \mathcal{F} \left( \bigcup_{\beta < \alpha} \varphi(\beta) \right), ~~~~~ (\forall \alpha \in (0,\omega_{1})) $$
Our purpose is to prove that the subset \( A_{M} = \bigcup_{\alpha < \omega_{1}} \varphi(\alpha) \subseteq X \) holds for any condition that is given. \( A_{0} = \varphi(0) \subseteq \bigcup_{\alpha < \omega_{1}} \varphi(\alpha) = A_{M} \) is clear. Besides \( \forall \alpha \in (0,\omega_{1}) \), the observation that we showed before and because of the inclusions \( \bigcup_{\beta < \alpha} \varphi(\beta) \subseteq \bigcup_{\beta < \omega_{1}} \varphi(\beta) = \bigcup_{\alpha < \omega_{1}} \varphi(\alpha) \) we can write \( \varphi(\alpha) = \mathcal{F} (\bigcup_{\beta < \alpha} \varphi(\beta)) \subseteq \mathcal{F}(\bigcup_{\alpha < \omega_{1}} \varphi(\alpha)) = \mathcal{F}(A_{M}) \) and by using the hypothesis \( \varphi(0) = A_{0} \subseteq \mathcal{F}(A_{0}) \subseteq \mathcal{F}(A_{M}) \), the result can be written like this,
$$A_{M}=\bigcup_{\alpha \in [0,\omega_{1})} \varphi(\alpha) \subseteq \mathcal{F}(A_{M})$$
Now, let us show that \( \mathcal{F}(A_{M}) = \bigcup_{f \in \mathcal{F}} f(A^{\omega}_{M}) \subseteq A_{M} \) holds. For this whatever is \( f \in \mathcal{F} \) is, all we have to show is \( f(A^{\omega}_{M}) = \lbrace f(\lbrace x_{n} \rbrace ^{\infty}_{n=1}) : \lbrace x_{n} \rbrace ^{\infty}_{n=1} \in A^{\omega}_{M} \rbrace \subseteq A_{M} \). When we take a serie like \( \lbrace x_{n} \rbrace ^{\infty}_{n=1} \subseteq A^{\omega}_{M} \), \( \forall n \in \mathbb{N} \) we can write \( x_{n} \in A_{M} = \bigcup_{\alpha < \omega_{1}} \varphi(\alpha) \) and with the help of \( \alpha_{n} < \omega_{1} \) we can write \( x_{n} \in \varphi(\alpha_{n}) \). In that case if we determine the already existent ordinal number \( \beta \) that holds \( \sup_{n} \alpha_{n} < \beta < \omega_{1} \), since \( \alpha_{n} < \beta \), \( \forall n \in \mathbb{N} \) and from here \( x_{n} \in \varphi(\alpha_{n}) \subseteq \bigcup_{\gamma < \beta} \varphi(\gamma) \). Because of this the following can be written,
$$ \lbrace x_{n} \rbrace ^{\infty}_{n=1} \in \left( \bigcup_{\gamma < \beta} \varphi(\gamma) \right)^{\omega} ~~ \textit{and} ~~ f( \lbrace x_{n} \rbrace ^{\infty}_{n=1} ) \in f \left( \left( \bigcup_{\gamma < \beta} \varphi(\gamma) \right)^{\omega} \right) \subseteq \bigcup_{f \in \mathcal{F}} f \left( \left( \bigcup_{\gamma < \beta} \varphi(\gamma) \right)^{\omega} \right) = \mathcal{F} \left(\bigcup_{\gamma < \beta} \varphi(\gamma) \right) = \varphi(\beta) \subseteq A_{M} $$
\( \therefore A_{M} \subseteq \mathcal{F}(A_{M}) \subseteq A_{M} \)