Question about the existence of sets

Question

If there is a sequence of set $a_{i}$ and $i \in I \wedge |I|>|\mathbb{R}|$, is the set $A$ which $\forall i (i \in I \rightarrow a_{i}\in A)$ and the set $\bigcup \limits_{i \in \mathbb{R}} a_{i}$ exists? And how to proof it by using the axioms of ZF(Zermelo-Frankel)C(axiom of choice)?

Answer 1

Since we assume that $(a_i)_{i\in I} = \{(i, a_i) \space|\space i \in I\} = \{\{\{i\},\{i,a_i\}\}\space|\space i\in I\}=: A_I$ is an existing set, we can construct $$X_1 := \bigcup A_I = \{\{i\},\{i,a_i\}\space |\space i\in I\}$$ by the axiom of union and likewise may construct $$ X_2 := \bigcup X_1 = \{i,a_i \space | \space i\in I\} $$

Now we already can construct your set $A$ as $$ A:= \{x\in X_2 \space | \space \exists i\in I.(\{\{i\},\{i,x\}\} \in A_I)\} = \{a_i \space | \space i \in I\}$$ by the axiom of separation and the axiom of pairing (for the sets $\{i\},\{i,x\}$ and $\{\{i\},\{i,x\}\}$).

If we interprete $\bigcup_{i\in \mathbb{R}}a_i$ as $\bigcup_{i\in \mathbb{R}\cap I}a_i$ we can write this similarly as $$ \bigcup \{x\in X_2 \space | \space \exists i\in (I\cap \mathbb{R}).(\{\{i\},\{i,x\}\} \in A_I)\} $$ again using the axiom of union and separation as well as the axiom of pairing.