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)?

  • $\begingroup$ Separation + Union? $\endgroup$ – Asaf Karagila Feb 5 at 8:10
  • $\begingroup$ We also need Infinity and Power set to guarantee there is a set of reals. (Power set is overkill, I know.) $\endgroup$ – Hanul Jeon Feb 5 at 9:12

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.

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    $\begingroup$ To the proposer: In the formal language of ZF there is no word "set". Things that satisfy conditions stated in the formal language are asserted to exist, or asserted to not exist. There are other foundational formulations in which you can say "$x$ is a set", but not in ZF. $\endgroup$ – DanielWainfleet Feb 5 at 11:34
  • $\begingroup$ What if the sequence $a_{i}$ is a collection? Can we still proof that? $\endgroup$ – Tzi yan Tschen Feb 5 at 11:34
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    $\begingroup$ To the proposer: "Collection" is an informal word, like "family", which for a set-theorist are just synonyms for "set". And in ZF, everything that exists is a set. $\endgroup$ – DanielWainfleet Feb 5 at 11:39

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