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This is Velleman 3.7, Problem 4

Below is the problem, verbatim.


Suppose $ \{ A_i \mid i \in I\}$ is a family of sets. Prove that if $\mathcal{P}(\bigcup_{i \in I} A_i) \subseteq \bigcup_{i \in I} \mathcal{P}(A_i)$, then there is some $ i \in I$ such that $\forall j \in I (A_j \subseteq A_i)$.


My question is not on the proof itself, but on the notation.

If a concrete instance of the family were $\{ \{2\}, \{2, 3\}\} $, are the following correct?

$ \bigcup A_i = \{2, 3\} $

$\mathcal{P}(\bigcup_{i \in I} A_i) = \{ \emptyset, \{2\}, \{3\}, \{2, 3\}\}$

$ \mathcal{P}(A_i) = \{ \emptyset, \{\{2\}\}, \{\{2, 3\}\}, \{ \{2\}, \{2, 3\} \} \} $

$ \bigcup_{i \in I} \mathcal{P}(A_i) = \{ \{2\}, \{2, 3\} \} $

I want to make sure I understand the problem first before attempting the proof.

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  • $\begingroup$ what is $A_i$? ? $\endgroup$
    – user 1
    Commented Jan 18, 2015 at 15:37
  • $\begingroup$ The problem statement only said that its a family of sets. $\endgroup$
    – Marty B.
    Commented Jan 18, 2015 at 15:42
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    $\begingroup$ The question is not clear. What is the relation between $A$ and $A_i$? Why did you choose such $A$ and what is the meaning of "then I believe $\cup A_i = \cdots$"? Please update the question. $\endgroup$
    – Krish
    Commented Jan 18, 2015 at 16:00
  • $\begingroup$ @Krish I updated the question. Hope it's clearer now. $\endgroup$
    – Marty B.
    Commented Jan 19, 2015 at 0:05
  • $\begingroup$ a possible duplicate of math.stackexchange.com/questions/1093822/… You can find an answer in this link. I hope it will help you. $\endgroup$
    – Krish
    Commented Jan 19, 2015 at 7:03

1 Answer 1

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I got the answer from Mathematica.

Clear[A, B, F];

A = Sort[{{2}, {2, 3}}];

Union[A] = {{2}, {2, 3}}

Subsets[Union[A]] = {{}, {{2}}, {{2, 3}}, {{2}, {2, 3}}}

Subsets[A] = {{}, {{2}}, {{2, 3}}, {{2}, {2, 3}}}

Union[Subsets[A]] = {{}, {{2}}, {{2, 3}}, {{2}, {2, 3}}}

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