Reading Velleman's "How To Prove It" I came across the following expression: $$ x \in\bigcup\{\mathscr P(A)\mid A\in \mathcal F\} $$ such that $\mathcal F$ is a family of sets, $A$ is a set, and $\mathscr P(A)$ is the power set of $A$. Now according to Velleman, it holds: $$ x \in\bigcup\{\mathscr P(A)\mid A\in \mathcal F\} \iff \exists A \in \mathcal F(x\in \mathscr P(A)).$$ The last term means "there is the set A which is an element of F, such that x is an element of the power set of A". Now, it seems to me, however, that the following equality also: $$ \exists A \in \mathcal F(x\in \mathscr P(A)) \iff x \in\{\mathscr P(A)\mid A\in \mathcal F\}.$$ Which implies:$$ x \in\bigcup\{\mathscr P(A)\mid A\in \mathcal F\} \iff x \in\{\mathscr P(A)\mid A\in \mathcal F\}.$$ Is that correct? Thanks in advance.
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1$\begingroup$ \cup is normally used in things like $A\cup B$ and $A_1\cup\cdots\cup A_n$, and \bigcup is normally used in things like $\displaystyle\vphantom{\frac\int\int}\bigcup_{x\in\mathcal X} A_x$, and I edited accordingly. Also notice that \mid provides proper spacing to its right and left and I changed "|" to \mid in several places. ${}\qquad{}$ $\endgroup$– Michael HardyJun 26, 2015 at 18:40
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$\begingroup$ Thank you, hopefuly someone can actually answer that $\endgroup$– EugeneJun 26, 2015 at 18:42
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$\begingroup$ It is not correct. $\endgroup$– CrostulJun 26, 2015 at 18:43
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$\begingroup$ What is $\mathcal{F}(x \in \mathscr{P}(A))$ supposed to mean? Also, I think you want your equality symbols to be $\iff$ symbols, no? $\endgroup$– Alex ProvostJun 26, 2015 at 18:44
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$\begingroup$ Three times you used "$=$" where it appears that you meant "$\Longleftrightarrow$". ${}\qquad{}$ $\endgroup$– Michael HardyJun 26, 2015 at 18:48
3 Answers
Let us simply analyze what the statement $x \in \bigcup \{ \mathscr P (A) \mid A \in \mathcal{F} \}$ means. For $x$ to be in a union means exactly that $x$ is in at least one of the members of the union. So we can rewrite this as
$$ \exists A \in \mathcal F : x \in \mathscr P(A)$$
(which is what I think you meant to write). However, your last equivalence is not true. Let's consider an example; take $\mathcal F = \{\emptyset, \{a\} \}.$ Now $\mathscr P(\emptyset) = \{ \emptyset \}$ and $\mathscr P(\{a\}) = \{ \emptyset , \{a\}\}.$ Thus
\begin{align} x \in \bigcup \{ \mathscr P (A) \mid A \in \mathcal{F} \} &\iff \exists A \in \mathcal F : x \in \mathscr P(A) \\ &\iff x \in \{\emptyset\} \text{ or } x \in \{ \emptyset , \{a\}\} \\ &\iff x = \emptyset \text{ or } x = \{ a \}. \end{align}
On the other hand, \begin{align} x \in \{ \mathscr P(A) \mid A \in \mathcal F\} &\iff x \in \{ \{ \emptyset \}, \{ \emptyset , \{a\}\}\} \\ &\iff x = \{ \emptyset \} \text{ or } x = \{ \emptyset , \{a\}\}. \end{align}
In other words: in the first case, $x$ is an element of some $\mathscr P(A)$, whereas in the second case, $x$ is equal to some $\mathscr P(A).$
No, not quite.
$x\in\{\mathscr P(A)\mid A\in\mathcal F\}$ if and only if there is some $A\in\cal F$, such that $x=\mathscr P(A)$.
It might be the case that $x\in\mathscr P(A')$ for a different $A'\in\cal F$. But just from the given information we cannot infer this. And since under the assumption that $\mathcal F=\{A\}$ we can show this to be false (meaning we can show that $\mathscr P(A)\notin\mathscr P(A)$), the equivalence you suggested is not true.
You're wrong:
$\exists A \in \mathcal F(x\in \mathscr P(A))$ means in plain English: $x$ is a subset of some $A$ in family $\mathcal F$, i.e. a member of the powerset of some $A$ in family $\mathcal F$.
while $\,x \in\{\mathscr P(A)\mid A\in \mathcal F\}$ is the powerset of some $A$ in family $\mathcal F$.
