2
$\begingroup$

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.

$\endgroup$
  • 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 Hardy Jun 26 '15 at 18:40
  • $\begingroup$ Thank you, hopefuly someone can actually answer that $\endgroup$ – Eugene Jun 26 '15 at 18:42
  • $\begingroup$ It is not correct. $\endgroup$ – Crostul Jun 26 '15 at 18:43
  • $\begingroup$ 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 Provost Jun 26 '15 at 18:44
  • $\begingroup$ Three times you used "$=$" where it appears that you meant "$\Longleftrightarrow$". ${}\qquad{}$ $\endgroup$ – Michael Hardy Jun 26 '15 at 18:48
1
$\begingroup$

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 $\mathscr P(A).$

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $\mathcal F$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.