Is the Number of Elements in a Set Equal to Its Power Set?

[Mr Davis 97]

Homework Statement

Show that $\bigcup \{\mathcal{P} X : X \in A \} \subseteq \mathcal{P} \bigcup A$

Homework Equations

The Attempt at a Solution

Suppose that $c \in \bigcup \{\mathcal{P} X : X \in A \}$. Then by definition this means that $\exists a \in A$ such that $c \in \mathcal{P} a$, or, equivalently, $\exists a \in A$ such that $c \subseteq a$, which implies that $c \subseteq \bigcup A$ which means that $c \in \mathcal{P} \bigcup A$.

Is this a correct proof? Also, why can't I just reverse the argument that that we have equality between sets and not just one being a subset of the other

 

[Dick]

Homework Statement

Show that $\bigcup \{\mathcal{P} X : X \in A \} \subseteq \mathcal{P} \bigcup A$

Homework Equations

The Attempt at a Solution

Suppose that $c \in \bigcup \{\mathcal{P} X : X \in A \}$. Then by definition this means that $\exists a \in A$ such that $c \in \mathcal{P} a$, or, equivalently, $\exists a \in A$ such that $c \subseteq a$, which implies that $c \subseteq \bigcup A$ which means that $c \in \mathcal{P} \bigcup A$.

Is this a correct proof? Also, why can't I just reverse the argument that that we have equality between sets and not just one being a subset of the other

It looks ok to me. As for why not the reverse, compare the power set of $\{1,2\}$ with the union of the power sets of $\{1\}$ and $\{2\}$.

 

[Mr Davis 97]

It looks ok to me. As for why not the reverse, compare the power set of $\{1,2\}$ with the union of the power sets of $\{1\}$ and $\{2\}$.

When would equality hold? It seems that it would hold iff $|A| = 1$, but I am not sure how to prove this.

 

[Dick]

When would equality hold? It seems that it would hold iff $|A| = 1$, but I am not sure how to prove this.

Count the elements in each set. If $|A|=n$ then the power set has $2^n$ elements.