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Power set
Set
context | $X$ … set |
definiendum | $ Y \in \mathcal{P}(X) $ |
postulate | $ Y\subseteq X $ |
Discussion
Here we define
$\mathcal{P}(X) \equiv \{Y\mid Y\subseteq X\}$
and want to claim that for each $X$, we have
$\exists! P.\,P = \{Y\mid Y\subseteq X\}$
or more formally
$\forall X.\,\exists! P.\,P = \{Y\mid Y\subseteq X\}$
which is short for
$\forall X.\,\exists! P.\,\forall Y.\,\left(Y\in P\Leftrightarrow Y\subseteq X\right)$
which is short for
$\forall X.\,\exists! P.\,\forall Y.\,\left(Y\in P\Leftrightarrow \forall Z.\,(Z\in Y\implies Z\in X)\right)$
Without the exclamation mark, this is exactly the Axiom of power set. Uniqueness follows from extensionality.
Examples
We can prove
$\forall Y.\,\left(Y\in \{\emptyset\}\leftrightarrow Y\subseteq \emptyset\right)$
Therefore, for $X$ being $\emptyset$, we can show that the job of $P$ is done by $\{\emptyset\}$. In other words
$\mathcal{P}(\emptyset)=\{\emptyset\}$ |
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Remarks
Generally, $\emptyset\in\mathcal{P}(X)$ for any $X$. Hence no power set is empty.
One also writes $ \mathcal{P}(X) \equiv 2^X\equiv\Omega^X $.
Reference
Wikipedia: Axiom of power set