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power_set [2015/10/08 14:22]
nikolaj
power_set [2015/10/08 14:24]
nikolaj
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 Here we define Here we define
  
-$\mathcal{P}(X) ​:= \{Y\mid Y\subseteq X\}$+$\mathcal{P}(X) ​\equiv ​\{Y\mid Y\subseteq X\}$
  
 and want to claim that for each $X$, we have and want to claim that for each $X$, we have
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 $\forall X.\,​\exists! P.\,\forall Y.\,​\left(Y\in P\Leftrightarrow \forall Z.\,(Z\in Y\implies Z\in X)\right)$ $\forall X.\,​\exists! P.\,\forall Y.\,​\left(Y\in P\Leftrightarrow \forall Z.\,(Z\in Y\implies Z\in X)\right)$
  
-The existence ​is granted from the [[https://​en.wikipedia.org/​wiki/​Axiom_of_power_set|Axiom of power set]] +Without the exclamation mark, this is exactly ​the [[https://​en.wikipedia.org/​wiki/​Axiom_of_power_set|Axiom of power set]]. 
- +Uniqueness ​follows from extensionality.
-$\forall X.\,\exists P.\,\forall Y.\,​\left(Y\in P\Leftrightarrow \forall Z.\,(Z\in Y\implies Z\in X)\right)$ +
- +
-and uniqueness ​follows from extensionality.+
  
 == Examples == == Examples ==
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