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--- empty_set [2015/10/08 20:33]
nikolaj
+++ empty_set [2015/10/09 15:39]
nikolaj
@@ Line 26: / Line 26: @@
 So is it true? Does our set theory permit the existence of such a set $y$?
-Existence is granted by the [[http://en.wikipedia.org/wiki/Axiom_of_empty_set|axiom of empty set (Wikipedia)]]:
+As $\bot\implies P$ is $\top$ for any $P$ and as $Q\land\top$ is logically equivalent to $Q$, the above is logically equivalent to
-$\exists y.\,\nexists x.\,x\in y$
+$\exists! y.\,\forall x.\,\left(x\in y\implies \bot\right)$
-which is equivalent to
+which is
-$\exists y.\,\forall x.\,\neg(x\in y)$
+$\exists! y.\,\forall x.\,\neg(x\in y)$
-which is short for
+or
-$\exists y.\,\forall x.\,\left(x\in y\implies\bot\right)$
+$\exists! y.\,\nexists x.\,x\in y$
-As $\bot\implies P$ is $\top$ for any $P$ and as $Q\land\top$ is logically equivalent to $Q$, we know
+Apart from the exclamation mark, this is exactly the [[http://en.wikipedia.org/wiki/Axiom_of_empty_set|axiom of empty set (Wikipedia)]].
-$\exists y.\,\forall x.\,\left((x\in y\implies \bot)\land(\bot\implies x\in y)\right)$
-which is also written
-$\exists y.\,\forall x.\,\left(x\in y\Leftrightarrow \bot\right)$
-which is what we wanted.
 Uniqueness is discussed, for example, in