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empty_set [2015/10/08 20:33] nikolaj |
empty_set [2015/10/09 15:40] nikolaj |
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| So is it true? Does our set theory permit the existence of such a set $y$? | 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 | Uniqueness is discussed, for example, in | ||
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| ----- | ----- | ||
| + | === Requirements === | ||
| + | [[Set theory]] | ||
| === Subset of === | === Subset of === | ||
| [[First infinite von Neumann ordinal]] | [[First infinite von Neumann ordinal]] | ||
