About Set Theory Axioms

Question

The axiom of Replacement Scheme implies separate axiom. I can not show this lemma. Does someone have any idea about it?

Also, http://en.wikipedia.org/wiki/Axiom_schema_of_specification#Relation_to_the_axiom_schema_of_replacement. — Martín-Blas Pérez Pinilla, Jan 13 '15 at 16:37

score 0 · Accepted Answer · answered Jan 13 '15 at 16:46

Let $P$ be a property of sets and let $a$ be a set.

If $c\in a$ then prescribe $F$ as an operation on sets by $x\mapsto x$ if $P\left(x\right)$ and $x\mapsto c$ otherwise.

Then $b:=\left\{ F\left(x\right)\mid x\in a\right\} $ satisfies $\forall x\left[x\in b\iff\left[x\in a\wedge P\left(x\right)\right]\right]$.

In special case $a=\emptyset$ you can do it with $b:=a=\emptyset$.

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