**Proof**: *(related to "Justification of Set Union")*

- Let $A$ and $B$ be sets.

- By the axiom of pairing there is a set $Z$ containing $A$ and $B$ as elements:

- By the axiom of separation there is a subset $Z’\subseteq Z$ containing exactly $A$ and $B$ as elements:

- By the axiom of union there is a set $Z^*$ containing the elements of $A$ or
^{1}the elements of $B.$

- By the axiom of separation there is a subset $Z^\dagger \subseteq Z^*$ containing exactly the elements of $A$ or
^{1}the elements of $B.$, i.e. $Z^\dagger =\{z\mid z\in A\vee z\in B\}.$

- This is the required set union $A\cup B:=Z^\dagger .$
- The axiom of extensionality ensures the uniqueness of such a set.
- Altogether, we have shown:

^{1} We mean the logical or operation, in the natural English language “and/or”.

q.e.d

| | | | Contributors: *bookofproofs* | References: [983]

(none)

[983] **Ebbinghaus, H.-D.**: “Einführung in die Mengenlehre”, BI Wisschenschaftsverlag, 1994, 3

FeedsAcknowledgmentsTerms of UsePrivacy PolicyImprint

© 2018 Powered by BooOfProofs, All rights reserved.

© 2018 Powered by BooOfProofs, All rights reserved.