1

While studying introductory analysis, I came across an exercise wherein we have to prove that if $A$ is a set, it can't contain itself. The hint indicates we're supposed to use the regularity axiom.

I want to find out the problem with defining $A$ as the set $\{A, x\}$, where $x$ is a non-set object. This definition doesn't violate the regularity axiom because of $x$, and such a set can exist because of the pair set axiom ($A$ and $x$ are objects, so there exists a pair set $\{A, x\}$ )

I tried looking up a proof online, but it only denotes a set containing itself as $A = \{A\}$. Why not $\{A, x\}$ or $\{A, x, y\}$, etc...?

Daniel R
  • 3,250
user9343456
  • 245
  • 2
    Sure, let $A = {A, x, y}$. Try applying the axiom of regularity to ${A}$. –  Mar 16 '16 at 06:15
  • @Rahul: in the text that I'm reading, the axiom states that if A is a non-empty set, there exists at least one element x of A which is either not a set, or is disjoint from A. So it seems that A satisfies the axiom, unless I'm missing something glaring – user9343456 Mar 16 '16 at 06:20
  • 2
    Try applying the axiom of regularity to ${A}$, not to $A$. –  Mar 16 '16 at 06:23
  • @Rahul: I see now. Thanks! The hint did also mention that the singleton set axiom was to be used. Pretty dense of me not to realize the significance.. – user9343456 Mar 16 '16 at 06:36

0 Answers0