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Consider the natural numbers $\mathbb{N}$ as a structure for NBG set theory. If we interpret the Axiom of Unions in this structure, we get the statement $(\forall a \in \mathbb{N})$ $(\exists w \in \mathbb{N})$ $(\forall x \in \mathbb{N})$ $((\exists y \in \mathbb{N})(x < y \land y < a) \Rightarrow x < w)$.

Is this statement true? I suspect so: for any $a$, just choose $w = a$. However, Mathematica disagrees with me, and I'm not entirely confident enough with first-order logic to trust myself.

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  • $\begingroup$ It is true. The union axiom asserts that for every set $x$, there is a set consisting precisely of the elements of elements of $x$. (Your version is weaker, since it is only one direction.) If you interpret sets as natural numbers and element of as $\lt$, then this translates in $\mathbb{N}$ to: for every natural number $n$, there is a number $m$ such that the numbers below $m$ are precisely the numbers that are below a number below $n$. This is true if we take $m=n-1$, using $m=0$ if $n=0$. Finally, this question may not be appropriate for MO, which is for research-level questions. $\endgroup$
    – Joel David Hamkins
    Commented Jun 20, 2010 at 3:49
  • $\begingroup$ Your argument that $w=a$ works is correct. You may be interested in this MO answer mathoverflow.net/questions/12584/…, which is about the similar situation using the reals instead of the natural numbers. $\endgroup$
    – Joel David Hamkins
    Commented Jun 20, 2010 at 3:53
  • $\begingroup$ Thanks for that! I do apologise if that was an inappropriate question; I'm new to this site. Had I phrased it as a general inquiry into the naturals as a model/structure for NBG rather than a specific "is this true?" question, would it have been more appropriate? $\endgroup$
    – rtg658
    Commented Jun 20, 2010 at 3:58

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