The real numbers and the axiom of foundation

Question

I am having a bit of confusion about the real numbers and ZF set theory (I asked a question about it a few days ago). I am a bit unsure as to why the real numbers can be in any model of ZF as they seem to contradict the axiom of foundation (regularity) i.e. that for every set $X$ and $Y\subseteq X$ $\exists a\in Y:\forall b\in Y ((b\neq a)\ b<a)$

Thanks very much for any help

Answer 1

The axiom of regularity states that $\in$ is well-founded. Namely if $A$ is a non-empty set, then $(A,\in_A)$ (the membership relation restricted to $A$) has a minimal element.

Unlike the natural numbers [read: finite ordinals], which are inherent objects to the universe of ZF, the ordering of the real numbers is not defined by $\in$. It can, however, be defined by $\subseteq$ which itself is definable.

However there is no requirement that $\subseteq$ is well-founded, and indeed it is not.

Answer 2

The axiom of foundation says that for each nonempty set $Y$ it holds that $\exists a\in Y:\forall b\in Y \neg(b\in a)$. This means that every set is well-founded with respect to the $\epsilon$-relation.

If the real numbers were well-founded with respect to the the $<$-relation, we would have for each nonempty set $Y$ of real numbers that $\exists a\in Y:\forall b\in Y \neg(b< a)$. This is indeed false as the example $Y=(0,1)$ shows.

Since the relations $\epsilon$ and $<$ do not have to agree on $\mathbb{R}$, this is in no way contradictory. Some relations are well-founded, others are not.