The real numbers and the axiom of foundation

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

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To my knowledge the Axiom of Foundation states that for every nonempty set $X$, there is an element $A\in X$ such that $A\cap X=\emptyset$. What does $((b\ne a)b<a)$ mean after all? – Hagen von Eitzen Oct 15 '12 at 20:26
I was say that wrt the ordering each set has minimal element but I got my ordering very confused, as you can see from the answer. I think what you have stated is equivalent though – hmmmm Oct 15 '12 at 20:48
The language of ZF has only one relational symbol: $\in$. There is no $<$, certainly not in axioms. – Ivan Kuckir May 9 '13 at 14:05

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.

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Thanks very much, just got a bit confused then – hmmmm Oct 15 '12 at 20:46
@hmmmm: No problems. It's not a bad question, so don't feel bad. – Asaf Karagila Oct 16 '12 at 5:57

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.

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