# Are non-standard models always not well-founded?

Are non-standard models of ZF set theory by definition always not well-founded? And it seems it is, because it must be.

But then, Wikipedia says that when there is a set that is a standard model of ZF, there exists smallest set $L_k$ that is standard model. So, this seems to confuse me a lot.

• User @Mikhail, please see the intended scope of tags before you apply them. The tag nonstandard-models is meant for models of arithmetic. It is a worthy point to consider expanding this to models of set theory as well, but you should bring this up on Mathematics Meta before you decide to take one-sided actions. – Asaf Karagila Jun 19 '18 at 11:31

By a standard model of ZF we mean a model $( M , E )$ of ZF where the relation $E$ is real membership on $M$, i.e., $x \mathrel{E} y \; \Leftrightarrow \; x \in y$. The relation $E$ is usually denoted either simply by $\in$, or by $\in_M$ to emphasise that we've restricted the domain.

Given any standard model $( M , \in_M )$ of ZF we can construct an isomorphic non-standard model $( M^* , E^* )$ quite easily by taking $M^* = \{ \{ x \} : x \in M \}$, and defining $\{ x \} \mathrel{E^*} \{ y \} \; \Leftrightarrow \; x \in y$. This model will be well-founded.

• Of course, by the Mostowski collapse lemma, if we have any model that is well-founded and extensional, that model is isomorphic to a model that is "standard" in the sense given here - and the collapse lemma also produces a transitive model, which is a stronger condition. So it would not cause any real difference to define a standard model to be one that, when viewed from the outside, has a well-founded membership relation (extensionality will follow automatically from the model satisfying ZF). – Carl Mummert Feb 21 '13 at 14:20
• @Carl: I agree that "nice" models of ZF are for all intents and purposes standard, but the literature does not seem to use standard in this wider sense. – user642796 Feb 21 '13 at 14:24
• !Arthur: my experience with the set theory literature is that "transitive model" is more common than "standard model" (although I am not a set theorist, so my experience is limited). Do Jech or Kunen define the term "standard model" in their texts? – Carl Mummert Feb 21 '13 at 14:32
• @CarlMummert No, neither book uses the term. Curious now about a printed reference. Anyway, standard means that the ordinals are standard, meaning isomorphic to an initial segment of the true ordinals. This is equivalent to saying that the model is well-founded. (In particular, Arthur's example $(M^*,E^*)$ is standard. It is also standard practice to replace well-founded parts by their transitive collapses, so it is common to abuse language and just say that a model is standard if its membership relation is true membership.) – Andrés E. Caicedo Feb 21 '13 at 15:28
• @Carl: As Andres mentions neither Kunen or Jech use the term. I've looked in a few other places and the only text I have been able to find it is in Just and Weese's Discovering Modern Set Theory (which is also the first set-theory text I ever studied). Google provided very few references to the definition I have used. The only positive seems to be that Wikipedia's Inner Model article uses this definition, which is where the OP's question originated. – user642796 Feb 21 '13 at 16:23

There are several different senses in which a set model of ZF set theory could be "standard". I have marked this answer as community wiki, so please feel free to add any other definitions that you think of.

1. The model could have an $\omega$ that is isomorphic to the standard $\omega$. These models are called $\omega$-models.

2. The membership relation of the model could be well founded when viewed from the outside. These models are called well founded models.

3. The model could use the standard membership relation $\in$ rather than some other membership relation. These models are sometimes called standard models.

4. The model could be a transitive set, and use the standard membership relation $\in$. A transitive set is a set $A$ such that, for all $B \in A$, every member of $B$ is also a member of $A$. These models are called transitive models.

For any given model, (4) implies (3) implies (2) implies (1), and none of the converse implications hold in general.

The Mostowski collapse lemma shows that every model of ZF satisfying (2) is isomorphic to a model satisfying (4).