# Adjoining new ordinals to a model — a question about one of Cohen's articles

The following is an excerpt from Paul Cohen's "The Discovery of Forcing", pp 1091, in which he explains why we do not want to add new ordinals to a countable transitive model $M$ when extending it using forcing:

Suppose $M$ were a countable model. Up until now we have not discussed the role countability might play. This means that all the sets of $M$ are countable, although the enumeration of some sets of $M$ does not exist in $M$. The simplest example would be the uncountable ordinals in $M$. These of course are actually countable ordinals, and hence there is an ordinal $I$, not in $M$, which is countable, and which is larger than all the ordinals of $M$. Since $I$ is countable, it can be expressed as a relation on the integers and hence coded as a set $a$ of integers. Now if by misfortune we try to adjoin this $a$ to $M$, the result cannot possibly be a model for ZF. For if it were, the ordinal $I$ as coded by $I$1 would have to appear in $M(a)$. However, we also made the rigid assumption that we were going to add no new ordinals. This is a contradiction, so that $M(a)$ cannot be a model. From this example, we learn of the extreme danger in allowing new sets to exist. Yet $a$ itself is a new set. How then can we satisfy these two conﬂicting demands?

1 I think this is a typo and he meant to write $a$.

What I understand: If $M$ is countable then it cannot contain all countable ordinals (since the set of all countable ordinals is itself uncountable) hence there is at least one countable ordinal not in $M$. Since it is countable it is a subset $a$ of $\omega$. If we adjoin $a$ to $M$ then in particular $a \in M(a)$ so that we have added an ordinal that was previously not in $M$.

What I don't understand:

Why can't $M(a)$ possibly satisfy ZF? If $M$ is a countable model of ZF and we add an ordinal $a$ not in $M$, why is it impossible for $M(a)$ to still satisfy ZF?

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Note that part of the working assumption is that passing from $M$ to $M(a)$ does not add ordinals; earlier on that page P. Cohen states:

An important decision is that no new ordinals are to be created.

and also in the text you quoted:

However, we also made the rigid assumption that we were going to add no new ordinals.

So $M(a)$ cannot satisfy ZF because it does not contain the ordinal that $a$ encodes.

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Thank you! I'm sorry but I still don't understand. My current understanding of Cohen's text is "Assume adjoining $a$ to $M$ does not add any new ordinals. Let $a$ be an ordinal not in $M$. Then $M(a)$ contains $a$ and hence an ordinal that was not previously in $M$ which is a contradiction to $M(a)$ does not contain any new ordinals. Hence $M(a)$ cannot satisfy ZF." What am I missing? –  Matt N. Jan 24 '13 at 12:19
@Matt: The text you quoted was really about some of the technical problems related to the discovery of forcing. Once he made the decision that ordinals should not be added he had to ensure that the sets he did add did not somehow encode new ordinals. But these new sets also had to be... new. This is (I believe) what he means by "these two conflicting demands." –  Arthur Fischer Jan 24 '13 at 12:36
@MattN. The way I read Cohen's text, $M(a)$ is the result of adding $a$ to $M$ in some unspecified way. What he wants from it is that it contains $a$ and has no ordinals that weren't already present in $M$. He then argues that if $M(a)$ satisfies all of this it cannot be a ZF model; if it were, we could decode $a$ and a new ordinal would appear. –  Miha Habič Jan 24 '13 at 12:37
@Matt: (This is a correction of my previous comment; lots of errors there.) Also note that $a$ is not an ordinal in the text quoted. As $I$ is assumed to be a countable ordinal, there is a well-ordering on $\omega$ of order-type $I$. We can then encode this well-ordering (via, say, a nice pairing function) as a subset of $\omega$. The new set $a$ is then this encoding in this example. –  Arthur Fischer Jan 24 '13 at 12:59
@Matt: If $a$ codes a well-ordering of $\omega$, then by Replacement any model of ZF containing $a$ must contain the order-type of the well-ordering it codes. –  Arthur Fischer Jan 24 '13 at 17:07