(related to Definition: Mostowski Function and Collapse)
We now take again the initial example and the other examples to make another observation. Remember that we can treat the contained relation $\in_X$ as a real relation between the elements of some given set $X.$ Moreover, please recall the introduced concept of an order embedding, in the case $X$ is ordered. Now, the relation $R$ used to define a Mostowski function $$\pi(x):=\{\pi(y)\mid y\in V\wedge yRx\}$$ has not necessarily to be an order relation, but it must be a well-founded relation. We want to observe, if $\pi$ is capable to produce embeddings of $R$ in the sense, that the following equivalence is fulfilled:
$$nRm\Longleftrightarrow \pi(n)\in_X\pi(m).\quad ( * )$$
Please note that the Mostowski function $\pi$ is right-unique, as every function is. The above equivalence is fulfilled in the cases, in which $\pi$ is, in addition, left-unique (i.e. injective).
After these preliminary considerations, we can now start to make our observations. In our case, the set $X,$ in which the contained relation $\in_X$ is defined, is the Mostowski collapse $X:=\pi([V]).$ The first two examples seem to fulfill the property $( * ),$ and we write below $\in$ instead of $\in_{\pi([V])}$ for readability reasons, but the reader should be aware that we mean the contained relation defined on the corresponding Mostowski collapse.
Two natural numbers $n$ and $m$ are in the strict order $n < m$ if and only if the corresponding image of the Mostowski function $\pi(n)$ is an element of the image $\pi(m).$ In other words, we have for all $n,m\in\mathbb N$ the relation embedding
$$n < m\Longleftrightarrow \pi(n)\in \pi(m).$$
It is left for the reader as an exercise to verify it for some special cases, for instance $n=1, m=2$ or $n=1, m=3.$
The natural number $m$ is the successor of the natural number $n$, if and only if the corresponding image of the Mostowski function $\pi(n)$ is an element of the image $\pi(m).$ In other words, we have for all $n,m\in\mathbb N$ the relation embedding
$$n+1=m\Longleftrightarrow \pi(n)\in \pi(m).$$
It is left for the reader as an exercise to verify it for some special cases, for instance $n=1, m=2$ or $n=2, m=3.$
In theses examples, there is no embedding possible and the property $( * )$ is only fulfilled in one direction $nRm\Longrightarrow \pi(n)\in_X\pi(m).$ The opposite direction is broken. For instance:
In the above examples, the new question arises, why $\pi$ in some cases does produce an embedding $( * )$ and why it does not in other cases. [Mostowski][mo] discovered that this has to do with the question, whether the underlying relation $R$ is extensional or not.
[mo][https://mathshistory.st-andrews.ac.uk/Biographies/Mostowski/]
Theorems: 1
