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Motivation: Observation 2: The Mostowski Function (Sometimes) Produces Relation Embeddings

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.

Ad Example 1

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 $n=1, m=3.$

Ad Example 2

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 $n=2, m=3.$

Ad Examples 3 and 4

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 example 3: $\pi(\{b\})\in\pi(\{a,c\})$ but $\{b\}\not\in \{a,c\}.$
  • In the example 4: $\pi(2)\in\pi(9)$ but $2$ does not divide $9.$

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 discovered that this has to do with the question, whether the underlying relation $R$ is extensional or not.

  • We have seen that strict orders are extensional, therefore, in example 1 there is an embedding possible.
  • In example 2, the successor relation $nRm:\Leftrightarrow n+1=m$ is also extensional, since from $\{n\in\mathbb N\mid n+1=m_1\}=\{n\in\mathbb N\mid n+1=m_2\}$ it follows that $m_1=m_2.$
  • It is left for the reader as an exercise to verify that the relations in the examples 3 and 4 are not extensional.

| | | | created: 2019-03-02 22:48:50 | modified: 2019-03-03 08:15:44 | by: bookofproofs | references: [8055]


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Bibliography (further reading)

[8055] Hoffmann, D.: “Forcing, Eine Einführung in die Mathematik der Unabhängigkeitsbeweise”, Hoffmann, D., 2018

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