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This question arose while learning nonstandard analysis.

The superstructure $V(X)$ of a nonempty set $X$ is defined recursively:

$$\begin{align*}V_0(X) &= X \\ V_{i+1}(X) &= V_i(X) \cup P(V_i(X)) \\ V(X) &= \bigcup_{i=0}^\infty V_i(X)\;, \end{align*}$$

where $i \in \mathbb{N}$ and $P$ is the powerset function. So it's a way to get every relation and function on $X$ that you could possibly want, by identifying the ordered pair $(a,b)$ with the set $\{\{a\},\{a,b\}\}$ or such. I'm great with this.

BUT what happens when there are relations in $X$ (as members of $X$)? I think that I need to distinguish between the relations that are formed by the superstructure construction and any relations that I might have started with in $X$. As an example problem, I want to prove (my book says it's true) that a relation $R$ is in $V(X)$ iff the domain and range of $R$ are in $V(X)$. This is not necessarily true if $R$ is in $X$. The members of $X$ that have members themselves don't necessarily empty out.

So are we supposed to distinguish between relations in $X$ and the relations that are formed from $X$?

Cheers, Rachel

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3 Answers 3

up vote 2 down vote accepted

I've seen the answer to this in several books now. The problem is dealt with simply by treating the members of X as "atoms". Even if you know (from the outside, being all-knowing) that X contains nonempty sets, you declare that members of X don't themselves have members in the same sense that the sets constructed from X have members. You can do this by distinguishing between membership relations or just pretending that members of X aren't sets.

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Just a note to complement Rachel's own answer: one formal way to sidestep this problem is to restrict attention to base sets -- sets $X\neq\emptyset$ such that for any $x\in X$, $x\cap V(X)=\emptyset$. One simply eliminates the problem of "extra membership" by always building superstructures over base sets.

Then the individuals relative to $V(X)$ are all elements in $X$, and those in $V(X)\setminus X$ are sets relative to $V(X)$.

This raises the question of how to replace one set by a base set of the same size. One method is to take a set $X$ of the same cardinality as our original set, but such that every element in $\bigcup X$ has the same rank $\alpha$, with $\alpha$ some fixed infinite ordinal. One can then show that for any $n$, if $x\in V_{n}(X)$, then $rank(x)\neq\alpha$. So no $V_{n}(X)$ can share any element with any $x\in X$, since each $y\in x\in X$ has rank $\alpha$. This means that $x\cap V(X)=\emptyset$ for any $x\in X$, and so $X$ is a base set.

This is how things are presented in the 2012 reprint of Chang & Keisler's Model Theory: see section 4.4.

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It's not clear what you mean by "a relation in $ X $". Do you mean a relation *on* $ X $? If so, a relation on a set $ X $ is usually defined as a subset of $ X \times X $, and using the definition for this that you gave yourself, namely the set of all $\{\{a\},\{a,b\}\}$ for all $ a, b \in X $, then any relation on $ X $ is, as a set of ordered pairs, a subset of the power set of the power set of $ X $, and its domain and range both are subsets of $ X $ and thus both a subset and an element of $ V(X) $.

If, on the other hand, you mean a relation $ R \in X $, then indeed, neither domain or range need to be an element or subset of $ V(X) $. A simple example would be $ R = S \times S $ for some set $ S $ and $ X = \{R\} $. Then clearly, $ dom(R) = range(R) = S \notin V(X) $ and $ S \not\subset V(X) $.

Lastly, if you mean $ R \subset X $, then, by definition, $ R \in P(X) \subset V(X) $, and of course for all $ r \in R \Longrightarrow r \in V(X) $, but in general, $ dom(R) \notin V(X) $. For example, take $ X = R = S \times S $ for some nontrivial set $ S $. Then $ dom(R) = S $ is neither an element nor a subset of $ X $ and thus not in $ V(X) $.

It is possible for both of those cases that domain and range are elements of or subsets of $ V(X) $, which depends on if they were elements or subsets of $ X $. For example, for any relation $ R $ we may just define $ X = R \cup dom(R) \cup range(R) $.

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Thanks for the response. To clarify, I mean "in" to denote set membership, so $R \in X$. Also, I mean an n-ary relation to generally be a set of n-tuples. So as we've now both said, if $R \in X$, the domain and range of $R$ are not necessarily in $V(X)$. So how is this situation dealt with? Any relations in $X$ are special wrt $V(X)$ -- they don't have the same properties as the other relations in $V(X)$. So what are they called or how are they treated? –  Rachel Mar 17 '12 at 5:54
    
I'm not really deeply familiar with nonstandard analysis, where you say this definition comes from. I merely observed the set properties here to confirm what you suspect. I would think that whenever this construction is used, X is chosen in a way that makes your question obsolete (such as discussing $V(\mathbb{R})$), but without knowing details on where you cite that from, I'm just guessing. –  Desiato Mar 18 '12 at 1:15

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