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Article Forcing in nonstandard analysis states in the page 266: "For an infinite ordinal $\alpha$, any set $X$ such that every element of an element of $X$ has rank $\alpha$ is a base set and hence any set can be easily replaced by a base set with the same size".

First I will give two definitions and one proposition concerning the first part of the above quotation. The definitions and proposition below are in the book "True infinitesimal differential geometry", written by Mikhail G. Katz.


Definition 1 (base sets): A set $X$ is a base set if $X\neq \emptyset$ and if $x\in X$ then $x\neq\emptyset$ and $x\cap V(X)=\emptyset$ in which $V(X)$ is the superstructure over the set $X$.

Definition 2: The von Neumann rank of a set $x$ is an ordinal $\text{rank} (x)$ defined so that $\text{rank} (\emptyset)=0$ and if $x\neq \emptyset $ then $\text{rank} (x)$ is the least ordinal strictly greater than all ordinals $\text{rank} (y)$, $y\in x$.

Definition 3: Let $\gamma$ be an ordinal. A set $X$ is a $\gamma$-base set if $X\neq \emptyset$, $X$ consists of non-emptyset elements, and we have $\text{rank}(a)=\gamma$ whenever $a\in x\in X$.

Proposition: If $\gamma$ is an infinite ordinal, and $X$ is a $\gamma$-base set, then $X$ is a base set.

Proof. See the page 161 of the book http://u.math.biu.ac.il/~katzmik/tidg.pdf


My question is: how do I prove the second part of that quotation using the definitions and proposition above? That is, how to prove that any set can be replaced by a base set with the same size?

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The point is that the number of sets of a given rank grows without bound, so by making $\gamma$ large enough, a $\gamma$-base set can have arbitrarily large cardinality. Indeed, the set of sets of rank $\gamma$ is $V_{\gamma+1}\setminus V_\gamma$. Since $|V_{\omega+\alpha}|=\beth_\alpha$, this means there are $\beth_{\alpha+1}$ sets of rank $\omega+\alpha$ for any ordinal $\alpha$. So, just pick $\alpha$ such that $\beth_{\alpha+1}\geq|X|$ (or even just $\beth_{\alpha}\geq|X|$), and then there will be at least $|X|$ different nonempty sets of sets of rank $\omega+\alpha$, so you can pick a set of of such sets with the same cardinality as $X$.

Or, for an explicit construction, let $X'=\{\{\{X,a\}\}:a\in X\}$. Then $X'$ is in bijection with $X$ by mapping $a$ to $\{\{X,a\}\}$, but every element of an element of $X'$ has the same rank (namely $\operatorname{rank}(X)+1$, since $a\in X$ so $a$ has smaller rank than $X$).

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