Existence of closures of sets

Question

I am reading Stillwell's Elements of Algebra. And in Chapter 1, he introduces the real quadratic closure of $\mathbb Q$ as

the set of the numbers obtainable form $\mathbb Q$ by square roots of positive numbers.

He also defines $\mathbb N$ as

the closure of the set $\{0\}$ under suucessor, that is, the intersection of all sets $S$ such that $0\in S$ and $n+1\in S$ whenever $n\in S$.

Though these "definitions" seem quite intuitive, I have the following issue with these definition:

1. For the first definition: Without assuming the existence of $\mathbb R$ (for defining the square roots of positives), how can we rigorously construct (that is prove the existence and uniqueness) the set?
2. For the second defintion: What are ''$0$'' and ''$1$''? Also, how is $+$ defined? And does the set of all such sets $S$ exist (and is it nonempty?) for us to allow to take the intersection?

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In general, how does one "close" a set, rigorously, with respect to a certain operation?

Answer 1

1. Add elements until you can't do it anymore.

   Try it! Define $S_0 = \mathbb Q$, we define the "formal square root" of a number $x$ as a symbol (not a number! it's just an arbitrarily chosen symbol with no meaning.) $\sqrt x$. Let $S_{i+1}$ be the set of all the different symbols you can get by adding, subtracting, square root, etc, modulo the relation generated by $(\sqrt x)^2 \sim x$. Then we can define $S_{\infty} = \bigcup_i S_i$. (You need to verify that the equivalence relation is respected.)

2. No, we don't define "1" and "+", instead we talk about the operation "+1". Also, however "+1" is defined, as long as it satisfies the Peano axioms (or pick any axiomatization of the natural numbers you like), the results will be the same. The usual definition is $x+1 = x \cup \{x\}$, with $0 = \varnothing$.

   Also, we don't need that "the set of all such sets" exist to take intersection. Depending on your axiomatization, you should instead go find a set $r$ such that $x \in r \Leftrightarrow \forall s. x \in s$, where $s$ runs over all the sets that you want to take intersection of.

Answer 2

I think your latter question isn't all that related to your first questions, but here we go:

Formally, if you have a way to generate $X_{n+1}$ from $X_n$, then you can generate $X = \cup X_i$ as follows:

• Define the nth "attempt" $f_n$ to be the function-class $f_n : \{0, \dots, n-1\} \to V$ (where $V$ is the universe of sets), given by $i \mapsto X_i$.
• Prove by induction that for any $n$, there is an attempt defined on $\{0, \dots, n-1\}$, and prove by induction that any two such attempts are equal.
• Define the union function-class $f$ of all the above function-classes (justified because we've just shown that they are all initial segments of each other, so if you want to know $f(i)$, you can just ask for $f_{i+1}(i)$.).
• Use the axiom of replacement to form the genuine set $U$ which is the image of $\mathbb{N}$ under the function-class $f$ defined above. (That is, $U = \{X_0, X_1, \dots\}$.)
• Use the union axiom to take the union of $U$.

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The "take the intersection of all sets with a certain property $p$" operation is a common one, and it's fine as long as you've shown there is some set $X$ with that property. Then you can form $\{ x \in X : \forall y, p(y) \to x \in y \}$. This is a genuine set by the axiom of comprehension (or subset selection, or separation, or whatever you call it). You just had to make sure you were restricting your attention to some small fragment of the universe (in this case, $X$) before you started.