How to produce uncountable ordinal from universe?

Question

A universe $U$ is defined as the following.

(1) If $x\in u\in U$, then $x\in U$.

(2) If $x\in U, u\in U$, then $\{x,u\}\in U,x\times u\in U$.

(3) If $x\in U$, then power set $\mathcal{P}(x)\in U,\cup_{x\in x}x\in U$.

(4) $N=\{0,1,2,\dots\} \in U$ is the set of all finite ordinals.

(5) If $f:a\to b$ is surjection s.t. $a\in U$, then $b\in U$.

From $(4),(1),(2)$, I can deduce that all finite ordinals $\{0,\dots, n\}\in U$ exists. From $(3)$, I can deduce $\mathcal{P}(N)\in U$ which is not countable.

$\textbf{Q:}$ How do I see usual sets of real numbers and related infinite sets in $U$? Consider $\{\pm 1\}\times N\times N\to Q$ via realization of rational numbers $Q$ as $\pm\frac{a}{b}$ form. This is surjection. The former $\{\pm 1\}\times N\times N\in U$. Hence $Q\in U$. How do I see $R$ constructed by power set operation and product operation? I can consider $2^N$ to define 2-adic expansion in $[0,1]$ but does this see the whole interval?

Answer 1

EDIT: You've misstated $(5)$ in a crucial way; the right version is

If there is a surjection from $a$ to $b$, $a\in U$, and $b$ is a subset of $U$, then $b\in U$.

This is much weaker than what you've written. In particular, according to your definition there is exactly one universe, namely the class of all sets $V$: if $U$ is a universe, then by your version of $(5)$ for each $a\in V$ we would have $\{a\}\in U$, and so $(1)$ would imply $a\in U$.

We will in fact only care about the following consequence of the condition above:

$(*)\quad$ Universes are closed under subsets: if $a\subseteq b\in U$, then $a\in U$.

To see this, there are two cases. If $a\not=\emptyset$ then fixing $x\in a$ we can define a surjection from $b$ to $a$ by sending $y\in b\setminus a$ to $x$ and $y\in a$ to $y$, and since $b\subseteq a$ we can apply the condition above. And $\emptyset\in U$ follows from $(1)$ + $(4)$ (every nonempty transitive set has $\emptyset$ as an element).

EDIT: Per Eric Wofsey's comments below, the right version of (5) is in fact massive overkill here; I'm using it to give you a sense of how it can be applied.

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Keep in mind that we're reasoning within some theory. E.g. in ZF, we can talk about universes, and we can ask what is true of all universes.

I claim the following (and I believe this is what MacLane is getting at):

ZF proves that $\mathbb{R}\in U$ for every universe $U$.

The proof is straightforward:

• We have $\mathbb{N}\in U$ by assumption.

• So $\mathcal{P}(\mathbb{N})\in U$.

• There is a surjection from $\mathcal{P}(\mathbb{N})$ to $\mathbb{R}$ (remember, we're working inside ZF here).

• So we know $\mathbb{R}\in U$ as well - if we can show that $\mathbb{R}\subseteq U$ (this isn't necessary with respect to the definition you've given, but see above).

• To do this, we need a formal definition of $\mathbb{R}$. (Any of the usual ones will work, but we do need to fix one.) Dedekind cuts are particularly nice. First, let's take for granted that $\mathbb{Q}$ (defined appropriately) is in $U$. From $(*)$ it follows that every individual Dedekind cut is in $U$, and applying $(*)$ a second time tells us that $\mathbb{R}$ - the set of all Dedekind cuts - is also in $U$.

• Now how do we get $\mathbb{Q}\in U$? Your argument uses your version of $(5)$, so doesn't work here. But it's easy to show that each specific rational is in $U$, so $(*)$ suffices.

Answer 2

How do I see usual sets of real numbers and related infinite sets in $U$?

We usually define everyone's favourite numbers in the following way, or a slight variant if it's convenient:

• Your (5) is called replacement and implies something else called comprehension, meaning you can form arbitrary subsets of sets. (For a proof, see here.) I'll use this to form whatever subsets I like as I go.
• The ordered pair $(a,\,b)$ is usually defined as $\{\{a\},\,\{a,\,b\}\}$; you can prove this is a set with your axioms, but you probably already knew that since you're happy with Cartesian products viz. $x\times u$ in (2). The constructions of interest herein frequently proceed as follows: form whatever ordered pairs we want, then form a set of them that's a subset of some Cartesian product.
• Given a set $A$, the relations on $A$ are the subsets of $A\times A$. See here if you don't know about a specific kind of relation called equivalence relations. I'll leave it to you to see why your axioms imply there is a set of equivalence relations on $A$.
• The set you called $N$ is normally denoted $\omega$, but let's split the difference and call it $\Bbb N$. An integer is identified with the ways to express it as a difference of non-negative integers. More formally, the integers $\Bbb Z$ are then the equivalence classes on $\Bbb N^2$ with $(a,\,b)\sim (c,\,d)$ defined as $a+d=c+b$, so e.g. $(1,\,2),\,(3,\,4)\in -1$.
• Similarly, the rationals $\Bbb Q$ are the equivalence classes on $\Bbb Z\times (\Bbb Z\backslash\{0\})$ with $(a,\,b)\sim (c,\,d)$ defined as $ad=cb$, so e.g. $(2\,4)\in\frac12$.
• There are actually two popular ways to go from $\Bbb Q$ to $\Bbb R$, with each real number either a Dedekind cut or an equivalence class of Cauchy sequences in $\Bbb Q$. The former identifies a real number with the set of rationals less than it; the latter with the set of rational Cauchy sequences of which it is the limit. Either way, you'll be using Cartesian products a lot.
• Formally, a Dedekind cut is an ordered pair $(A,\,B)$ of nonempty subsets of $\Bbb Q$ of union $\Bbb Q$ and empty intersection, such that each element of $A$ is less than each element of $B$. For example, we could take $A=\{x\in\Bbb Q|x<0\lor x^2<2\}\,B=\{x\in\Bbb Q|x>0\land x^2>2\}$, so our cut corresponds to $\sqrt{2}$.
• You can think of a sequence in $S$ as a surjection from $\Bbb N$ or a subset thereof to $S$ or a subset thereof, so there is clearly a set of sequences in $\Bbb Q$ (and by comprehension a set of Cauchy sequences, on which we can construct equivalence relations/classes; our interest is in calling sequences $a_n,\,b_n$ equivalent if $\lim_{\to\infty}(a_n-b_n)=0$).

I can consider $2^N$ to define 2-adic expansion in $[0,1]$ but does this see the whole interval?

So your strategy gets a binary representation of the elements of $(0,\,1)$, from which we can get the rest of $\Bbb R$ from your (5) with a bijection such as $x\mapsto\arctan\pi\left(x-\frac12\right)$.

On an unrelated note, your title and original edit had an interest in uncountable ordinals and cardinals in general. For the former, see here. The latter is problematic because cardinals are usually defined in a way that depends on the axiom of choice or an axiom I mentioned in a comment - the axiom of regularity - which we're not meant to use here. Yes, there are two ways it can work; see here for both of them.

If you find this or anything else confusing, don't worry; the universe definition used here is almost identical to the most popular set theory. Your (2) gives us the axiom of the unordered pair, your (3) the power set and union axioms, your (4) the axiom of infinity, and as aforesaid your (5) replacement and hence comprehension. With extensionality (which you accepted in a comment), you've already committed to ZF-, which is just the standard "Zermelo-Fraenkel set theory" except for regularity (its omission is what the - means). So most set theory proofs you'll find anywhere fit your universe nicely, unless they really can't be done without the axiom of regularity (and possibly something else an individual set theory may or may not embrace, such as the axiom of choice).

Note: as has been pointed out in comments, (1)-(5) aren't so much set-theoretic axioms as defining properties of universes. However, the procedure for proving e.g. $\Bbb R\in U$ works in pretty much the same way. Similarly, (5) isn't quite the same as the usual replacement but, again, the usual proofs in ZF adapt well.