In ZFC, the axiom of union is the statement that the union of any set exists. Suppose we omit this axiom, and instead add the axiom that binary unions exist. Can we still derive the full union axiom?
Can the union axiom be weakened
set-theory
Related Solutions
Nice question!
It turns out that in $T=\mathsf{ZFC}-\mathrm{Union}$, we can prove that unions of finite sets exist. There is a (remarkably recent) paper devoted to this issue. What follows is a (very) brief account of my reading of (section 3 of) the paper:
Greg Oman. On the axiom of union, Arch. Math. Logic, 49 (3), (2010), 283–289. MR2609983 (2011g:03122).
(Unfortunately the paper is behind a paywall, and the author does not have a version in their page or in the arXiv. Contact me if you find difficulties accessing it.)
Of note is that the argument uses choice in essential ways, so the question remains whether unions of finite sets can be proved to exist in $\mathsf{ZF}-\mathrm{Union}$. (Similarly, but perhaps less surprisingly, replacement also plays a key role.)
First, note that $T$ proves that if $x,y$ are sets, then $x\cup\{y\}$ exists: Map each singleton $\{z\}$ in $\mathcal P(x)$ to $z$, and every other subset to $y$. Call this Lemma 1.
The basic theory of ordinals can be carried out in $T$ (e.g., any two distinct ordinals are comparable by $\subset$ and by $\in$), and (crucially) $T$ proves that every set is in bijection with an ordinal. Call this Theorem 1.
This argument requires some care for two reasons: First, the usual approach uses transfinite recursion, which uses the union axiom. Second, arguing about the existence of functions requires some care, since at this stage it is not even clear that Cartesian products exist. However, we have a way of circumventing the second issue, since some functions do exist, since we have the axiom of choice (stated in the form: Choice functions exist).
Oman's argument (which is really a careful reworking of Zermelo's original approach) goes as follows: Given $X$, begin with a choice function $G$ on the nonempty subsets of $X$. Call $f$ an approximation iff its domain is an ordinal, its range is contained in $X$, and for each $\alpha\in\mathrm{dom}(f)$, $$ f(\alpha)=G(X\setminus\{f(\beta)\mid \beta<\alpha\}). $$ Now check that approximations are compatible, and that every element of $X$ is in the range of some approximation. For the latter, letting $Y$ be the subset of $X$ consisting of those elements of $X$ that appear in the range of some approximation, note that for each $y\in Y$ there is a unique $\alpha_y$ such that there is some approximation $f$ with $\alpha_y\in\mathrm{dom}(f)$ and $f(\alpha_y)=y$. It follows that $F=\{(\alpha_y,y)\mid y\in Y\}$ exists, and one readily verifies that it is itself an approximation.
One completes the argument by checking that the range of $F$ is $X$. But if it is not, then letting $\gamma=\mathrm{dom}(F)$, we have that $F\cup\{(\gamma,G(X\setminus Y))\}$ is also an approximation (thanks to Lemma 1!). This is a contradiction, since the definition of $Y$ then implies that $G(X\setminus Y)\in Y$, which is absurd as $G$ is a choice function. This completes the proof of Theorem 1.
By the way, if you are not familiar with this argument, you may want to read the nice survey by Kanamori and Pincus,
Akihiro Kanamori, and David Pincus. Does $\mathsf{GCH}$ imply $\mathsf{AC}$ locally?. In Paul Erdős and his mathematics, II (Budapest, 1999), pp. 413–426, Bolyai Soc. Math. Stud., 11, János Bolyai Math. Soc., Budapest, 2002. MR1954736 (2003m:03076).
(Theorem 3.3.c is false, but this does not affect their exposition on Zermelo's argument.)
Using Theorem 1, Oman argues that (finite) Cartesian products exist: Given $x,y$, find ordinals $\alpha,\beta$ in bijection with them. If $\gamma=\max\{\alpha,\beta\}$, then it is easy to check that $\alpha\times\beta$ exists, since it is a subset of $\mathcal P(\mathcal P(\gamma))$. Replacement completes the proof.
It follows immediately that the collection of functions from $x$ to $y$ exists. Using this, we can now prove
Theorem 2. If $x$ is a set and $\{|y|\colon y\in x\}$ is a bounded set of ordinals, then $\bigcup x$ exists.
Note that the boundedness assumption is obviously necessary, and it is immediate in the presence of the union axiom. This could therefore be seen as the key feature that the union axiom gives us. As an immediate corollary, if $x$ is finite, then $\bigcup x$ exists, answering your question affirmatively.
Let's conclude by sketching the proof of Theorem 2: We may assume that no element of $x$ is empty. Let $\gamma$ be a bound for the cardinalities of the elements of $x$. For $y\in x$, let $S_y$ be the set of all surjections from $\{y\}\times\gamma$ onto $y$, and note that $S_y\ne\emptyset$. Now consider a choice function $G$ on $\{S_y\mid y\in x\}$, so for each $y\in x$, $G(y):\{y\}\times\gamma\to y$ is onto. Replacement now implies that $\bigcup x=\{G(y)(y,\alpha)\mid (y,\alpha)\in x\times\gamma\}$ is a set, and we are done.
Of course, if we have a complete ordered field $F$ we can define a copy of $\Bbb N$ inside it: $0$ and $1$ are the ones from the field, and we can define $\Bbb Z$ as the smallest subring that contains $1$ and $\Bbb N$ are just the elements $\ge 0$ in that $\Bbb Z$. I don't see any issue with that and it's well-known that from $\Bbb N$ we can construct a complete ordered field. So assuming one or the other is equivalent. But $\Bbb N$ is more in line with a minimality ideal: assume the simplest object as an axiom (Occam's razor like). And we have the tradition of starting with Peano's axioms in foundational theory as well. So it's never done in "your" order.
Best Answer
No. Binary unions are a theorem of the Axiom of Replacement:
I couldn't personally find a proof of this. With choice it's relatively easy: Assume w.l.o.g. $|A| \geq |B|$. From powerset plus specification we have a set with $2 |A|$ elements. For ease of language we'll label our two images of $A$ we've picked with $0$ and $1$. From the inclusion $B$ into $A$ (call it $f$), map $(0, a)$ to $a$, and $(1, a)$ to the unique preimage of $a$ under $f$, if it exists or $\emptyset$ otherwise. Applying axiom of replacement on our set, we get a superset of the union $A \cup B$.
I needed choice to get $|A| \leq |B|$ or $|B| \leq |A|$ for the first step.