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I've been reviewing the axioms of ZFC, and I'm trying to make sense of why the "Axiom of Union" was put in place. While the existence of the intersection (of two) sets seems to be a "Theorem" we can prove from the Axiom of Separation, at first I had trouble seeing the gap between the Axiom of Separation and the construction of the union.

Is the idea that we have no way to express the construction of an arbitrary union (of subsets) by appropriate formula $F(x)$ involving a general set of variables, as we do in the construction of the intersection?

I can see how the Axiom of Separation gives the existence of the union for a finite number of (chosen) subsets, but I guess that using the Axiom of Separation is the strongest construction?

Kenta S
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AF3
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  • I've added the "axioms" tag. – Carl Mummert Sep 28 '14 at 01:16
  • I am guessing that if you assume the existence of arbitrary intersections, you can simply define $\bigcup \mathcal C=\bigcap {C:A\subseteq C\forall A\in\mathcal C}$. Indeed, there are many sets that contains $A$ and $B$, but we want the smallest such set. – Pedro Sep 28 '14 at 01:24
  • @PedroTamaroff You can't do that, because ${C:A\subseteq C}$ is not a set. – Thomas Andrews Sep 28 '14 at 01:37
  • Oh. I did say "there are many sets", right @ThomasAndrews? – Pedro Sep 28 '14 at 01:40
  • But you didn't say "too many" :) @PedroTamaroff – Thomas Andrews Sep 28 '14 at 01:43
  • It seems to me like the idea of the union axiom is that we will be able to express a given set as some kind of partition -- is that one the real powers of the axiom? – AF3 Sep 28 '14 at 01:47
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    I suppose that you only need the axiom to say that there exists a set $U$ such that for each $A\in \mathcal C$, $A\subseteq U$. Essentially, you need to assert an upper bound. The existence of a least upper bound follows from other axioms. – Thomas Andrews Sep 28 '14 at 02:07
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    Possible duplicate of Can we prove the existence of $A\cup B$ without the union axiom? – Bunny Aug 09 '17 at 00:08
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    @Pedro No, the issue is not that there are too many sets $C $ as in your proposal. The problem is the opposite: you cannot prove that there are any such sets at all in general without the axiom. We wouldn't need the axiom otherwise, by separation. (Of course, if we assume separation, there is no need to assume the existence of arbitrary intersections: We can just prove it.) – Andrés E. Caicedo Aug 09 '17 at 01:10
  • @AndrésE.Caicedo. Precisely. If we can prove there exists $ C$ such that $\forall A\in \mathcal C;(A\subset C)$ then we can use the other axioms to obtain $\cup \mathcal C.$.... But we can't find such a $C$ without Union...... – DanielWainfleet Aug 09 '17 at 05:48

2 Answers2

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No, the problem is that union gives a bigger set. Given $A$ and $B$ what set contains $A\cup B$ ?

On the other hand $A$ contains $A \cap B$.

Rene Schipperus
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  • I thought this was kind of built in to the Axiom of Separation. If were to use this Axiom to construct the union of two (arbitrary sets), wouldn't this be: (All x)(Exists y)(x e y <==> x e z ^ (xeA or xeB) ? – AF3 Sep 28 '14 at 01:56
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    What is $z$? There is a no $z$, just $A,B$, so your formulation doesn't make sense. @AF3 – Thomas Andrews Sep 28 '14 at 02:09
  • Anyway, it would have to be of the form $$ \exists y:\forall x: x\in y\iff (x\in A\lor x\in B)$$ which is not a separation axiom formula. On the other hand, intersection is: $\exists y:\forall x:x\in y\iff (x\in A\land x\in B)$. This is constructible via the axiom of separation. @AF3 – Thomas Andrews Sep 28 '14 at 02:12
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First let's take a look at the entire formal statement which gives the Axiom Schema of Separation:

$$\forall w_1, \dots ,w_n \forall X \ \exists Y \ \forall x(x\in Y \iff [x \in X \ \land \ \phi(x, X,w_1, \dots ,w_n)])$$

What Thomas Andrews has written in the comment below the previous answer is that there can be no such formulation of $\phi$ where $(x\in A\lor x\in B)$ because there would have to be a set $X$ that already contained $A \cup B$. Note that the Axiom Schema of Separation implies that $Y$ be a subset of $X$ but we can't say that the existence of such a $X$ is assured and therefore $\phi$ cannot be formulated to satisfy the statement of the axiom schema.

Now we can prove that the unions of finite sets exist without the union axiom but that requires the use of the Power Set Axiom. Please see Can we prove the existence of $A\cup B$ without the union axiom? if the questions aren't marked as a doubles yet.

Bunny
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