Talk:Axiom of union

Intersection of the empty set
I added a brief note about the reason that the "intersection of the empty set" makes no sense in ZF; but I'm no set theorist, so please feel free to revert if I'm talking crap. Dmharvey 20:28, 21 April 2006 (UTC)


 * Your statement seems correct, assuming an axiom of intersection which goes like: the intersection of a set is the set consisting of all sets which are contained in members of A, or $$\forall A, \exist B, \forall C: C \in B \iff (\forall D: (C \in D \land D \in A) \lor D \notin A)$$. But you could patch up the axiom to exclude the empty set: $$\forall A, \exist B, \forall C: C \in B \iff (\forall D: (C \in D \land D \in A) \lor D \notin A) \land \exist E: C \in E \land E \in A$$ or equivalently $$\forall A, \exist B, \forall C: C \in B \iff (\forall D: (C \in D \land D \in A) \lor D \notin A) \land \exist E: E \in A$$ which makes the intersectionof the empty set the empty set. Of course this should probably be interpreted as that there is no "intersection of the empty set", but the article now reads as if that is the reason that there cannot be an axiom of intersection, which I think is not right. --MarSch 09:19, 24 April 2006 (UTC)


 * I've made a few changes to address your concerns... is that an improvement? I think it's an important point to mention, because anyone with any familiarity with "naive set theory" who is reading about ZF for the first time will see the axiom of union and wonder "why isn't there an axiom of intersection", and pretty much the reasons are: (1) ZF is designed so you don't need it, and (2) if you have an unrestricted axiom of intersection, you run into all the sort of "universe" problems that ZF is specifically trying to avoid. Dmharvey 11:01, 24 April 2006 (UTC)

actually, the intersection of an empty set makes perfect sense with the stated definition, since the intersection is defined as a subset of the sum, and therefore, for an empty set, it is empty.--78.8.74.38 (talk) 02:24, 14 March 2010 (UTC)


 * To 78.8.74.38: No. You should not just make arbitrary definitions. Your "definition" of intersection is not what is meant by that term. So I reverted you. JRSpriggs (talk) 03:01, 15 March 2010 (UTC)

Interpretation section error?
The first line in the interpretation section is:

"What the axiom is really saying is that, given a set A, we can find a set B whose members are precisely the members of the members of A."

Shouldn't this be something like the following given the formal statement $$\forall A\, \exists B\, \forall c\, (c \in B \iff \exists D\, (c \in D \land D \in A)\,)$$ ?

"What the axiom is really saying is that, given a set A, we can find a set B whose members are contained in members of A." — Preceding unsigned comment added by Zair.M (talk • contribs) 00:04, 11 December 2014‎ (UTC)

Infinite Intersection
An intersection axiom can be made even more redundant if we set E = union X, we then have for the infinite intersection:


 * inter X = { y in union X | forall x (x in X -> y in x) }

Jan Burse (talk) 20:17, 6 March 2017 (UTC)

Independence of the axiom of union
For background, see Zermelo–Fraenkel set theory, Von Neumann universe, Beth number, and Transitive model.

To show that the Axiom of union is independent of the other axioms of ZFC, we construct a transitive model which satisfies all the other axioms (that is: Axiom of extensionality, Axiom of regularity, Axiom of empty set, Axiom of pairing, Axiom of power set, Axiom schema of specification, Axiom schema of replacement, Axiom of infinity, and Axiom of choice) but does not satisfy the axiom of union. If the axiom of union could be proved from the other axioms, then this model would be impossible. So we will have proved independence.

For any ordinal &alpha;, define
 * $$ W_\alpha := \bigcup_{\beta < \alpha} \{ X : X \subseteq W_\beta \land \vert X \vert < \beth_\omega \} $$

by transfinite induction.

Then one can show
 * $$ \alpha < \beta \Rightarrow W_\alpha \subseteq W_\beta $$

and $$ W_\alpha $$ is a transitive set.

Furthermore
 * $$ W_{\beth_\omega{}^{+} + 1} = W_{\beth_\omega{}^{+}} $$

so that this hierarchy eventually becomes constant. [The superscript plus sign + means to take the successor cardinal. This will be a regular cardinal since we are assuming the axiom of choice holds in the universe in which we construct the model.]

We take
 * $$ W := W_{\beth_\omega{}^{+}} $$

to be our model. It satisfies the other axioms because none of them can take sets in W (which thus have cardinality less than $$ \beth_\omega $$) and make a set outside of W.

That it fails to satisfy the axiom of union is apparent from the fact that
 * $$ \{ \beth_n : n < \omega \} \in W, $$

but
 * $$ \bigcup \{ \beth_n : n < \omega \} = \beth_\omega \notin W . $$

OK? JRSpriggs (talk) 08:16, 11 August 2019 (UTC)

Here are some more interesting details. W is a transitive set.
 * $$ W_\alpha = W \cap V_\alpha $$.
 * $$ W = \{ X : X \subseteq W \land \vert X \vert < \beth_\omega \} $$.

Any transitive model, such as W, will automatically satisfy the axioms of extensionality and regularity.
 * $$ V_{\omega + \omega} \subseteq W $$
 * $$ \{ V_{\omega + n} : n < \omega \} \in W $$
 * $$ \bigcup \{ V_{\omega + n} : n < \omega \} = V_{\omega + \omega} \notin W $$

We can give a uniform verification for the axioms of separation and thus avoid dealing with the relativization to W of the quantifiers of the differentia. Similarly for the axioms of replacement.


 * $$ \vert X \vert < \beth_\omega \Rightarrow \exists n < \omega \, ( \vert X \vert \leq \beth_n ) $$
 * $$ \vert X \vert \leq \beth_n \Rightarrow \vert \mathcal{P} (X) \vert \leq \beth_{n + 1} < \beth_\omega $$

So the powerset will not take us out of W.

It is also interesting to note that finite unions stay inside W.
 * $$ ( A \in W \land B \in W ) \Rightarrow ( A \cup B ) \in W $$
 * $$ ( A \in W \land \vert A \vert < \aleph_0 ) \Rightarrow \bigcup A \in W $$

OK? JRSpriggs (talk) 00:39, 13 August 2019 (UTC)