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In Naive Set Theory By Halmos he states the axiom of unions as:

For every collection of sets there exists a set that contains all the elements that belong to at least one set of the given collection.

Isn’t it better to say replace at least one with any? I actually think *at least one * is wrong because if my collection had sets {a, b, c} but my union contained all the elements of a then I’ve satisfied the definition but that’s not our typical understanding of the union.

How do I reconcile this?

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  • $\begingroup$ Of course the union contains all the elements of $a$. Does it contain only the elements of $a$? It does if $b \subseteq a$ and $c \subseteq a$, so that each element of $a$ is an element of one, two, or three sets in the collection... $\endgroup$ Commented Apr 28 at 9:29
  • $\begingroup$ Alternatively, perhaps you are thinking that we might write $a \cup b \cup c = a$ by ignoring the $b$ and $c$. But then you haven't correctly applied the "all" from "contains all the elements". The elements of $b$ and $c$ are also elements that are contained in at least one of the sets and you have to include all elements that do that. $\endgroup$ Commented Apr 28 at 9:30
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    $\begingroup$ It definitely is not incorrect. You're wrongly equating the statement "a set that contains all the elements that belong to at least one set of a given collection" to "a set whose elements all belong to at least one set of a given collection." Your counterexample only works for the second definition, which is not the one given. The difference is that the hypothetical set in question (the union) must indeed contain every element belonging to the collection, whereas the second definition refers to any subset of the union. $\endgroup$ Commented Apr 28 at 9:35
  • $\begingroup$ Thank you i see my error $\endgroup$
    – Darby Bond
    Commented Apr 28 at 9:37

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