# Why is it important to have disjoint sets in a union for the union to make sense w.r.t the order types?

This question has been bugging me for quite some time now.

Say we have some $\beta$ smaller than some $\gamma$ and a sequence

$\beta$$\epsilon : \epsilon smaller than cf(\beta) cofinal in \beta and say we have some sets An\epsilon and each of these An\epsilon has order type less than \gamma$$n$.

Now $\forall n$ in $\omega$ let $B$n= $\cup$ $A$n$\epsilon$ for all $\epsilon$ < $\gamma$ and suppose in the end I can write $\beta$ as the union of all the $B$n (but that is not really my problem here)

Why can I deduce that $B$n has order type less than $\gamma$$n+1 only if all my sets An\epsilon are disjoint and do not overlap? (since we have a union of less then \gamma sets each of which is of order type less than \gamma$$n$)

Why can't I still guarantee that the $B$n will still have order type $\gamma$$n+1$ if all the $A$n$\epsilon$ are not disjoint?

I know that I need to take the $A$n$\epsilon$ to be [$\epsilon$,$\epsilon+1$) so that they are disjoint.

But why does everything in the union have to be in order?

I hope I conveyed my question clearly. Thanks in advance for any help.

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You didn't say so, but you are speaking of ordinals here and ordinal exponentiation.

The problem is that if you take the union of sets not in order, then you can't control the order type. Let me give an easy example to illustrate the point. Suppose that An for each n < ω consists of a single ordinal. If the ordinal of An is below the ordinal of Am whenever n < m, then the union set Un An will clearly have order type ω. But if I drop that requirement, then I can get any countable ordinal at all! That is, every set is the uion of its singletons. In particular, if α is a countable ordinal, then α = U { {β} | β < α} is the union of countably many one point orders. Indeed, if you go to the non-ordinal context, then every countable order type, such as Q, is the union of countably many singletons.

A similar problem arises in your specific question. You can't guarrantee that the union of the Anε has small order type, if you allow them to get mixed up all together.

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Thank you for your answer. Let me see if I understood it. Let's choose in disorder some of your $A_n$ sets: $A_1$ is $\omega+1$ , $A_2$ is 5, $A_3$ be $\omega$. When I take the union the o.t is $\omega$+6+$\omega$ . Is this true? Same for the $A_n^{\epsilon}$ : if I do not take them in order, I can choose $\gamma^{2n}$ for the first one (1 element in this set so its o.t is 1!), 5 for the second one, $\gamma$ for the 3rd one, but when I take the union I get to something way bigger than $\gamma^{n+1}$. But the bound is $\gamma^+$. – Carlo Von Schnitzel Feb 12 '10 at 21:29
Not quite. With your A1, A2 and A3, the union is just omega+1, since your A2 and A3 are subsets of A1. You would get omega+6+omega (which is = omega+omega, or omega*2) if you had A1=omega, A2=[omega,omega+5], and A3=[omega+5,omega+omega). This would be the disjoint case, and where the sets appear in disjoint intervals that are spaced out in the same order as the indices. In my answer, I had very tiny order types for each An, but the union had huge order type. But you are right in your last sentence, that gamma+ is still a bound, and this is a cardinality calculation. – Joel David Hamkins Feb 12 '10 at 23:37
I see. With your choice the o.t of $A_1$ is $\omega$, that of $A_2$ is 6 ($A_2$ has $\omega$, $\omega$+1, $\omega$+2, $\omega$+3, $\omega$+4, $\omega$+5, so 6 elements) and $A_3$ has $\omega$ elements. But say I had $A_1$ is {$\omega$}, $A_2$ is {$\omega$, $\omega$+5} and $A_3$ is {$\omega$+5, $\omega$+$\omega$+}, these sets respectively have 1 element, 2 elements and 2 elements, but their union has 3 elements so it is {$\omega$, $\omega$+5, $\omega$+$\omega$} so the o.t is 3. It seems to me that we would just be adding all these elements by juxtaposition. – Carlo Von Schnitzel Feb 13 '10 at 1:16
I was thinking that I had to lay down on the table $\omega$ things and then $\omega$+5 in front of it and then $\omega$+$\omega$ things in front of what I had put in the table at the previous step, but I was wrong...It would have been the case if that was the width of some intervals. – Carlo Von Schnitzel Feb 13 '10 at 1:16
I don't quite follow you. The union operation on sets is commutative; it doesn't matter what the order is. Perhaps you have in mind the idea of ordinal addition, where one puts a copy of each set into a separate block, in order to make them disjoint, and then takes the disjoint union of these sets (that is, the ordinary union of disjoint copies of the sets). When one defines alpha+beta, it means a copy of alpha followed by a copy of beta, and it is in this sense that is used. In contrast alpha union beta is simply the larger of the two ordinals. – Joel David Hamkins Feb 13 '10 at 3:41