# 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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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
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