Showing $\sum A_i$ for $i\in I$ is an order if $I$ is and each $A_i$ is I'm currently working through Vaught's 'Set Theory: An Introduction' and was stuck on Chapter 5 q1.1:
'Show $$ \sum_{i=\underline{I}}^n \underline{A_i } $$ is an order if $\underline{I}$ is and $\underline{A_i }$ is.'
From my understanding, $\underline{A_i }$ is an algebraic structure s.t. $\underline{A_i } = (A, R, •)$, where R is the binary relation over $A$ where $R \subseteq A \times A$ and • is the binary operation over $A$.
As per Vaught, an order would be $\underline{A_i }$ = (A, $\leq$) and $\underline{I}$ = (I, $\leq$) s.t. the summation known as the ordered sum is defined by $\cup_{i\in I}A_i$
I'm honestly not entirely sure a) what an order is and b) how to approach this problem.
Thanks in advance for the help!
 A: What Vaught calls an order is usually called a linear order or a total order. A partial order $\le$ on $A$ is a binary relation on $A$ that is reflexive, transitive, and antisymmetric. (These terms are defined at the beginning of the section.) It is a linear order — what he calls an order — if it is also connected, meaning that for all $x,y\in A$, either $x<y$, $x=y$, or $y<x$. I will use the term linear order rather than his order.
Each $\underline{A_i}$ is a linear order $\langle A_i,\le_i\rangle$: $A_i$ is some set, and $\le_i$ is a relation on $A_i$ that is a linear order on $A_i$. $\underline{I}$ is a linear order $\langle I,\le_I\rangle$. And what you’re asked to compute, at least according to what I can see in the Amazon Look Inside for the second edition of the book, is
$$\sum_{i\in\underline{I}}\underline{A_i}\,.$$
The underlying set, $A$, of this ordered sum is the disjoint union of the sets $A_i$:
$$A=\bigcup_{i\in I}\big(\{i\}\times A_i\big)\,.$$
The order relation, $\le$, on $A$ is defined as follows. Each element of $A$ is an ordered pair $\langle i,a\rangle$, where $i\in I$ and $a\in A_i$. If $\langle i,a\rangle,\langle j,a'\rangle\in A$, then $\langle i,a\rangle\le\langle j,a'\rangle$ if and only if either $i<_Ij$, or $i=j$ and $a\le_ia'$. What you’re to prove is that this relation $\le$ really is a linear order on the set $A$: that it’s reflexive, transitive, antisymmetric, and connected.
For instance, for antisymmetry we must prove that if $\langle i,a\rangle\le\langle j,a'\rangle$ and $\langle j,a'\rangle\le\langle i,a\rangle$, then $\langle i,a\rangle=\langle j,a'\rangle$. From the definition of $\le$ we know that if $\langle i,a\rangle\le\langle j,a'\rangle$, then $i\le_Ij$, and if $\langle j,a'\rangle\le\langle i,a\rangle$, then $j\le_Ii$, and we’re told that $\underline{I}$ is a linear order. By definition, then, $\le_I$ is antisymmetric, so $i=j$. But then $\langle i,a\rangle\le\langle i,a'\rangle$, which by definition means that $a\le_ia'$, and also $\langle i,a'\rangle\le_i\langle i,a\rangle$, meaning that $a'\le_ia$. And $\le_i$ is a linear order, hence antisymmetric, so $a=a'$, and therefore $\langle i,a\rangle=\langle j,a'\rangle$, as desired.
The other three properties are proved similarly.
