"Given two ordinals $\alpha$ and $\beta$, let $A = (\alpha$ x {0}) $\cup$ $(\beta$ x {1}).
Then, define a well ordering on A by:
$(v, i) <_{A} (\tau, j) \iff (i \lt j) \lor (i = j$ $\land$ $v \lt \tau)$
Then $\alpha + \beta$ is the ordinal isomorphic to (A, $\lt_{A}$)."
I was having trouble understanding this, so I tested the definition with 2 and 3.
Let $\alpha = 2$ and let $\beta = 3$.
Since 2 = {0,1}, the ordered pairs I got from ($\alpha$ x {0}) are (0, {0}) and (1, {0}). Similarly for, 3, the ordered pairs I got from ($\beta$ x {1}) were (0, {1}), (1, {1}), and (2, {1}).
So, A = {(0, {0}), (1,{0}), (0, {1}) , (1, {1}), (2, {1})}
Since $(A, \lt_A)$ is a woset, it is a subset of A x A.
I Got A x A to be (using [] brackets to differentiate between elements):
{
[(0, {0}), (0, {0})], [(0, {0}), (1, {0})], [(0, {0}), (0, {1})], [(0, {0}), (1, {1})], [(0, {0}), (2, {1})],
[(1, {0}), (0, {0})], [(1, {0}), (1, {0})], [(1, {0}), (0, {1})], [(1, {0}), (1, {1})], [(1, {0}), (2, {1})],
[(0, {1}), (0, {0})], [(0, {1}), (1, {0})], [(0, {1}), (0, {1})], [(0, {1}), (1, {1})], [(0, {1}), (2, {1})],
[(1, {1}), (0, {0})], [(1, {1}), (1, {0})], [(1, {1}), (0, {1})], [(1, {1}), (1, {1})], [(1, {1}), (2, {1})],
[(2, {1}), (0, {0})], [(2, {1}), (1, {0})], [(2, {1}), (0, {1})], [(2, {1}), (1, {1})], [(2, {1}), (2, {1})]
}
Using the definition of the well ordering I got the set $(A, \lt_A)$ to be:
{
[(0, {0}), (1, {0})], [(0, {0}), (0, {1})], [(0, {0}), (1, {1})] , [(0, {0}), (2, {1})],
[(1, {0}), (0, {1})], [(1, {0}), (0, {1})], [(1, {0}), (1, {1})], [(1, {0}), (2, {1})],
[(0, {1}, (1, {1})], [(0, {1}), (2, {1})],
[(1, {1}), (2, {1})]
}
But, if this is 2 + 3, it should intuitively have 5 elements and yet, it has 11.
Is this a correct interpretation of the definition? Did I go wrong somewhere?
First, $$2\times\{0\}=\{\langle 0,0\rangle,\langle 1,0\rangle\}\;,$$ not $$\{\langle 0,\{0\}\rangle,\langle 1,\{0\}\rangle\}\;.$$ Similarly,
$$3\times\{1\}=\{\langle 0,1\rangle,\langle 1,1\rangle,\langle 2,1\rangle\}\;,$$
so
$$A=\{\langle 0,0\rangle,\langle 1,0\rangle,\langle 0,1\rangle,\langle 1,1\rangle,\langle 2,1\rangle\}\;.$$
If you check the definition of $<_A$, you’ll see that I listed the elements of $A$ in increasing $<_A$-order. In fact, the set $<_A$ as a subset of $A\times A$ is
$$\begin{align*} &\Big\{\big\langle\langle 0,0\rangle,\langle 0,0\rangle\big\rangle,\big\langle\langle 0,0\rangle,\langle 1,0\rangle\big\rangle,\big\langle\langle 0,0\rangle,\langle 0,1\rangle\big\rangle,\big\langle\langle 0,0\rangle,\langle 1,1\rangle\big\rangle,\big\langle\langle 0,0\rangle,\langle 2,1\rangle\big\rangle,\\ &\big\langle\langle 1,0\rangle,\langle 1,0\rangle\big\rangle,\big\langle\langle 1,0\rangle,\langle 0,1\rangle\big\rangle,\big\langle\langle 1,0\rangle,\langle 1,1\rangle\big\rangle,\big\langle\langle 1,0\rangle,\langle 2,1\rangle\big\rangle,\\ &\big\langle\langle 0,1\rangle,\langle 0,1\rangle\big\rangle,\big\langle\langle 0,1\rangle,\langle 1,1\rangle\big\rangle,\big\langle\langle 0,1\rangle,\langle 2,1\rangle\big\rangle,\\ &\big\langle\langle 1,1\rangle,\langle 1,1\rangle\big\rangle,\big\langle\langle 1,1\rangle,\langle 2,1\rangle\big\rangle,\\ &\big\langle\langle 2,1\rangle,\langle 2,1\rangle\big\rangle\Big\}\;. \end{align*}\tag{1}$$
There are $\binom52+5=10+5=15$ ordered pairs in $<_A$, since each pair of distinct members of $A$ must appear once (in the order $\langle\text{smaller},\text{larger}\rangle$), and each member of $A$ must appear once with itself.
Here I’m assuming that $<_A$, despite the use of $<$ instead of $\le$, is a well-order in the usual sense, which includes reflexivity. If your definition of well-order is actually of a strict well-order, so that $a\not<_A a$ for any $a\in A$, then you should delete the $5$ ordered pairs in the first column of $(1)$.
The definition really is just a way of combining $\alpha$ and $\beta$ so that $\beta$ follows $\alpha$. The problem is that $\alpha$ and $\beta$ aren’t disjoint, so we can’t simply move $\beta$ after $\alpha$. Instead, we make a copy, $\alpha\times\{0\}$, of $\alpha$, and a copy, $\beta\times\{1\}$, of $\beta$ in such a way that the copies are disjoint, and then we can put the copy of $\beta$ after the copy of $\alpha$.
Within $\alpha\times\{0\}$ we put $\langle \xi,0\rangle<\langle\eta,0\rangle$ if and only if $\xi<\eta$; with this ordering $\alpha\times\{0\}$ really does ‘look just like’ $\alpha$, except that each element is dragging along the label $0$. Similarly, in $\beta\times\{1\}$ we put $\langle\xi,1\rangle<\langle\eta,1\rangle$ if and only if $\xi<\eta$ in $\beta$, so that $\beta\times\{1\}$ ‘looks just like’ $\beta$ with a label $1$ glued to each element.
Then we order $A$ by ordering first on the basis of the labels, so that everything with a $0$ label (i.e., everything in $\alpha\times\{0\}$) goes before everything with a $1$ label (i.e., everything in $\beta\times\{1\}$). Within the label blocks we order according to the original $\alpha$ and $\beta$ orders. Thus, $A$ ends up ordered like this:
$$\underbrace{\longrightarrow}_{\alpha\times\{0\}}\underbrace{\Longrightarrow}_{\beta\times\{1\}}\;,$$
where the ordering with the $\longrightarrow$ is isomorphic to $\alpha$, and the ordering within the $\Longrightarrow$ is isomorphic to $\beta$.