Csdrhrt

Here is Prob. 8, Sec. 10, in the book Topology by James R. Munkres, 2nd edition:

Problem 8 (a):

Let $A_1$ and $A_2$ be disjoint sets, well-ordered by $<_1$ and $<_2$, respectively. Define an order relation on $A_1 cup A_2$ by letting $a < b$ either if $a, b in A_1$ and $a <_1 b$, or if $a, b in A_2$ and $a <_2 b$, or if $a in A_1$ and $b in A_2$. Show that this is a well-ordering.

Problem 8 (b):

Generalize (a) to an arbitrary family of disjoint well-ordered sets, indexed by a well-ordered set.

I think I'm clear as to the proof required in Prob. 8 (a).

My Attempt at Prob. 8 (b):

Let $J$ be a (non-empty) well ordered set; let $$left{ A_alpha colon alpha in J right}$$ be a collection of non-empty, (pairwise) disjoint well-ordered sets indexed by set $J$; and let $$ A colon= bigcup_{alpha in J} A_alpha. $$
For each $alpha in J$, let $<_alpha$ denote the well-ordering relation on the set $A_alpha$.

For any two elements $a, b in A$, let us define $a < b$ to mean the following:

Either $a, b in A_alpha$ for some $alpha in J$ and $a <_alpha b$, or $a in A_alpha$, $b in A_beta$ for some $alpha, beta in J$ such that $alpha <_J beta$.

Then the set $A$ is a well-ordered set.

Is this statement correct?

If so, then here is my proof:

Let $S$ be a non-empty subset of the set $A = bigcup_{alpha in J} A_alpha$. Let $J_0$ be the following subset of $J$.
$$ J_0 colon= left{ alpha in J colon S cap A_alpha neq emptyset right}. $$
Then the set $J_0$ is non-empty, and as such it has a smallest element, say $alpha_0$. Then $alpha_0 in J$ and $S cap A_{alpha_0}$ is a non-empty subset of $A_{alpha_0}$ and so has a smallest element $a_{alpha_0}$, which is also the smallest of $S$ with respect to the order relation on $A$.

Is this proof correct? If so, then is my presentation accessible enough? If not, then where have I erred?

Yes, your statement and proof are correct, although I agree with drhab that you might want to say a few words explaining why $a_{alpha_0}$ actually is the smallest element of $S$. For this, I would replace the last sentence with:

Then $alpha_0in J$ and $Scap A_{alpha_0}$ is a non-empty subset of $A_{alpha_0}$, and so has a smallest element $a_{alpha_0}$. This is because, given any other element $sin S$, either $sin A_{alpha_0}$, $sin A_beta$ for some other $beta$. In the first case $a_{alpha_0}<s$ by definition; in the second, $alpha_0<_Jbeta$ by definition of $alpha_0$ and so $alpha_0<s$.