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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?

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    $\begingroup$ The statement is correct. Further you described correctly how to find the least element of $S$. It should be followed by proving that it really is the least element of $S$: just let $s\in S$ and prove that indeed $a_{\alpha_0}\leq s$. $\endgroup$ – drhab Jan 1 '18 at 11:39
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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 $\alpha_0\in J$ and $S\cap A_{\alpha_0}$ is a non-empty subset of $A_{\alpha_0}$, and so has a smallest element $a_{\alpha_0}$. This is because, given any other element $s\in S$, either $s\in A_{\alpha_0}$, $s\in A_\beta$ for some other $\beta$. In the first case $a_{\alpha_0}<s$ by definition; in the second, $\alpha_0<_J\beta$ by definition of $\alpha_0$ and so $\alpha_0<s$.

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  • $\begingroup$ This answer exists to remove the question from the Unanswered queue; please upvote or give Best Answer to complete this process. $\endgroup$ – aleph_two Dec 24 '18 at 4:18

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