In Topology, Munkres gives the following construction for a
minimal uncountable ordered set. Such a set has a largest element; the section obtained from this largest element is uncountable but any other section is countable:
Start with an uncountable set $B$ that is well-ordered under some order
$<$Then consider $A = \{1,2\} \times B$ in the lexicographical order (I can define this if anybody wants). Any $A_{\Omega}$, a section of $A$ (which is defined to be $\{x \in A | x < \Omega \}$), is
uncountable for $\Omega$ of the form $(2,y)$ where $y \in B$.
Here is the
non-sequitur troubling me:
There exists a smallest $\Omega$ such that $A_{\Omega}$ is uncountable. In other words, for $\beta < \Omega$, $A_{\beta}$ is countable.
How do we know that a smallest $\Omega$ exists?
The set we desire is $A_{\Omega} \cup \{\Omega\}$. It has a largest element $\Omega$ such that $A_{\Omega}$ is uncountable but any other section is countable.
Best Answer
Since the set $\{1,2\}\times B$ is well ordered, any non-empty subset of this set has a smallest element.
Now consider the subset of $\{1,2\}\times B$ consisting of all those elements whose section is uncountable, clearly, it's non-empty and hence there exists a smallest element, denote it by $\Omega$, hence the result follows!