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.

  • 2
    $\begingroup$ The smallest uncountable well-ordered set is $A_\Omega,$ not $A_\Omega\cup\{\Omega\}.$ $\endgroup$ – bof May 21 '17 at 10:33

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!


My opinion of Munkres' method is not high, although he is not illogical.

If $B$ is uncountable and $<$ is a well-order on B then let $C=\{x\in B: \{y|y<x\}$ is uncountable$\}.$ Then

(1) If $C\not =\phi$ then let $x_C=\min C.$ By def'n of $C$ and of $x_C,$ if $y<x_C$ then $\{z|z<y\}$ is countable. So $\{y\in B:y\leq x_0\}$ is a minimal uncountable order.

(2) If $C=\phi$ take $d\not \in B$ and extend $<$ to a well-order on $D=B\cup \{d\}$ by letting $b<d$ for all $b\in B.$ Then $\{y\in D: y<b\}$ is countable for each $b\in B$ but $\{y\in D:y<d\}=B$ is uncountable.

In Set Theory: An Introduction To Independence Proofs, by K. Kunen, we have (not verbatim) : Let $W$ be the set of well-orders that are subsets of $\omega\times \omega.$ (Where $\omega$ is the least infinite ordinal. Recall that any binary relation on a set $S$ is jut a subset of $S\times S.$) Now any well-order is isomorphic to the $\epsilon$-order on a unique ordinal, so we can obtain the set $B$ of all and only those ordinals that are isomorphic to members of $W.$ Then $B$ is in fact the set of all countable ordinals, so $B=\omega_1,$ the least uncountable ordinal .... This is done without AC (the Axiom of Choice). So we do not need AC to prove the existence of an uncountable well-order.


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