Prob. 6 (c), Sec. 10 in Munkres' TOPOLOGY, 2nd ed: Set of elements having no immediate predecessors in the minimal uncountable well-ordered set

Let $S_{\Omega}$ be the minimal uncountable well-ordered set. Let $X_O$ be the subset of $S_{\Omega}$ consisting of all elements $x$ such that $x$ has no immediate predecessor.

Then how to show that $X_O$ is uncountable?

My effort:

Suppose that $X_O$ is at most countable. Since $S_{\Omega}$ is uncountable, there exists an element $a \in S_{\Omega} - X_O$. Then $a$ has an immediate predecessor, call it, $a_1$.

Since the subset $S_{\Omega} - X_O$ of $S_{\Omega}$ is non-empty, it has a smallest element, say, $u$. Let $u_1$ be the immediate predecessor of $u$. Then $u_1 \not\in S_{\Omega} - X_O$. So $u_1$ has no immediate predecessor. Thus $u_1 \in X_O$. So $X_O$ is non-empty.

Let $x_0$ be the smallest element of $X_O$.

What next?

How to proceed from here?

A non-empty subset $A$ of $S_{\Omega}$ is (at most) countable if and only if $A$ is bounded above in $S_{\Omega}$.

Since $S_{\Omega}$ is well-ordered, it has a smallest element, and this element has no immediate predecessor and is thus in $X_O$; so $X_O$ is non-empty.

Suppose that $X_O$ is countable. Then $X_O$ has an upper bound $a$; let $b$ denote the immediate successor of $a$, which exists since $S_{\Omega}$ is well-ordered and has no largest element. Since $b$ has an immediate predecessor, namely $a$, therefore $b \not\in X_O$. So $X_O \subset S_b$.

Moreover, every well-ordered set has the least-upper-bound property. Let $c$ be the supremum of $X_O$. Then $c \leq a$.

What next?

For $a \in S_{\Omega}$ i will denote by $a+1$ the successor of $a$, $a+2$ the successor of $a+1$, etc...

You should use the following useful fact:

Let $A\subset S_{\Omega}$ any subset. Then $A$ is bounded above if and only if $A$ is countable.

Now, suppose that $X_0$ is bounded above (and countable). Then, there exists $a \in S_{\Omega}$ such that $X_0 < a$. Consider the set $$A=\{ a, a+1, a+2, \dots, a+n, \dots, \} = \{ a+n :n \in \Bbb{Z}_{\ge0}\}$$ $A$ is clearly countable, so it is bounded above. Let $\psi = \sup A$. Obviously, $\psi \notin A$, since $A$ has no maximum.

I claim that $\psi$ has no predecessor. Otherwise, suppose $\psi = \alpha +1$ for some element $\alpha \in S_{\Omega}$. Then $\alpha$ is not an upper bound of $A$, hence there exists come $n \ge 0$ such that $\alpha \le a+n$. Then $\psi = \alpha +1 \le a+(n+1) < a+ (n+2)$: a contradiction.

We proved that $\psi$ has no predecessor, i.e. $\psi \in X_0$: a contradiction.

So $X_0$ cannot be a countable set.

• Can you spell out why $A\subset S_\Omega$ countable implies bounded above? – Ben Blum-Smith Jun 7 '15 at 21:46
• Let $A$ be countable, and suppose it has no maximum. Then $$A\subseteq \bigcup_{a \in A} \{ x \in S_{\Omega} : x < a \} = B$$ B is a countable ordinal since it is a countable union of countable sets. But now, this means that it is an ordinal smaller than $\Omega$, so $B$ is bounded in $S_{\Omega}$. – Crostul Jun 7 '15 at 21:57
• – Crostul Jun 7 '15 at 22:02
• @Crostul, what do you mean by $a+1$? You see, we haven't got any operator of addition here. – Saaqib Mahmood Jun 8 '15 at 4:22
• @Crostul, by $a+1$, do you mean the successor of $a$, which exists because $S_{\Omega}$ is well-ordered and has no largest element? – Saaqib Mahmood Jun 8 '15 at 4:24