Vtyjkyuk

Theorem. There exists a well ordered set $A$ having a largest element $Omega$ such that the section $S_Omega$ of $A$ by $Omega$ is uncountable but every other section of $A$ is countable.

Proof. Consider an uncountable well ordered set $B$, consider $C=1,2times B$ with the lexicographic order and let $Omega $ be the smallest element of $C$ for which $S_Omega= x<Omega$ is uncountable. Let $A=S_OmegacupOmega$
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I'm trying to show that if $X_0$ is a subset of $S_Omega$ of all elements that have no immediate predecessors, then $X_0$ is uncountable.

Assume the converse: $X_0$ is countable. Any countable subset of $S_Omega$ has an upper bound in $S_Omega$. Let $Gamma in S_Omega$ be this bound: $xle Gamma$ for all $xin X_0$. Now $S_Omega$ has no largest element, so every element in $S_Omega$ has an immediate successors. I believe I have to apply this to elements of $X_0$ and find connection with immediate predecessors. So for every $x$ in $X_0$, there is an immediate successor $s(x)in X_0$. But how to establish connection between immediate predecessors and get a contradiction?

If you're allowed to use the fact that any countable set is bounded in $S_Omega$ (which is true, but not quite trivial from what you've mentioned so far unless I'm missing something), then you're on the right track with letting $Gamma$ be an upper bound for $X_0.$

What you want to look at then is the set $Z=xin S_Omega : x ge Gamma.$ In this set, by definition, every element except possibly $Gamma$ has an immediate predecessor. And it's easy to see that for every element of $Z$ except $Gamma,$ the element's immediate predecessor is in $Z.$ So $Z$ is a well-ordered set with least element $Gamma$ such that every other element has an immediate predecessor. This means $Z$ is order isomorphic to the natural numbers with their usual ordering, and hence is countable, which gives the required contradiction.

To see that it is isomorphic to the natural numbers, take any $z_0in Z$ (other than $Gamma$). Then let $z_1$ be the immediate predecessor of $z_0,$ and $z_2$ the predecessor of $z_1$ if $z_1ne Gamma,$ and so on. This sequence must hit $Gamma$ at some point, because otherwise the sequence would give a set with no least element. Thus every element of $Z$ is the $n$-th successor of $Gamma$ for some $n.$

Alternatively, you can prove that $X_0$ is unbounded (and hence uncountable) directly, as Carl suggested in the comments. Let $ain S_Omega.$ Well ordering implies every element has an immediate successor, so define recursively $a_0=a,$ $a_n+1=s(a_n),$ and $A=a_n :nin mathbb N$

You know that the initial segment $S_a$ of elements $< a$ is countable, and that that $A$ is countable. You can also show that $S_acup A$ contains the initial segment of each of its elements and the successor of each of its elements.

$S_acup A$ is countable, hence not all of $S_Omega,$ so there is a least element $bin S_Omega$ larger than all of its members. In fact, we have $S_acup A=S_b.$ We can see that $b$ does not have an immediate predecessor since any element $c<b$ is in $S_acup A,$ so $s(c)in S_acup A$ so is not equal to $b.$