This question already has an answer here:

As the title of the question suggests, if $\langle A\,,\,<\rangle$ is an infinite linearly ordered set such that for each $a\in A$, the initial section $\text{sec}(a,A,<)$ is a finite set, ¿is it true that it must be isomorphic to $\langle\mathbb{N}\,,\,\text{<}_{\mathbb{N}}\rangle$?

Obviously, every well-ordered set that verifies this property is isomorphic to $\langle\mathbb{N}\,,\,\text{<}_{\mathbb{N}}\rangle$, for if $\langle A\,,\,<\rangle$ is a well-ordered set, there exists a unique ordinal $\alpha$ such that $\langle A\,,\,<\rangle$ is isomorphic to $\langle\alpha\,,\,\in_{\alpha}\rangle$. In that case, every initial section of the first turns into an ordinal less than $\alpha$, and if every initial section of the first is finite, then every ordinal less than (or in other words, contained in) $\alpha$ is finite, so $\alpha$ is equal to $\omega$, the set of all natural numbers.

However, since a set that is not well-ordered cannot be isomorphic to one that is well-ordered, like $\langle\mathbb{N}\,,\,\text{<}_{\mathbb{N}}\rangle$, we must prove that if every initial section of $\langle A\,,\,<\rangle$ is finite, then the set is well ordered. How shall we proceed?

Thanks in advance.


marked as duplicate by Asaf Karagila elementary-set-theory Apr 25 at 13:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ If $(A,<)$ is not well-ordered, certainly it can't be isomorphic to $\mathbb{N}$? I guess you have to show that if every initial section of $A$ is finite, then $A$ is well-ordered. $\endgroup$ – Ehsaan Apr 25 at 12:57

A linearly ordered set $(A, \le_A)$ whose initial segments are all finite is necessarily well-ordered.

To see this, let $\varnothing \ne U \subseteq A$. Then:

  • For each $a \in U$, the set $D_a = \{ x \in A \mid x \le_A a \}$ is finite by assumption;
  • For each $a,b \in U$ we have $a \le_A b$ if and only if $D_a \subseteq D_b$; and
  • For each $a,b \in U$, we have $D_a \subseteq D_b$ or $D_b \subseteq D_a$, since $A$ is linearly ordered.

Letting $a \in U$ be such that $|D_a|$ is least, it follows that $a$ is a minimal element of $U$.

Now you know that $(A, \le_A)$ must be well-ordered, you should be able to prove without too much trouble that $(A, \le_A) \cong (\mathbb{N}, \le)$.


Not the answer you're looking for? Browse other questions tagged or ask your own question.