# In an infinite linearly ordered set every initial section is finite. ¿Is it isomorphic to $\langle\mathbb{N}\,,\,\text{<}_{\mathbb{N}}\rangle$? [duplicate]

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?

## marked as duplicate by Asaf Karagila♦ elementary-set-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 25 at 13:05

• 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. – 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 $$(A, \le_A) \cong (\mathbb{N}, \le)$$.