I'm trying to find a way to visualize the first uncountable ordinal $\omega_1$. This is rather difficult, as the visualization tactic that I often use for countable ordinals - namely, the "matchstick" representation - fails for $\omega_1$. This is because the matchstick representation is effectively an order-preserving embedding of the ordinal into $\mathbb{R}$, but it can be shown that no such embedding exists for $\omega_1$.

I'm wondering though, is there some other simple, but non-Archimedean group that you could embed $\omega_1$ into - specifically something simple which is the direct sum or product of $\Bbb R$'s and $\Bbb Z$'s, ordered lexicographically?

For instance, could you perhaps embed it into the direct sum $\Bbb R \times \Bbb R$, ordered lexicographically? Or could you perhaps embed it into $\Bbb R \times \Bbb Z$, ordered either left-to-right or right-to-left lexicographically?

I'm sure that it's possible to contrive a trivial example with uncountably many R's or Z's that handles the case, but I'm ideally interested in seeing what can be done with small finite direct sums of a few $\Bbb R$'s or $\Bbb Z$'s. It would be nice if some sort of non-Archimedean matchstick representation made it a bit easier for us to get our heads wrapped around uncountable ordinals.

  • $\begingroup$ $\mathbb{N}$ is not a group. And what does it mean for a group to be archimedean or not? $\endgroup$
    – Zhen Lin
    Mar 10, 2014 at 10:05
  • $\begingroup$ Well, there's always the long line. en.wikipedia.org/wiki/Long_line_%28topology%29 $\endgroup$ Mar 10, 2014 at 10:13
  • $\begingroup$ Zhen Lin: sorry, that was cruft from an earlier edit - changed it to R and Z. As for archimedean groups - en.wikipedia.org/wiki/Archimedean_group $\endgroup$ Mar 12, 2014 at 23:40
  • $\begingroup$ Since $\omega_1$ is a well-ordering, the answer about $\Bbb N$ and about $\Bbb Z$ is the same answer. $\endgroup$
    – Asaf Karagila
    Mar 12, 2014 at 23:48
  • $\begingroup$ Yeah, but N isn't a group, and I chose to make all N's Z's rather than change "archimedean group" to "archimedean monoid" and have to define the new term. It was cruft. $\endgroup$ Mar 13, 2014 at 0:30

2 Answers 2


All you have to do is note that if $A$ and $B$ are two linearly ordered sets which satisfy the following property $(*)$, then their lexicographic product satisfies the same property as well:

$(*)$ Every interval has a countable cofinality (i.e. there is a sequence converging to the endpoint of the interval).

Clearly $\Bbb R$ and $\Bbb N$ both satisfy that. So in order to find a linear order that doesn't we have to venture outside. For example $\Bbb R\times\omega_1$. But that feels a bit "trivialized", after all we're not looking for $\omega_1$ explicitly. But here's the problem, if we want $\omega_1$ to embed into the order, then we need to have an interval whose endpoint cannot be reached by a countable sequence. This means that there is an embedding of $\omega_1$ into that order to begin with. So in order to have $\omega_1$ embed into some lexicographic product of $\Bbb R$, we first have to have it embed into the other summand.

To see that $(*)$ for both $A$ and $B$ imply that $A\times B$ has $(*)$, note that given an interval $U$ in $A\times B$ then either there is some $a\in A$ and $(b_1,b_2)\subseteq B$ such that $\{a\}\times(b_1,b_2)$ is cofinal in $U$, in which case we can pick a cofinal sequence from $(b_1,b_2)$.

If there is no such point $a$, then there is an interval $(a_1,a_2)\subseteq A$ such that $(a_1,a_2)\times B$ is cofinal in $U$. In which case we can construct a cofinal sequence from $(a_1,a_2)$ to witness the cofinality of $U$.


You can't embed $\omega_1$ into the $\Bbb{R}^n$ or $\Bbb{N}^n$ (in the sense of topological embedding.) Because if $A$ is a subset of $\Bbb{R}^n$ then it has only countably many isolated point, but $\omega_1$ has $\aleph_1$ many isolated points. For the same reason, you cannot embed $\omega_1$ into $\Bbb{N}^n$, even you cannot embed it into $\Bbb{R}^\omega$ or $\Bbb{N}^\omega$ with product topology!

Addition. You cannot embed $\omega_1$ into $\Bbb R^2$ with lexicographical order. Since $\Bbb{R}^2$ with lexicographical order is metrizable. So if $\omega_1$ is embedded in $\Bbb{R}^2$ with lexicographical order, then $\omega_1$ is metrizable, which is not.


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