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The usual "matchstick" representation of an ordinal number can be thought of as an order-preserving injection of that ordinal into the interval [0,1]. For example, here's a representation of $\omega^2$:

One can use this technique to visualize even larger and larger ordinals; for instance, $\omega^3$ would look the same as the above, but each individual matchstick would be replaced with yet another infinite triangle leading to the next matchstick, so that there are three layers of infinite triangles. So long as there's an order-preserving injection from your chosen ordinal into $[0,1]$, this technique will work to help you visualize it.

So my question is: what is the supremum of all ordinals for which an order-preserving injection exists in this way? Is it the first uncountable ordinal? The initial ordinal of $2^{\aleph_0}$? Something smaller? Something larger?

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Mike Battaglia
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  • I'm pretty sure that I answered this at least twice. – Asaf Karagila Feb 06 '14 at 06:38
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    A bit late to the party, but I might want to mention: The set of all numbers that can be written as $0.\underbrace{11\dots11}_a$ (with $a\ge0$) is isomorphic to $\omega$. Similarly, the set of numbers that can be written as $0.\underbrace{11\dots11}_a0\underbrace{11\dots11}_b$ (with $a,b\ge0$) is isomorphic to $\omega^2$ (since the function that maps $0.\underbrace{11\dots11}_a0\underbrace{11\dots11}_b$ to $a\omega+b$ is order-preserving), the set of $0.\underbrace{11\dots11}_a0\underbrace{11\dots11}_b0\underbrace{11\dots11}_c$ is isomorphic to $\omega^3$, etc. – Akiva Weinberger Jan 13 '15 at 01:48

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$\mathbb Q$ is a universal countable linear order: Any countable linear order embeds into it. This is easily proved by enumerating the countable order, and defining the embedding by recursion, using that $\mathbb Q$ is dense and has no end-points.

Since $\mathbb Q$ is isomorphic to $\mathbb Q\cap(0,1)$, we have that every countable ordinal embeds into $\mathbb Q\cap(0,1)$. If we want the embedding $f$ to be continuous, replace $\mathbb Q\cap(0,1)$ with $[0,1]$, and redefine $f(\lambda)$ for each limit $\lambda$ in the domain of $f$ as the supremum of the previous $f(\beta)$.

[With some extra work, we can arrange that the embedding into $\mathbb Q\cap(0,1)$ itself is continuous, if this is preferable.]

Finally, $\omega_1$ itself does not embed into $\mathbb R$: If $f:\omega_1\to\mathbb R$ is strictly increasing, there is an injection of $\omega_1$ into $\mathbb Q$: Send $\alpha$ to some rational in $(f(\alpha),f(\alpha+1))$. This is of course impossible.

[We can prove a stronger result here as well, namely, any increasing $f:\omega_1\to\mathbb R$ is eventually constant.]

Andrés E. Caicedo
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