4
$\begingroup$

The ordinal number $\omega^2$ can be visualized as $\omega$-many copies of $\omega$. Likewise, the ordinal number $\omega^3$ can be visualized as $\omega^2$-many copies of $\omega$, arranged as appropriately (= lexicographically) ordered rows in the cube $\omega^3$ (rows being sequences of cells parallel to the x-axis):

enter image description here

It "only" takes the ability to visualize higher-dimensional (hyper-)cubes to visualize arbitrary countable ordinals of the form $\omega^n$ as $\omega^{n-1}$ appropriately ordered copies of $\omega$.

But this ability comes to an end when thinking of the countable ordinal number $\omega^\omega$, since the $|\omega|$-dimensional hypercube $\omega^{|\omega|}$ consists of uncountably many cells and cannot represent a countable ordinal.1

While (presumably) the countable ordinal number $\omega^\omega$ cannot be visualized as a countable number of copies of $\omega$ (however carefully arranged) – can it be visualized in a similarly easy visual way, only "slightly" advanced? (For example, by replacing $\omega$ by some other (higher) ordinal number?)

If there were such a "similarly easy way": Up to which limit would it lead? Up to $\epsilon_0$? And which "similarly easy way" would come beyond that limit?


1 Do $|\omega|$-dimensional vector spaces – e.g. Hilbert spaces – pose such serious problems?

$\endgroup$
7
  • $\begingroup$ Hilbert spaces are NEVER of dimension $|\omega|$. Every Banach space (and in particular Hilbert spaces) are either finite dimensional, or have dimension $\geq2^{\aleph_0}$. Moreover, dimension is cardinality, so ordinals are not the right way of thinking about it. Finally, the ordinal $\omega^\omega$ correspond to ordering the polynomials in $\Bbb N[x]$ first by length, and then lexicographically at each length. $\endgroup$
    – Asaf Karagila
    Jan 4, 2014 at 19:46
  • 1
    $\begingroup$ @AsafKaragila:yeah I learnt Hilbert space from physics too, and isn't Fourier basis an infinite countable base (for a bounded interval $I$, in Hilbert space $L^{2}(I)$)? $\endgroup$
    – Gina
    Jan 4, 2014 at 20:27
  • 4
    $\begingroup$ @Gina and Hans, Schauder bases are different from Hamel bases, and dimension is really the cardinality of a Hamel basis. Hans, ordinals are not impervious to permutations, cardinals are. This means that infinite ordinals don't make a good bedrock for dimension. There will be automorphism which don't preserve dimension. $\endgroup$
    – Asaf Karagila
    Jan 4, 2014 at 21:41
  • 2
    $\begingroup$ On the other hand, there are algebraic settings where ordinal "dimensions" crop up. These are usually connected in some way to lengths of certain chains, e.g. the Krull dimension of a ring etc. $\endgroup$ Jan 4, 2014 at 21:46
  • 2
    $\begingroup$ Relevant previous stackexchange question: Intuition for ${\omega}^{\omega}$ $\endgroup$ Jan 6, 2014 at 22:50

3 Answers 3

5
$\begingroup$

$\omega^\omega$ can be visualized, in what I think is a fairly nice way in a static 2D image featured on the wikipedia page for ordinal number:

image from wikipedia

Also, if you're willing to allow dynamic visualizations, then Stephen Brooks's transfinite number line goes well past $\epsilon_0$ (to $\Gamma_0$), as well as providing a more linear (if colorful) look at $\omega^\omega$: omega to the omega

$\endgroup$
4
  • $\begingroup$ @Marc: How can one be sure that in the first picture only countably many copies of $\omega$ are present? $\endgroup$ Jan 5, 2014 at 10:06
  • $\begingroup$ @HansStricker Each turn goes to the next power of $\omega$, and there are countably many turns (you could make up a sequence of radii). $\endgroup$
    – Mark S.
    Jan 6, 2014 at 3:54
  • $\begingroup$ But the $n$-th turn contains $|\omega|^{n-1}$ copies of $\omega$, so there would be $|\omega|^{|\omega|}$ copies of $\omega$ below $\omega^\omega$. Doesn't this imply that the picture is somehow misleading? $\endgroup$ Jan 6, 2014 at 7:17
  • $\begingroup$ @HansStricker $|\omega|^n=|\omega|$ for all $n$, so apriori, in the limit we have at most $|\omega_1|$ copies. But we can do better, by observing $\omega^\omega<\omega_1$, or saying "the union of countably many countable sets (with explicit bijections to $\mathbb N$) is countable", etc. I feel I'm falling short of addressing your objection, but maybe this will help you make it clearer. $\endgroup$
    – Mark S.
    Jan 7, 2014 at 0:22
0
$\begingroup$

It is wrong to think of $\omega^\omega$ as a hypercube of actually infinite dimension, which would correspond to the set of all infinite sequences of natural numbers, which is uncountable. One may think - if at all - of $\omega^\omega$ only as a hypercube of potentially infinite dimension, e.g. as $\bigcup_{n<\omega}\omega^n$, i.e. the functions $f:\omega \rightarrow \omega$ with finite support.

Especially: Other than $\omega^n$ – which is a well-ordered set of disjoint copies of $\omega$ – the ordinal $\omega^\omega$ cannot be thought of this way, even though the hypercube or the spiral representation might suggest it.

$\endgroup$
3
  • $\begingroup$ First, you probably shouldn't answer your own question :) You can edit your original question or comment on others' answers. But I'm confused by what you've written: You say it's wrong to think of $\omega^\omega$ as $\bigcup \omega^n$, and yet you seem to think of it that way. And, yes, it's wrong: only countably many sequences of integers are eventually zero. $\endgroup$ Jan 7, 2014 at 1:33
  • $\begingroup$ @TedShifrin: I think Hans is saying the ordinal $\omega^\omega$ is the (countable) union (lub) of the (countable) $\omega^n$, which is what you need it to be (as a limit ordinal) to get well-ordering out of the lexicographic ordering. The protest regarding potential vs. actual infinity of dimension confused me, but setting that aside, his Answer helps explain what difficulty he encounters in "visualizing" $\omega^\omega$. $\endgroup$
    – hardmath
    Jan 7, 2014 at 6:31
  • $\begingroup$ @TedShifrin.I dk what you mean by saying that it's wrong to think of $\omega^{\omega}$ as $\cup \omega^n.$ The ordinal $\omega^{\omega}$ is equal to $\cup_{n\in \omega}A(n)$ where $A(n)$ is the ordinal $\omega^n.$ $\endgroup$ Jul 21, 2018 at 5:03
0
$\begingroup$

I recently asked myself the same question. A google search led me to this topic. Although Asaf Karagila gave the answer in his comment, I would like to point a lecture notes which considered this - https://www.dpmms.cam.ac.uk/~tf/fundamentalsequence.pdf . Look at page 18, the last paragraph (under Exercise 4, continues on the next page). There is also given intepretation of $\omega^{\omega^{\omega}}$ and of $\varepsilon_0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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