# How do you find the smallest of homeomorphic ordinals?

I am trying to get a better feel for the topology of ordinals and just received a great answer to this question where the Cantor-Bendixson rank and degree turn out to be a complete homeomorphism invariant of countable compact spaces. Now I am curious about what is going on within homeomorphism classes of ordinals. In particular, I want to know how to characterize countable ordinals $\alpha$ which are minimal among all ordinals homeomorphic to $\alpha$.

Certainly finite ordinals and $\omega$ are easy. Next, if $\alpha\cong \omega+1$, then it seems like we would have to have $\omega+1\leq \alpha<\omega +\omega$ since there can only be one limit point in $\alpha$. Things become a little fuzzier for me with ordinals that have infinitely many limit points.

For instance, suppose $\alpha$ is an ordinal homeomorphic to $\omega^{\omega}$. Is it necessarily true that $\alpha\geq \omega^{\omega}$? Is there a general rule for knowing when an ordinal is the smallest among all ordinals homeomorphic to it?

• Your last statement looks true to me. The limit points of $\omega^\omega$ are isomorphic to $\omega^\omega$, but for smaller ordinals, if you keep taking limit points, the process eventually terminates. Om the other hand I don't immediately see for example whether $\omega^\omega\cong\omega^\omega\cdot 2$. – Cheerful Parsnip May 14 '11 at 11:02
• @Jim: They are not. The $\omega$=th iteration of the process of taking limit points is empty for $\omega^\omega$ but not for $\omega^\omega\cdot 2$. I mention this in the partial answer below. – Andrés E. Caicedo May 14 '11 at 14:16
• @Andres: Thanks. I didn't think of looking at the $\omega$th iteration. – Cheerful Parsnip May 14 '11 at 16:12

[Edit: I apologize for the multiple corrections. The full result is now stated, and there are links to the details not provided here.]

Very nice question. The following is an answer for successor ordinals:

A non-zero ordinal $$\alpha$$ is (additively) indecomposable iff whenever $$\alpha=\beta+\gamma$$, then $$\gamma=\alpha$$ or $$\gamma=0$$. This is equivalent to saying that $$\alpha=\omega^\delta$$ for some $$\delta$$, where the exponentiation is in the ordinal sense.

Let $$t(\alpha)$$ denote the order type of the limit points of $$\alpha$$. The following is proved by a straightforward induction:

Fact 1. Let $$0.

1. $$t(\omega n)=n-1$$.
2. If $$\alpha$$ is a limit ordinal, $$t(\alpha+\beta)=t(\alpha)+1+t(\beta)$$.
3. $$t(\omega^{m+1} n)=\omega^m n$$ for any $$0.
4. $$t(\omega^\delta n)=\omega^\delta n$$ for $$\delta$$ infinite.

Using Fact 1, we can prove:

Claim. If $$\alpha$$ is infinite and successor, then the following are equivalent:

1. No $$\beta<\alpha$$ is homeomorphic to $$\alpha$$.
2. $$\alpha=(\gamma n)+1$$ for some indecomposable $$\gamma$$, and some finite $$n>0$$.

Moreover, $$\alpha=(\gamma n)+1$$ as in item 2 is homeomorphic to $$\alpha+\beta+1$$ for any $$\beta<\gamma$$.

Let's prove the claim. First, note that if $$\alpha=(\gamma n)+1$$ and $$\beta<\gamma$$ then $$\beta+1+\gamma=\gamma$$ so $$\alpha=\beta+1+\gamma n+1$$ and $$\alpha+\beta+1=\gamma n+1+\beta+1$$ are homeomorphic. This is because $$\alpha$$ is the disjoint union of two open sets, one of order type $$\alpha$$, and one of type $$\beta+1$$.

(This is why we need the $$1$$s at the end, to ensure both pieces are open.)

This proves that 1. implies 2. To see the converse, use Fact 1 to argue by induction that if $$t(\alpha)$$ denotes the order type of the set of limit points of $$\alpha$$, and if $$\gamma$$ is an infinite indecomposable ordinal, then no ordinal smaller than $$\gamma n+1$$ has set of limit points homeomorphic to $$t(\gamma n+1)$$, and this completes the proof of the claim.

The above deals with successor ordinals. Limit ordinals are different, and I do not currently have a complete answer. For one thing, limit ordinals are not compact, while successor ordinals are. Using Fact 1, it follows immediately that

• No ordinal other than $$\omega n$$ itself is homeomorphic to $$\omega n$$ for $$0.

Recall Cantor's normal form theorem, that any non-zero ordinal can be written uniquely as $$\alpha=\omega^{\beta_0}n_0+\dots+\omega^{\beta_k}n_k$$ where $$k\in\omega$$, the $$\beta_i$$ are strictly decreasing, and the $$n_i$$ are non-zero naturals.

The key to continue is a refinement of the idea of examining order-types: The concept of Cantor-Bendixson derivative. The derivative of a space $$X$$ is the collection of its limit points. The $$\alpha$$-th derivative of $$X$$ is the result of iterating this operation $$\alpha$$ times:

• $$X^{(0)}=X$$.
• $$X^{(\alpha+1)}=(X^{(\alpha)})'$$ is the space of accumulation points of $$X^{(\alpha)}$$.
• $$X^{(\lambda)}=\bigcap_{\alpha<\lambda}X^{(\alpha)}$$ for $$\lambda$$ limit.

By an induction that elaborates the proof of the Fact above, we have:

Fact 2. If $$\delta>0$$, the $$\delta$$-th derivative of $$\omega^\delta$$ is empty, and it is its first empty derivative. Moreover, $$(\omega^{\delta} n)^{(\delta)}=\{\omega^{\delta}m\mid 0 for any $$0.

It follows that $$n_0$$ in the Cantor normal form of $$\alpha$$ as above can be recovered by iterating the derivative operation. In particular,

• $$\alpha=\omega^{\beta_0}n_0+\dots+\omega^{\beta_k}n_k$$ cannot be homeomorphic to any ordinal whose Cantor form does not begin as $$\omega^{\beta_0}n_0$$. No two ordinals of the form $$\omega^\delta n$$ are homeomorphic, and no ordinal other than $$\omega^\delta$$ is homeomorphic to $$\omega^\delta$$.

The observations above almost settle the problem completely. I did some online digging, and found that the problem is completely solved in a technical report of the Institute for Language, Logic and Computation of the University of Amsterdam, the short paper "A classification of ordinal topologies" by Vincent Kieftenbeld and Benedikt Löwe, as of this writing available here.

Let $$\omega^{\beta_0}n_0+\dots+\omega^{\beta_k}n_k$$ be the Cantor normal form of an ordinal $$\alpha$$. Write $$lc(\alpha)=\beta_0$$ and $$c(\alpha)=n_0$$; finally, set $$p(\alpha)=0$$ if $$\alpha=\omega^{\beta_0}n_0\ge\omega$$, and $$p(\alpha)=\omega^{\beta_k}$$ otherwise. ($$lc,c,p$$ stand for "limit complexity", "coefficient", and "purity", in the notation of the Kieftenbeld-Löwe paper.)

Theorem (Kieftenbeld-Löwe). $$\alpha$$ and $$\beta$$ are homeomorphic iff $$lc(\alpha)=lc(\beta)$$, $$c(\alpha)=c(\beta)$$, and $$p(\alpha)=p(\beta)$$.

It may be worth remarking that the argument holds in ZF, i.e., no use of the axiom of choice is required. The details of the argument can be seen in the paper, but (once we have the right statement), it is actually a nice exercise to finish the proof from the argument started above.

Let me note that as a descriptive set theorist, there is a natural follow up question, namely, when is it that the Borel structures of two ordinals $$\alpha$$ and $$\beta$$ are Borel isomorphic. This problem turns out to be more delicate. It is solved in a paper by Su Gao, Steve Jackson, and Vincent Kieftenbeld, "A classification of ordinals up to Borel isomorphism", Fundamenta Mathematicae 198 (2008), 61-76. (As of this writing, the paper is available from Fundamenta's page under subscription, or from Kieftenbeld's page.) Their result uses choice in an essential way, and the problem is far from settled in ZF, even under additional assumptions such as the axiom of determinacy. Sketching the proof would take us far from the current question, but let me give the (AC) statement:

Given $$\alpha$$, let $$\kappa(\alpha)=0$$ if $$|\alpha|$$ is a singular cardinal or countable. Otherwise, let $$\kappa(\alpha)$$ be the largest cardinal such that $$|\alpha|\cdot\kappa(\alpha)\le\alpha$$.

Theorem (Gao-Jackson-Kieftenbeld). $$\alpha$$ and $$\beta$$ are Borel isomorphic iff $$|\alpha|=|\beta|$$ and $$\kappa(\alpha)=\kappa(\beta)$$.

• So am I wrong in saying that $\beta=\omega+1$ and $\alpha=\omega 2=\omega + \omega$ are homeomorphic? It seems this is implied by your claim. – J.K.T. May 14 '11 at 4:20
• Hi J.T. My claim was originally stated incorrectly, I needed to add ones at the end of everything, I just noticed I left them out. Note your $\beta$ and $\alpha$ are not homeomorphic: $\beta$ is compact but $\alpha$ is not. I do not currently seem to have a full answer for limit ordinals, but the above deals with all successor ordinals. – Andrés E. Caicedo May 14 '11 at 6:00
• I thought the derivative might show up somewhere! It is very interesting that the $\omega^{\delta}$, $\delta>0$ are only homeomorphic to themselves. Thanks for such a detailed treatment Andres. – J.K.T. May 14 '11 at 15:13
• It is intriguing how in both questions of the OP, both you and Joel mistook Bendixson's name to be Bendixon, despite providing a wikipedia link. :-) – Asaf Karagila May 14 '11 at 15:46