Question regarding disjoint unions, sequential compactness, and Dedekind-finiteness

I have proved the following two results:

• $[\mathsf{ZF}]$ The disjoint union of a Dedekind-finite family of sequentially compact topological spaces is again sequentially compact.

• $[\mathsf{ZF}+\text{Countable Choice}]$ If the disjoint union of a family of topological spaces is sequentially compact, then the family is finite, and each member of the family is sequentially compact.

I've further been able to show that the latter fails if there is an infinite, Dedekind-finite set--if $X$ is such a set, topologize $X$ discretely, which is the same as topologizing $X$ as the disjoint union of an infinite, Dedekind-finite family of singleton spaces, each of which is sequentially compact, and since sequences of points of $X$ will necessarily have only finitely-many distinct points by Dedekind-finiteness, we have that $X$ is also sequentially compact.

I'd like to be able to weaken the second result to the following

Claim: If the disjoint union of a family of topological spaces is sequentially compact, then the family is Dedekind-finite, and each member of the family is sequentially compact.

Ideally, I'd like to do it in a setting without Countable Choice, but I just don't see a way out of it. Showing that each member of the family is sequentially compact is simple, and I can readily conclude that a sequence of points of the larger space must lie strictly outside all but finitely-many members of the family of smaller spaces, since otherwise, we can build a new sequence with no two points lying in the same member of the family of smaller spaces, which can have no convergent subsequence. However, while Dedekind-finiteness of the family is sufficient to ensure that this doesn't happen, I cannot see why it should be necessary. If the family is Dedekind-infinite, then we have a countably-infinite subfamily, but without a choice function, I see no way to choose a single point from each member of the subfamily to make a sequence without convergent subsequence. (Of course, I could fix this if I added the hypothesis that the family of topological spaces had a choice function, but I'd like to avoid this, if possible.)

Edit: It is consistent with $\mathsf{ZF}$ that there is a countably-infinite set $X$ of $2$-element sets whose union is Dedekind finite. Considering the two-element sets in the indiscrete topology, each of them is readily sequentially compact, and the disjoint union is sequentially compact by virtue of its Dedekind-finiteness. Thus, my claim clearly need not hold in $\mathsf{ZF}$ alone.

Is there some weaker Choice principle than Countable Choice that will allow me to prove it? (It occurs to me that this might be a better fit at Math.Overflow, but I thought I'd try it here, first.)

• You can give those two-element sets the discrete topology instead so that the space $\hspace{1.7 in}$ will also be Hausdorff (since it will be discrete). $\;$ – user57159 Sep 21 '13 at 8:39
• @Ricky: True enough. – Cameron Buie Sep 21 '13 at 9:45
• By the way, I have a feeling that these results are not new. You should dig the papers about topology and choice and see if you can find it there (I'll be surprised if the answer is negative). There was a lot of investigation in this direction that certain sums of certain forms of compactness are such and such are statement equivalent to countable choice, or whatever. – Asaf Karagila Sep 21 '13 at 16:32
• @Asaf: I've actually been looking at a bunch of such papers recently. Haven't seen that result yet, but I suspect I'll come across it eventually. – Cameron Buie Sep 21 '13 at 16:34

Note that if there is an infinite Dedekind-finite set at all, then there is one which can be mapped onto $\omega$.

Suppose that $A$ is a Dedekind-finite set which can be mapped onto $\omega$, and fix $f$ to be a surjection. This gives us a partition of $A$ into $A_n=\{a\in A\mid f(a)=n\}$. Giving each of the $A_n$'s a discrete topology, as you did in your example, is immediately a sequentially compact set (every sequence must have only finitely many distinct elements). Now their sum is a countable sum of Dedekind-finite sets which is Dedekind-finite itself, so again trivially it has to be sequentially compact.

So in order to have that, you must have that there are no infinite Dedekind-finite sets at all. Which is strictly weaker than the axiom of countable choice, by the way.

Now if we only assume there are no infinite Dedekind-finite sets, the result still doesn't hold. Take a countable family $A_n$ of non-empty sets without a choice function, and topologize each of them with the trivial topology. These sets are now compact, and sequentially compact, for obvious reasons. Their sum, is too sequentially compact, because a sequence will have to concentrate on finitely many $A_n$'s and therefore it will have a convergent subsequence; otherwise write $x_k$ for the sequence and let $n(k)=\min\{k\mid x_k\in A_n\}$ or $\infty$ otherwise. Since the sequence meets infinitely many $A_n$'s taking $a_n=a_{n(k)}$ when $n(k)<\infty$ is a partial choice function; so the axiom of countable choice holds (because it is equivalent to partial choice, as you know).

The above can be used to show that if the sum of countably many sequentially compact space is sequentially compact, then the axiom of countable choice fails. Therefore your wanted result is equivalent to the axiom of countable choice to begin with and we can't do much better.

• How is your first sentence shown? $\;$ – user57159 Sep 21 '13 at 7:59
• @Ricky: It's a theorem of Kuratowski that $\omega<2^A\iff\omega\leq^*A$, so if $A$ is an infinite Dedekind-finite set which cannot be mapped onto $\omega$, its power set is Dedekind-finite, and mapping infinite sets to $0$ and finite sets to their cardinality is a map from $\mathcal P(A)$ onto $\omega$. – Asaf Karagila Sep 21 '13 at 8:01
• I think you need to topologize the sequence of partial products, rather than just the sets $A_n$ themselves, to get that sequences must concentrate. $\;$ – user57159 Sep 21 '13 at 8:13
• @Ricky: Yes, of course. You're right. It might be the case that we can choose a finite subset from each $A_n$ instead. Thank you. – Asaf Karagila Sep 21 '13 at 8:55
• Just what I was looking for. Cheers! – Cameron Buie Sep 21 '13 at 15:29