I know that there is a model M for ZF such that. for an uncountable set S in this model and for every collection $\{ (X_s, d_s): s\in S\}$ of metric spaces in this model, their product $\prod_{s\in S} X_s$ in M is metrizable in M. In particular, for an uncountable set S in M, the product $\mathbb{R}^S$ is metrizable, however, I have not found this result in the literature so far. I would be grateful if you could tell me whether you have located it in the literature. If your answer is YES, please, tell me where I can find this result. I know how to prove the result.
Dear Eliza,
Your result about metrizability of the product R^S sounds very interesting and I have never seen it before.
So you're claiming that any argument proving non-metrizability of uncountable products of lines must use choice. Have you checked anything by B. Banaschewski? I know he's been keenly interested in uses of AC.
I thank you for your comments. I am sure that I proved my result very carefully. I am just typing its proof to show it to my colleague from Cracow. I shall ask him to check it most carefully although I've already checked it twice. Of course, my result is in ZF plus the negation of AC. I am glad to inform you that my long paper, co-authored with A. Piekosz, "Compactness and compactifications in generalized topology" has been just accepted for publication in "Topology and Its Applications". The paper contains some of our discoveries also surprising. Among them equivalents of the ultrafilter theorem. The whole work is in ZF, most results proved without AC. We work in ZF. We are interested in how to avoid AC when this is possible and what can happen with known theorems in models for ZF plus the negation of AC.
Since you're interested in \neg AC, have you looked into topological consequences of full determinacy (AD)?
Dear Professor Bankston,
Yes, I have looked at AD but, unfortunately, still not deeply enough. However, I hope that I will know more about AD and its consequences in a not-too-distant future. I've just sent to my colleague my proof that $\mathbb{R}^J$ is metrizable for some uncountable set $J$ in a model for $ZF+\neg CC(fin)$. Let me wait for his checking whether my proof is correct. I hope that it is correct but, since people make mistakes and are unable to notice all their own mistakes, checking correctness by someone else can be even necessary. Eliza
I've just explained it to myself that, in ZF, the statement that R^J is metrizable for some uncountable set J is an equivalent of the negation of CC(fin). I hope that I am not wrong. Anyway, my colleague will check the details. Regards, Eliza
I showed the following theorem of mine to David Fremlin:
Theorem. In ZF, the following conditions are all equivalent:
(i) the negation of CC(fin); (ii) there exists an uncountable set J such that R^J is metrizable;
(iii) there exists an uncountable set J such that {0, 1}^J is metrizable;
(iv) there exists an uncountable set J such that {0, 1}^J is first-countable.
The answer of D. Fremlin sent to me via e-mail was:
OK. I see how (i)-(iv) link together, that's very nice, well done!
Regards, Eliza
Since some researches have asked me via e-mail how I deduce that R^J can be metrizable for an uncountable set J in a model for ZF, let me give to all concerned the following theorem:
Theorem. Suppose that a non-empty set J is a countable union of finite sets and that $\{ (X_j, d_j): j\in J\}$ is a collection of (quasi)- metric spaces. If each $X_j$ is equipped with the topology induced by d_j, then the product $\prod_{j\in J} X_j$ is (quasi)-metrizable.
Now, when M is a model for ZF+negation of CC(fin), there exists in M an uncountable set J such that J is a countable union of finite sets , so, for such a set J, R^J is metrizable in M and, moreover, for example, S^J is quasi-metrizable in M where S is the Sorgenfrey line. I was also surprised when I discovered these facts, They have looked strange in my opinion but they do not look strange now. They are relatively simple results.