It is known to me that, in some models for ZF, for an uncountable set $J$, the Cantor cube {0,1}^J can be metrizable. Unfortunately, I still do not know whether there is a model for ZF in which, for some uncountable set $J$ the Cantor cube {0,1}^J is simultaneously non-compact and metrizable. Perhaps, there are mathematicians in the world who know. I would be grateful to them for useful hints.
Dear Eliza,
I assume that the uncountable set J satisfying 2^J is metrizable can be expressed as a countable union of a family A of finite non-empty sets without a choice function. If this is the case then 2^J cannot be compact because otherwise A would have to have a choice function. So any ZF model witnessing the failure of the countable axiom of choice for finite sets, such as Cohen's second model, will do the job.
The product of compact sets is compact. So, there is no such thing as you say.
Dear Professor Keremedis,
Thank you for your answer. I have proved with details that, in every model for ZF+negation of CC(fin), a Cantor cube can be simultaneously metrizable and non-compact. However, my proof is still not sufficiently elegant. I shall try to improve it. Do you know an article in which the non-compactness of {0,1}^J for an uncountable J such that J is a countable union of finite sets has been published?
Dear Zingang Pan, Products of compact spaces are compact in every model for ZFC but some products of compact spaces are non-compact in some models for ZF,
Regards, Eliza
To be perfectly precise and honest, I must confess that I can prove that, in every model for ZF+the negation of CC(fin), there is an uncountable set $J$ such that the Cantor cube ${0, 1}^J$ is both metrizable and non-compact. Unfortunately, I still have not proved to myself with satisfactory details that, for every uncountable set $J$ such that $J$ is a countable union of finite sets, the Cantor cube ${0, 1}^J$ is non-compact.
Eliza
Dear Eliza,
To the best of my knowledge, the only closest result to your question published is Corollary 2 in
http://www.jstor.org/stable/20799296?seq=5#page_scan_tab_contents
which implies that whenever 2^J is compact then for every n∈N, every countable family of n-element subsets of J has an infinite subfamily with a choice set.
All the best,
Kyriakos
Now, I am sure that, in ZF, CC(fin) is equivalent to the statement:: Every metrizable Cantor cube is compact. Therefore, in every model for ZF+the negation of CC(fin) there exist metrizable Cantor cubes that are non-compact.
Eliza
I shall give a lecture about Cantor cubes that are both metrizable and non-compact at toposym in Prague on July 28, 2016. In Cohen's second model there are Cantor cubes that are simultaneously metrizable, not second-countable and non-compact; however ,$2^{\mathbb{R}}$ is never metrizable and, as Kyriakos Keremedis proved, $2^{\mathbb{R}$ is non-compact in this model. My slides will be available after the conference. Eliza
Dear Prof. Eliza Wajch, kindly upload your this lecture after the toposym..
Ok, I shall do it. I am just typing an extended article about it. I shall submit the article as soon as it is completed.
Regards, Eliza Wajch
- Badges
- Science topic
- Similar topics
- Mathematics