Fix . Prove (or maybe disprove) that is not -furcus for any .
Solution
It is indeed not -furcus. Notice that if then it is obviously not -furcus, so we assume .
Let be two (nonempty) closed subset of , we claim that is also closed. To do this, BWOC, there is a real number such that is a limit point of . Then there exist infinite sequences such that as . As , after taking a subsequence we can assume as . Then also as . Because both and are closed we see , so , a contradiction.
BWOC, is -furcus. Take , which is a closed set in . Also notice that is countable. Therefore the set ( times) is also closed and countable. This shows that it cannot be dense in (otherwise because it is closed we see it is equal to , which is not countable). Thus there exist positive reals such that . Now there exists such that . Take . Then from assumption there exists a factorization where , which implies . However, , which has empty intersection with , a contradiction.