Re: Cardinality of real topology without choice

Rotwang wrote:

On 22 Jan, 01:06, "Stephen J. Herschkorn" <sjhersc...@xxxxxxxxxxxx>
wrote:


In ZF without choice, can we establish the cardinality of the standard
topology on the reals? I'm guessing the answer is no.



I think it's "yes". Since (0,r) is open for every real r, the
cardinality is at least c, that of the continuum. On the other hand
note that the topology has a countable base X (e.g. X = {B_{1/n)(q)
for rational q, natural n}). Since every open set is a union of
elements of B, the topology has cardinality at most 2^|X|, which is c
(I think).



I think there is a problem here. Let B be a countable base for the topology T. (B is your X.) You have shown there is a surjection from P(B) to T. However, without choice, we cannot conclude there is an injection from T to P(B).

--
Stephen J. Herschkorn sjherschko@xxxxxxxxxxxx
Math Tutor on the Internet and in Central New Jersey

.

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