Re: Countably infinite Hausdorff topology?

From: David C. Ullrich (ullrich_at_math.okstate.edu)
Date: 09/15/04 

Date: Wed, 15 Sep 2004 06:33:08 -0500

On Tue, 14 Sep 2004 14:04:39 GMT, "shedar" <nobody@nonesuch.com>
wrote:

>"Robert Israel" <israel@math.ubc.ca> wrote in message
>news:ci50j1$hsf$1@nntp.itservices.ubc.ca...
>> In article <fkl1d.6375$G03.1882402@news4.srv.hcvlny.cv.net>,
>> Stephen J. Herschkorn <herschko@rutcor.rutgers.edu> wrote:
>> >A standard exercise is to show that any infinite sigma-field has
>> >cardinality at least 2^omega. That got me thinking about topologies.
>>
>> >Does there exist a Hausdorff space whose topology is countably infinite?
>>
>> Suppose X is a Hausdorff space with an infinite topology. In particular,
>> X is infinite. Then, with at most one exception, each point of X has
>> a neighbourhood whose complement is infinite (i.e. if there were
>> two points x and y whose neighbourhoods all had finite complements, they
>> could not have disjoint neighbourhoods, violating the Hausdorff
>> requirement). So let x_1 be a point of X with an open neighbourhood U_1
>> whose complement is infinite. Now X \ U_1 is also an infinite Hausdorff
>> space, and the same reasoning applies to it: there is x_2 in X \ U_1
>> with an open neighbourhood U_2 such that X \ U_1 \ U_2 is infinite.
>> Moreover, again using the Hausdorff property we may assume that x_1
>> is not in U_2. Using induction
>
>[See comment (*) by Shedar below.]
>
>> we get sequences of points x_j of X
>> and open sets U_j such that x_i is in U_j iff i = j. Then for any
>> subset S of N, U_S = union {U_j: j in S} is an open set which contains
>> x_i iff i is in S. These constitute a family of distinct open
>> sets of cardinality c.
>>
>
>Comment (*): Correct me if I wrong, but I think "induction" (or technically,
>recursion on the least limit ordinal)

I don't think "or, technically" is quite right here - there's nothing
imprecise or informal about just saying "by induction"; induction
on the natural numbers is a perfectly respectable thing.

>would not "cut it". I think one needs
>to invoke AC (or at least the axiom of countable choice) to get the
>sequences mentioned above.

Seems to me that you may well be right that some sort of AC is
required for the argument as stated. Seems like a slightly
silly thing to point out, because people use AC without
mentioning it this way all the time.

Seems particularly silly in this case because it's trivial
to convert the argument into one that does not use AC
(at the expense of converting it into a proof by contradiction):
Assume the topology is countable and fix an enumeration of
the open sets. Now each time we need to choose an open
set with a certain property choose the one with that
property that comes first in the enumeration.

("Or technically", use the fact that the first infinite
ordinal is well-ordered...)

>Shedar
>

************************

David C. Ullrich