Re: Countably infinite Hausdorff topology?

From: David C. Ullrich (ullrich_at_math.okstate.edu)
Date: 09/17/04 

Date: Fri, 17 Sep 2004 08:44:07 -0500

On Fri, 17 Sep 2004 12:04:04 GMT, "shedar" <no_one@nonesuch.com>
wrote:

>"David C. Ullrich" <ullrich@math.okstate.edu> wrote in message
>news:ir9gk0p7vvkua3budg8ems9n17ch4tqtg6@4ax.com...
>> 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...)
>>
>
>1. On the Use of AC:
>
>I agree that by re-casting the previous exposition in some form of Reductio
>Ad Absurdum (RAA), one may be able to get away from the need to use AC. It's
>just that I feel whenever AC is invoked, it should be explicitly mentioned,
>regardless of whether it is being used in its "most intuitive" form (e.g.,
>Russell's MP), or in any of its other forms--it's only fair.

Well, whether you think this or not, the fact is it's _not_ always
explicitly mentioned. In fact it's often used without mention in
a context where the author _appears_ to agree that it _should_ be
mentioned (for example every book on analysis that I know mentions
that it's needed for things like the Hahn-Banach theorem but
uses the fact that a countable union of countable sets is
countable without mentioning that this depends on AC), hence
it seems _very_ likely to me that people other than set
theorists often use AC without being aware of it.

The fact that this happens _so_ often is why it seemed
silly to me for you to point it out in this one case...

>[...]
>
>(b) Transfinite Recursion or AC "to-the-rescue":
>
>It is then clear that when mathematicians define a function "by induction",
>they are TECHNICALLY appealing to the Recursion Theorem (the proof of which,
>incidentally, requires transfinite induction).

Question. When one defines a function from N (or if you prefer
omega) to somewhere by recursion does this require the recursion
theorem for ordinals larger than omega?

>[...]
>What do you think?

What _I_ think is that worrying about all this in the present
context is silly. Of course that's just me.

> Shedar
>

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

David C. Ullrich

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