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## Topology Q+A Board

From: Henno Brandsma
Date: Mar 22, 2006
Subject: Jones' lemma

In reply to "Re: Re: Re: pictorial idea of radial plane", posted by Even_Distribution on Mar 21, 2006:
>In reply to "Re: Re: pictorial idea of radial plane", posted by Henno Brandsma on Mar 21, 2006:
>>In reply to "Re: pictorial idea of radial plane", posted by sTudent on Mar 21, 2006:
>>>In reply to "pictorial idea of radial plane", posted by no name on Mar 21, 2006:
>>>>Can someone give me a pictorial idea of a radial place?
>>>>A subset of R^2 is radially open iff it contains an open line segment in each direction about each of its points.
>>>>The plane with this kind of topology is called the raidal plane.
>>>>How is it different from the usual topology on R^2?
>>>>How could it be not normal?
>>>>
>>>>thanks
>>>>
>>>
>>>Evidently the radial topology is finer than the euclidean one. To see that
>>>it is strictly finer, note that if a set A has at most two points on every
>>>line of the plane then R^2 - A is radially open so A is closed in the
>>>radial topology. As every subset of A also has this property, every subset
>>>of A is also closed hence A is closed discrete. For example, a
>>>circumference is closed discrete in the radial topology (but evidently
>>>not discrete in the euclidean topology).
>>>
>>>Moreover, the radial plane is separable: the points with both coordinates
>>>rational form a dense set. Indeed, let G be radially open with (x,y) in G.
>>>Then, by radiality, there is a rational r such that (r,y) is in G. Then,
>>>again by radiality, there ia a rational s such that (r,s) is in G.
>>>
>>>So the radial plane is separable and contains a closed discrete set of
>>>size continuum. Hence, by Jones's lemma, it is not normal.
>>>
>>>BTW, I am not able to prove that it is regular. Is this true? It is
>>>Hausdorff because its topology is finer than the euclidean one.
>>
>>Mm, I recall a paper by Kunen about a similar topology: the plus-topology
>>(U is open iff for each (x,y) there is some small +-sign centered at (x,y)
>>that stays inside U), also called the topology of separate continuity.
>>(the radial topology is basically the topology so that functions are continuous
>>iff they are continuous on every line). This plus-topology is not regular, as
>>it is separable but has weight > c and a regular space that is separable has weight <= c.
>>Perhaps something similar is the case here.
>>
>>Henno
>>
>
>
>Is there another way of showing that the radial plane is not normal? Without introducing weight, etc.,? I have not found one, but I do believe it is quite possible.

Well, the Jones' lemma (or the case we used anyway) can be shown without using weight and all that.
Suppose that X is a normal space, and suppose that X has an infinite dense set D and a closed and discrete
subspace C. Then 2^|C| <= 2^|D|, where |C| and |D| are the cardinalities of C and D and 2^|C|
is the cardinality of the power set of C etc.

Proof: for every subset A of C we have note that A and C\A are both closed in C (as C is discrete as a subspace)
and so both closed in X (as C is closed). So we have 2 disjoint closed sets in a normal space and so
we have a Urysohn function f_A : X --> R such that f_A[A] = {0} and f_A[C\A] = {1}.
Note that if A != A', say x in A\A' , then f_A(x) = 0 and f_A'(x) = 1, so f_A != f_A', and
different subsets of C give different continuous functions from X to R.
Hence there are at least 2^|C| distinct continuous functions from X to R.

Now consider the set C(X,R) of all continuous functions from X to R. As R is Hausdorff and D is dense
any continuous functions is determined by its values on D: f|D = g|D then f = g everywhere.
So f --> f|D defines an injection from C(X,R) to all functions from D to R, and this latter
set has cardinality <= |R|^|D| = (2^|N|)^|D| = 2^(|N| x |D|) (standard set theory)
= 2^|D| (as D is infinite).
So |C(X,R)| <= 2^|D| and 2^|C| <= |C(X,R)| by the first part.
Hence 2^|C| <= 2^|D|.

Now see what happens for C = a circle in R^2 in the radial plane: this has size |R| and is
discrete, as remarked by sTudent.
Also Q x Q is countable and dense.
So if X were normal, we could apply Jones' lemma and we would have
2^|R| <= 2^|N| = |R|, which contradicts Cantor's theorem: 2^|R| > |R|.
So X is not normal.
So normality enables a counting argument on continuous functions, and
then set theory does the rest.

The same proof idea also works to show non-normality of the Niemitzki/Moore plane and
the square of the Sorgenfrey line.

Henno