Re: Accumulation points in metric space
- From: "Dave L. Renfro" <renfr1dl@xxxxxxxxx>
- Date: 21 Feb 2007 15:14:15 -0800
Dave L. Renfro wrote (in part):
If (X,d) is not separable, then there exists an uncountable
subset having no accumulation point in X. In fact, I believe
even more is true: If (X,d) is not separable, then there
exists an uncountable subset that is uniformly isolated
(i.e. there exists epsilon > 0 such that the distance
between each pair of distinct points in the uncountable
subset is at least epsilon).
I happen to be at home now, where my notes are. Here's
the theorem and proof.
THEOREM: If (X,d) is a non-separable metric space, then
there exists epsilon > 0 and an uncountable
subset E of X such that, for each pair of distinct
points x & y in E, we have d(x,y) > epsilon.
PROOF: I first claim there exists a positive integer n
such that for all countable subsets B of X we have
(for all x in X)(for all b in B)( d(x,b) > epsilon ).
If not, then for all positive integers k there exists
a countable subset B_k of X such that the distance
between each pair of points, one point chosen from X
and one point chosen from B_k, is less than or equal
to epsilon. This contradicts the non-separability of
X, since UNION(k=1 to oo) B_k is a countable dense
subset of X. I will now construct an appropriate set
E, of cardinality aleph_1, by transfinite induction
over the ordinals less than omega_1.
Choose any x in X and denote this point by x_1.
Let alpha > 0 be a countable ordinal and assume that,
for each ordinal beta < alpha, we have chosen a point
x_beta such that the distance between each pair of
distinct points in the set {x_beta: beta < alpha} is
greater than 1/n. Since {x_beta: beta < alpha}
is a countable subset of X, and there exist points
in X not belonging to this countable set (it trivially
follows from the non-separability of X that X is
uncountable), it follows from what I proved earlier
that there exists x_alpha in X such that each of the
distances between x_alpha and the points belonging
to {x_beta: beta < alpha} is greater than 1/n.
The set E = {x_beta: alpha < omega_1} is a set of
cardinality aleph_1 such that any two of its distinct
points are a distance of more than epsilon = 1/n
from each other.
NOTE 1: There does not exist a positive lower bound on epsilon
that uniformly works for all non-separable metric
spaces. This is because there exist non-separable
metric space of arbitrarily small diameter.
[Pick an uncountable set and define the distance
between each pair of distinct points to be an
arbitrarily small specified positive value.]
NOTE 2: In the case of the sequence space elle^infinity,
we can find a concrete example of cardinality c
with epsilon = 1: The collection of all sequences
whose entries are either 0 or 1.
NOTE 3: This theorem was given in a topology class I had
under Peter J. Nyikos in Spring 1987. A few years
ago (2001 or 2002), I came across the result in
Sierpinski's "General Topology" [1956 edition
translated by C. Cecilia Krieger and reprinted
by Dover Publications in 2000], Section 52,
Lemma, pp. 116-117. The proof Nyikos presented
and the proof in Sierpinski's book are the
same as the proof above.
Dave L. Renfro
.
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