some properties of uncountable subsets
of the real numbers∗
sauravbhaumik†
2013-03-21 22:00:33
Let S be an uncountable subset of R. Let A := {(x, y) : (x, y)∩S is countable}.
For
Lindelöff, there is a countable subfamily A 0 of A such that
S R0 is hereditarily
S
A = A . For the reason that
of members of A 0 has a countable
S each
S in0
tersection with S, we have that ( A ) ∩ S is countable. As the open set A 0
can be expressed uniquely as the union of its components, and the components
are countablySmany, we label the components as {(an , bn ) : n ∈ N}.
See that ( A 0 ) ∩ S is precisely the set of the elements of S that are NOT
the condensation points of S.
Now we’d propose to show that {an , bn : n ∈ N} is precisely the set of the
points which are unilateral condensation points of S.
Let x be a unilateral (left, say) condensation point of S. So, there is some
r > 0 with (x, x + r) ∩ S countable. So, there is some (an , bn ) such that
(x, x + r) ⊆ (an , bn ). See, if x ∈ (an , bn ), then x is NOT a condensation point,
for x has a neighbourhood (an , bn ) which has a countable intersection with S.
But x is a condensation point; so, x = an . Similarly, if x is a right condensation
point, then x = bn .
Conversely, each an (bn , resp) is a left (right, resp) condensation point. Because, for each
S ∈ (0, bn − an ), we have (an , an + ) ∩ S countable. And as no
an , bn isSin A 0 , an , bn are condensation points.
So, A 0 is the set of non-condensation points - it is countable; and {an , bn }
are precisely the unilateral condensation points. So, all the rest are bilateral
condensation points. Now we see, all but a countable number of points of S are
the bilateral condensation points of S.
Call T the set of all the bilateral condensation points that are IN S. Now,
take two x < y in T . As x is a bilateral condensation point of S, (x, y) ∩ S is
uncountable; and as T misses atmost countably many points of S, (x, y) ∩ T is
uncountable. So, T is a subset of S with in-between property.
∗ hSomePropertiesOfUncountableSubsetsOfTheRealNumbersi
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We summarize the moral of the story: If S is an uncountable subset of R,
then
1. The points of S which are NOT condensation points of S, are at most
countable.
2. The set of points in S which are unilateral condensation points of S, is,
again, countable.
3. The bilateral condensation points of S, that are in S, are uncountable;
even, all but countably many points of S are bilateral condensation points
of S.
4. The set T ⊆ S of all the bilateral condensation points of S has got the
property: if ∃x < y ∈ T , then there is also z ∈ T with x < z < y.
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