Countability of subset of R?
sg...@hotmail.co.uk
uncountable subset of R" to which I didn't get a reply, probably
because my post was one of many similar ones in that thread. Sorry to
ask it again, but it seems like quite an interesting question in its
own right to me and I would be interested to hear the answer if anyone
knows it, since I have been frustrated by my own impotent attempts to
answer it.
Suppose that X is a subset of the unit interval with the property that
every element of X is isolated to the right, i.e. for all x in X there
exists an e > 0 such that (x,x+e) n X is empty. X is clearly
countable. Is the same true of its closure?
-Rotwang
Robert Sheskey
No. The closure of such a set can have the cardinality of the reals.
Suppose f:R -> (0,1) is a strictly increasing function that is
left-continuous everywhere and right-discontinuous at the rationals.
Then f(Q) has the properties we want: for any rational q,
lim_{x -> q+} f(x) > f(q), so we have that f(q) is right-isolated,
and for any real r, lim_{q -> r-} f(q) = f(r), where the limit is
taken over rational q, so we also have that closure(f(Q)) = f(R).
Such an f can be obtained by, say, letting f(x) = sum_{q_k < x} 1/2^k,
where q_1, q_2, ... is an enumeration of the rationals.
Robert Sheskey
Dave L. Renfro
> Suppose that X is a subset of the unit interval
> with the property that every element of X is
> isolated to the right, i.e. for all x in X there
> exists an e > 0 such that (x,x+e) n X is empty.
> X is clearly countable. Is the same true of its closure?
The closure of such a set does not have to be countable.
In fact, much more is possible. Let C be any nonempty
perfect nowhere dense subset of [0,1] (i.e. a "Cantor set")
and let X be the collection midpoints of the bounded
complementary intervals of C (i.e. the midpoints of
the connected components of [0,1] - C). Then X is
countable but the closure of X is equal to X union C,
which has cardinality c. Indeed, the closure of X
can have Lebesgue measure arbitrarily close to 1.
Sets with countable closure are rather "small" as
countable sets go. The so-called scattered subsets
of the reals can be somewhat larger, being
characterized as the countable G_delta sets,
but even the scattered sets are quite small.
Scattered sets not only fail to be dense in every
interval (i.e. they are nowhere dense in the reals),
scattered sets are even nowhere dense relative to
every nonempty perfect set. Note that a slight
modification of the example I gave in the previous
paragraph (choose the left endpoints of the complementary
intervals) is dense in C, and hence is far from
being nowhere dense relative to every perfect set.
(Nowhere dense relative to a perfect set P means
that the closure of the set contains some nonempty
intersection of an open interval with P. With this
example, the closure is identical to the perfect
set C.)
Dave L. Renfro
sg...@hotmail.co.uk
> In article <1171461464.927661.11...@v45g2000cwv.googlegroups.com>, s...@hotmail.co.uk says...
Nice counterexample, thanks.
-Rotwang
David C. Ullrich
No. For example, if S is the set of all right endpoints
of "retained" intervals in the usual construction of the
usual Cantor set, then S is right-isolated but the
closure of S is the Cantor set.
>-Rotwang
************************
David C. Ullrich