Cantor (-) ternary sets
____________________
Let be a family
of intervals defined as follows:
and
with ,
i.e. from every interval of
-th
generation remove an open interval of length
centred at the mid-point of to get two new (closed) intervals of next generation.
Let
.
Picture for
.0____________________.____________________.____________________.1
.0____________________.
. ___________________.1
.0_______.
. _______.
. _______.
.
______.1
.0__. . __. . __. . __. . __. . __. . __. .__.1
The sets are
compact (because closed and bdd) and we have
.
Thus (by intersection of compact sets thm) the set
is a compact set.
If , clearly it
contains a sequence of points
(which are the end pts of the intervals ).
In fact it contains uncountably many pts (see later).
Note that for any
the set
is nowhere
dense, which means for any open
interval , the
intersection
is not
dense in
.
(It is easy to see this from the construction of : For some
there is
and
we remove from it an open interval.)
Remark: The nowhere dense sets and their countable unions are called the sets
of the 1st category.
One can characterize the Cantor ternary set ,
, as a set of numbers of the form
with . (To avoid
nonuniqueness we discard representation for which
for some
.) Using this
representation we see that
Theorem
The Cantor ternary set is uncountable.
Proof: It is sufficient to show that there is function which is onto.
For this we note that every real number can be represented as
follows:
with tn = 0 or 1. (This representation may be not unique.)
The desired function is obtained by setting
with
Proposition
The total length of removed intervals equals .
Proof: The number of intervals with generations equals
. We remove from each of them an interval of
length
. Thus sum of
the lengths of all removed intervals equals
.
Remark In particular for , the sum of lengths of all removed intervals equals
,
i.e. it is equal to the length of the entire interval .
The sets are
Lebesgue measurable and
.
In particular
(similarly as for , although
is uncountable)!