Problem 2.14 in Probability and Measure by Billingsley requires
knowing the definition of nowhere-dense sets. Appendix 15 provides a
definition of nowhere-dense sets and sets of the first category with
respect to :
"The set is, by definition, dense in the set if for
each and each open interval containing ,
meets . This is the same thing as requiring (the closure of ). The set is, by
definition, nowhere dense if each open interval contains
some open interval that does not meet . This makes sense:
if contains an open interval that does not meet ,
then is not dense in ; the definition requires that
be dense in no interval ."
The usual topological definition of density is as follows: Let be a metric space. Let . We say that is
dense in if and only if . The topological
definition of nowhere-dense is: is nowhere-dense in if
. We show that these
definitions are consistent with the ones given in the textbook when .
Let's show that the definition of density is consistent. First, we
suppose that is dense in if and only if for every
and every open interval containing , . We show that this means . Let's show
that . Let . We show that . There exists open interval , , such that . If , then
(and so we're done). Otherwise, if , then . Since this is true for every
open interval containing , (where denotes the
limit points of ). Therefore, . Now we show
that . Let . For every
open interval , . This means ; if not, then would not be dense in (because for every and every open interval containing , ). So, .
Now, let's suppose that . Then, we show that for
every and every open interval containing , . Let . Then, . So, or . Choose an arbitrary open interval containing
. If , then because . Otherwise, if , then . This immediately implies .
This proves that the definitions of density are equivalent (for ). Now, we show that the definition of nowhere-dense is
consistent when . First, let's clarify the definition
given by Billingsley: is nowhere-dense if for every open interval
, there exists open interval so that . This is the same thing as saying is not dense in any
open interval . To see this, note that if is not dense in any
(nowhere-dense), then for some and some open interval
containing , . Then, for any open
interval , note that . So, if we
take this interval small enough, we can find a so that
.
Now, we show that the topological definition is consistent. First,
suppose that Billingsley's definition is true. We show that
. Suppose for contradiction
that . Then, there exists open
interval such that and . Since
is nowhere-dense, there exists open interval such
that . But . Since , that means . Consider any . Then, . Since is open, . This implies that , which is a
contradiction. So, .
Next, let's prove the other direction: suppose that
. Suppose there exists open
interval such that for every open interval , , . Choose any . For every open containing
, . Then, either or . This implies that . Since was
arbitrary, this means that . For any ,
is an open neighborhood contained in , so
. But this contradicts the
hypothesis, so we must have that for every open interval , there
exists an open interval , , such that .
This shows that the topological definition of nowhere-dense is the
same as the definition given by Billingsley when we take .