Dense and Nowhere Dense Sets
Dense Sets in a Topological Space
Definition: Let
be a topological space. The set
is said to be Dense in
if the intersection of every nonempty open set with
is nonempty, that is,
for all
.
Given any topological space
it is important to note that
is dense in
because every
is such that
, and so
for all
.
For another example, consider the topological space
where
is the usual topology of open intervals. Then the set of rational numbers
is dense in
. If not, then there exists an
such that
.
Since
we have that
for some open interval
with
and
. Suppose that
. Then we must also have that:
![{\displaystyle {\begin{aligned}\quad \mathbb {Q} \cap U=\mathbb {Q} \cap (a,b)=\emptyset \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ab5f1cb41335658997f8664af70e97f3963a210a)
The intersection above implies that there exists no rational numbers in the interval
, i.e., there exists no
such that
. But this is a contradiction since for all
with
there ALWAYS exists a rational number
such that
, i.e.,
. So
for all
. Thus,
is dense in
.
We will now look at a very important theorem which will give us a way to determine whether a set
is dense in
or not.
Theorem 1: Let
be a topological space and let
. Then
is dense in
if and only if
.
- Proof:
Suppose that
is dense in
. Then for all
we have that
. Clearly
so we only need to show that
.
Nowhere Dense Sets in a Topological Space
Definition: Let
be a topological space. A set
is said to be Nowhere Dense in
if the interior of the closure of
is empty, that is,
.
For example, consider the topological space
where
is the usually topology of open intervals on
, and consider the set of integers
. The closure of
,
is the smallest closed set containing
. The smallest closed set containing
is
since
is open as
is an arbitrary union of open sets:
![{\displaystyle {\begin{aligned}\quad \mathbb {Z} ^{c}=...(-2,-1)\cup (-1,0)\cup (0,1)\cup (1,2)\cup ...\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/80110ecae734dec0376627f4540c7a88723ed535)
So what is the interior of
? It is the largest open set contained in
. All open sets of
with respect to this topology
are either the empty set, an open interval, a union of open intervals, or the whole set (the union of all open intervals). But no open intervals are contained in
and so:
![{\displaystyle {\begin{aligned}\quad \mathrm {int} ({\bar {\mathbb {Z} }})=\emptyset \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f790a43c4341931507814b05848490a8b3d7ca3d)
Therefore
is a nowhere dense set in
with respect to the usual topology
on
.
The Baire Category Theorem
Lemma 1: Let
be a topological space and let
. If
is a nowhere dense set then for every
there exists a
such that
.
Theorem 1 (The Baire Category Theorem): Every complete metric space is of the second category.
- Proof: Let
be a complete metric space. Then every Cauchy sequence
of elements from
converges in
. Suppose that
is of the first category. Then there exists a countable collection of nowhere dense sets
such that:
![{\displaystyle {\begin{aligned}\quad X=\bigcup _{i=1}^{\infty }A_{i}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/18711850c706ea8d20983afc07502e01ca7dc46b)
- Let
. For each nowhere dense set
,
there exists a set
such that
.
- Let
be a ball contained in
such that
. Let
be a ball contained in
whose radius is
and such that
. Repeat this process. For each
let
be a ball contained in
whose radius is
and such that
and such that
.
- The sequence
is Cauchy since as
gets large, the elements
are very close. Since
is a complete metric space, we must have that this Cauchy sequence therefore converges to some
, i.e.,
.
- Now notice that
for all
because if not, then there exists an
such that
for all
. Hence
is open and so there exists an open ball
such that
but then
because
for all
.
- Since
for all
then since
we must have that then ![{\displaystyle x\not \in A_{n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9ae948e4ce01359d4aeee23664eb00cb0e514e3b)
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