Metric Topology
Contents
3.2. Metric Topology#
In this section, we shall assume
3.2.1. Topology#
(Topology)
For a given set
Both the empty set
and are elements of .Any union of elements of
is an element of .Any intersection of a finitely many elements of
is an element of .
In other words,
The elements of a topology
There are several properties of geometrical objects which don’t depend on the exact shape of an object and are preserved under continuous deformations like stretching, twisting, crumpling and bending. Some of these properties include the dimension, compactness, connectedness, etc..
A metric
3.2.2. Balls#
(Open ball)
Given a point
is called an open ball at
This definition is a generalization of the concept of neighborhood on a real line.
(Closed ball)
Given a point
is called a closed ball at
(Open ball closed ball containment)
In a metric space
We note that while this property is valid for metric spaces, it is not in general valid for arbitrary topological spaces.
Proof. Open ball contains closed ball.
Let
and .Let
be an open ball with radius .Let
be any number such that . E.g., we can pick .Consider the closed ball
.For every
, .Thus,
.
Closed ball contains open ball.
Let
be a closed ball.Consider the open ball
.For every
, . Thus, .Thus,
.
3.2.3. Open Sets#
(Open sets)
A subset
In other words, there exists an
Every open ball is an open set.
Proof. Let
for some
Let
Consider the open ball:
For any
Thus,
Arbitrary unions of open sets are open sets.
Proof. We prove this by showing that for every element in an arbitrary union of open sets, there exists an open ball around it entirely contained inside the union.
Let
be a family of open subsets of .Let
.Let
.Then, there exists an
such that .Since
is open, there exists an open ball .Hence,
is open.
Finite intersections of open sets are open sets.
Proof. We prove this by showing that for every element in a finite intersection of open sets, there exists an open ball around it entirely contained inside the intersection.
Let
be a finite collection of open subsets of .Let
.Let
. Then .Thus, for each
, there exists an open ball .Let
. Since , hence .Thus, we can construct an open ball
at .Note that
.In other words,
.Thus,
.Thus, for every
, there exists an open ball at entirely contained in .Thus,
is open.
The empty set
(Metric topology)
Let
Proof. By Remark 3.1,
By Theorem 3.4, arbitrary union of open sets is also an open set.
By Theorem 3.5, a finite intersection of open sets is also an open set.
Thus, the family of open sets determined by a metric
Hence, it is a topology.
3.2.4. Closed Sets#
(Closed sets)
A subset
(Trivially closed and open sets)
Proof.
(Singletons are closed)
Every singleton is a closed set.
Proof. Let
Let
Let
Let
.Then, the open ball
doesn’t contain .Thus,
.Thus,
is open.
is open in . is closed in . is neither open nor closed in .
The set of natural numbers
Proof. We prove it by showing that its complement is open.
The set
Each open interval is an open set.
The arbitrary union of open sets is open.
Thus,
is open.Thus,
is closed.
Arbitrary intersections of closed sets are closed sets.
Proof. We proceed as follows:
Let
be a family of closed subsets of .Then
is a family of open subsets of .Then, their union
is an open set.Hence, its complement
is a closed set.But, by De Morgan’s law:
.Thus,
is a closed set.
Finite unions of closed sets are closed sets.
Proof. We proceed as follows:
Let
be a finite collection of closed subsets of .Then,
are open.Then, their union
is open.Then, its complement
is closed.By De Morgan’s law:
.Thus,
is a closed set.
3.2.5. Interior#
(Interior point)
Let
By definition an interior point of a set belongs to the set too.
(Interior point in terms of closed balls)
Let
Then,
This result shows that interior points of a set
We emphasize that this interpretation of interior points is not applicable for all topological spaces but is perfectly valid for metric spaces.
Proof. Let
Then, there exists an open ball
.By Theorem 3.2, there is a closed ball
with some .Thus,
.
For the converse, let there be a closed ball
Then,
.Thus,
.Thus,
is an interior point of .
(Interior)
Let
Note that
Let
In words, every open set that is contained in
Proof. Since
(Interior is set of interior points)
The interior of a set
Proof. Let
We first show that
For the converse, we proceed as follows.
This result is simply a different characterization of the
interior of
Proof. Assume,
Assume
If
Proof. We have:
But
Alternate proof.
Let
be an interior point of .Then
is an interior point of too since .Thus,
.Thus
.
3.2.6. Closure#
(Closure point)
A point
In other words:
Every point in
Proof. Let
(Closure)
Let
Let
In words, every closed set that contains
Proof. Consider the set
(Closure is set of closure points)
The closure of a set
Proof. Let
We first show that
Let
(i.e. ).Then, there exists an open ball
such that .For any
, then, there exists an open ball such that (since is an open set).Thus,
.Hence,
. In other words, the open ball doesn’t contain any closure point of .Thus,
.Thus, for every point
, there exists an open ball .Thus,
is open.Consequently,
is closed.
Next, we show that
Since every point of
is a closure point, hence .Since
is a subset of any closed set containing , hence .
Last, we show that
Let
be a closed set such that .Then,
is open.For any
, there exists an open ball .Thus,
.In particular,
(since ).Thus,
is not a closure point of (i.e. ).Thus, no point in
is a closure point.Thus, every closure point belongs to
.Thus,
.We have established that
is a subset of any closed set that contains .In particular,
since is closed and (by definition).
Together,
This result is simply a different characterization of the
closure of
Proof. Assume,
Assume
A closed ball is a closed set.
Proof. Let
be a closed ball. We proceed by showing that its complement is open:
Let
. Then .Let
.Consider the open ball
.For any
:Thus,
, i.e. .Thus,
.In other words, for every point in
, there exists an open ball contained in .Thus,
is an open set.Thus,
is closed.
Let
In words, complement of interior is the closure of the complement.
Proof. We proceed as follows:
Let
, i.e., is not an interior point of .Then, for every
, is not a subset of .Thus, for every
, is not empty.Thus,
is a closure point of .Thus,
.
The same logic for the converse:
Let
, i.e., is a closure point of .Thus, for every
, is not empty.Thus, for every
, is not a subset of .Thus,
is not an interior point of .Thus,
.
Together, the equality is established.
If
Proof. We have:
Alternate proof.
Let
be a closure point of .Then
is a closure point of too since .Thus,
.Thus
.
Closure of union contains union of closures.
Proof. Let
Let
Let
.Then
for some .Then
is a closure point of some .Then
is a closure point of .Thus,
.
We arrive at the result:
Closure of a finite union is equal to the union of closures.
Proof. Let
Let
Let
From the previous result, we have established that
We now seek to prove that
are closed. is a finite union of closed sets.Hence,
is closed due to Theorem 3.10. .Thus,
.Thus,
.Thus,
due to Proposition 3.6.
3.2.7. Boundary#
(Boundary point)
A point
In other words,
(Boundary)
The boundary of a set
Proof. Let
for every .Thus
is a closure point of . for every .Thus, there is no
such that .Thus,
is not an interior point of .Thus,
.Thus,
.
Let
is a closure point.Thus,
for every . is not an interior point.Thus, there is no
such that .Thus, for every
, .Combining,
is a boundary point.Thus,
.
Boundary of a set is closed.
Proof. By definition:
is open. Hence is closed. is closed.Thus,
is an intersection of two closed sets.Thus,
is closed (Theorem 3.9).
3.2.8. Frontier#
(Frontier point)
A frontier point of a set
(Frontier)
The set of all frontier points of a set
Note
Some authors treat boundary points as frontier points.
They treat boundary and frontier as synonymous.
There is no consistent terminology for the set
Proof.
If
For a closed set, the frontier and boundary are same.
Proof. Every boundary point belongs to the closed set.
The frontier of a closed set is closed.
Proof. For a closed set
3.2.9. Accumulation#
(Deleted neighborhood)
Let
(Accumulation point)
A point
In other words, every deleted neighborhood of
Note that an accumulation point need not belong to the set
Every accumulation point is a closure point.
Although, every closure point need not be an accumulation point.
(Derived set)
The set of accumulation points of a set
(Isolated point)
A point
In other words, there exists an
A closure point is either an accumulation point or an isolated point.
Proof. Let
We need to show that
Since
is not an accumulation point, there exists such that .Since
is a closure point, hence is not empty.Then,
must be .Thus,
.Finally, since
, is an isolated point of .
This is a restatement of the previous result.
A set is closed if and only if it contains all its accumulation points.
Proof.
Recall that
If
Proof. Let
If
then .So consider the case where
.For every
, there exists such that .Thus,
for every .At the same time,
.Thus,
.
3.2.10. Interior II#
Interior of a finite intersection is the intersection of interiors.
Proof. Define:
From Theorem 3.15:
In this derivation, we have made use of the facts that:
complement of interior equals closure of complement (Theorem 3.15).
closure of a finite union is a union of the closures (Theorem 3.18).
3.2.11. Dense Sets#
(Dense subsets)
A subset
A set
Proof. Assume
Let
.There exists
such that .Since
, .Thus,
is not empty.
Assume
Let
. If then is a closure point of . So assume .Let
and consider the open ball .Since
is nonempty and open, hence .Thus,
is a closure point of .Thus,
is dense in .
Complement of a dense set has an empty interior.
Dense sets can be characterized in terms of convergent sequences. This is discussed in Theorem 3.36.
3.2.12. Equivalent Metrics#
(Equivalent metrics)
Let
(Balls inside balls)
Let
Then, for every
Similarly, for every
Proof. Let
Consider the open ball
.Then,
is an open set of .Since both metrics determine same topology, hence
is also an open set of .Since
, hence is an interior point of in .Thus, there exists
such that
We proceed similarly for the other way round.
Consider the open ball
.Then,
is an open set of .Since both metrics determine same topology, hence
is also an open set of .Since
, hence is an interior point of in .Thus, there exists
such that
We can actually prove a stronger result.
(Metric equivalence characterization as balls inside balls)
Let
The two metrics
for every
and
for every
Proof. One part of this result was proved in Theorem 3.25.
We now assume that
for every
Let
be an open set of .Let
.Then,
is an interior point of in .There exists an open ball
.Then, there exists an open ball
.Thus,
is an interior point of in too.Thus,
is an open set in too.Similarly, we can show that if
is an open set in then it is an open set in too.
Thus, both metric spaces determine same topology. Hence, they are equivalent.
(Equivalent metrics as equivalence relation)
Let
Let
Proof. [Reflexivity]
Let
be an arbitrary metric on .Then,
since it determines same topology.
[Symmetry]
Let
and be two metrics on .If they are equivalent, then they introduce same topology.
Thus,
implies .
[Transitivity]
Let
and .Then, all three
determine the same topology on .Thus,
holds too.
(Strongly equivalent metrics)
Let
Two strongly equivalent metrics are equivalent.
Proof. Let
Let
be an open set of .Let
be an interior point.Then, there exists an open ball
.Consider the open ball
.For any point
Thus,
.Thus,
.Thus,
is an interior point of in .Thus,
is an open set in .Similarly, if
is an open set in then it is an open set in .Thus, both metric spaces determine the same topology of open sets.
Thus, they are equivalent.
3.2.13. Connectedness#
(Connectedness)
A metric space
(Connected vs non-connected spaces)
is connected. is not connected.
(Connected subsets)
Let
3.2.14. Counter Examples#
This subsection is a collection of some examples which illustrate some salient aspects of topology.
(Empty Interior)
Consider the set