3.2. Metric Topology#

In this section, we shall assume (X,d) to be a metric space; i.e., X is a set endowed with a metric d:X×XR.

3.2.1. Topology#

Definition 3.7 (Topology)

For a given set X, let T be a family of subsets of X. The family T is called a topology if

  1. Both the empty set and X are elements of T.

  2. Any union of elements of T is an element of T.

  3. Any intersection of a finitely many elements of T is an element of T.

In other words, T contains and X, is closed under arbitrary union and is closed under finite intersection.

The elements of a topology T are called open sets of the topology. Their complements are called closed sets. It is possible to introduce different topologies to the same set X. A set equipped with a topology is called a topological space. A topological structure enables us to define all kinds of continuity.

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 d imposes a topological structure on a set X. This section develops the topological structure of metric spaces.

3.2.2. Balls#

Definition 3.8 (Open ball)

Given a point xX and r>0, the set

B(x,r){yX|d(x,y)<r}

is called an open ball at x with radius r in X.

This definition is a generalization of the concept of neighborhood on a real line.

Definition 3.9 (Closed ball)

Given a point xX and r>0, the set

B[x,r]{yX|d(x,y)r}

is called a closed ball at x with radius r in X.

Theorem 3.2 (Open ball closed ball containment)

In a metric space (X,d), every open ball contains a closed ball. Similarly, every closed ball contains an open ball.

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.

  1. Let xX and r>0.

  2. Let B(x,r) be an open ball with radius r.

  3. Let t be any number such that 0<t<r. E.g., we can pick t=r2.

  4. Consider the closed ball B[x,t].

  5. For every yB[x,t], d(x,y)t<r.

  6. Thus, B[x,t]B(x,r).

Closed ball contains open ball.

  1. Let B[x,r] be a closed ball.

  2. Consider the open ball B(x,r).

  3. For every yB(x,r), d(x,y)<r. Thus, d(x,y)r.

  4. Thus, B(x,r)B[x,r].

3.2.3. Open Sets#

Definition 3.10 (Open sets)

A subset S of X is said to be open in X if for every xS there exists an “open ball” entirely within S.

In other words, there exists an r>0 such that B(x,r)S.

Theorem 3.3

Every open ball is an open set.

Proof. Let xX and consider the open ball

A=B(x,ϵ)={yX|d(x,y)<ϵ}

for some ϵ>0. We need to show that for every point in A, there exists an open ball entirely contained in A.

Let pA. Let r=ϵd(x,p). By definition r>0.

Consider the open ball:

S=B(p,r)={yX|d(p,y)<r}.

For any yS,

d(x,y)d(x,p)+d(p,y)<d(x,p)+r=ϵ.

Thus, yA. Thus, SA. Thus, there exists an open ball around p which is entirely contained in A.

Theorem 3.4

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.

  1. Let {Ai}iI be a family of open subsets of X.

  2. Let A=iIAi.

  3. Let xA.

  4. Then, there exists an iI such that xAi.

  5. Since Ai is open, there exists an open ball B(x,r)AiA.

  6. Hence, A is open.

Theorem 3.5

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.

  1. Let {A1,A2,,An} be a finite collection of open subsets of X.

  2. Let A=i=1nAi.

  3. Let xA. Then xAi1in.

  4. Thus, for each 1in, there exists an open ball B(x,ri)Ai.

  5. Let r=min{r1,r2,,rn}. Since ri>0, hence r>0.

  6. Thus, we can construct an open ball B(x,r) at x.

  7. Note that B(x,r)B(x,ri)Ai.

  8. In other words, B(x,r)Ai1in.

  9. Thus, B(x,r)A.

  10. Thus, for every xA, there exists an open ball at x entirely contained in A.

  11. Thus, A is open.

Remark 3.1

The empty set and X are both open.

is vacuously open since it contains no elements. Consequently, the requirement that an open ball surrounding every element of be contained within is vacuously true.

X is open since for every xX, any open ball B(x,r) is entirely contained within X by definition.

Theorem 3.6 (Metric topology)

Let (X,d) be a metric space. Then, the family of open sets determined by the metric d satisfies all the requirements of a topology. This topology is known as the metric topology determined by the metric d on the set X.

Proof. By Remark 3.1, and X are open.

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 d is closed under arbitrary union and finite intersection.

Hence, it is a topology.

3.2.4. Closed Sets#

Definition 3.11 (Closed sets)

A subset S of X is said to be closed in X if XS is open in X.

Theorem 3.7 (Trivially closed and open sets)

and X are both open and closed subsets of X.

Proof. is vacuously open since it has no points thus every point in has an open ball that lies within . Thus, X is closed.

X is open since for every point xX, every open ball at x lies entirely in X (by definition). Thus, is closed.

Theorem 3.8 (Singletons are closed)

Every singleton is a closed set.

Proof. Let xX and consider the singleton set A={x}.

Let B=XA. We show that B is open. Then A is closed.

  1. Let yB

  2. Let r=d(x,y).

  3. Then, the open ball B(y,r) doesn’t contain x.

  4. Thus, B(y,r)B.

  5. Thus, B is open.

Example 3.8

  • (0,1) is open in R.

  • [0,1] is closed in R.

  • (0,1] is neither open nor closed in R.

Proposition 3.2

The set of natural numbers N is closed in R.

Proof. We prove it by showing that its complement is open. The set RN can be written as a union of open intervals:

RN={(,1),(1,2),(2,3),,(n,n+1),}.
  1. Each open interval is an open set.

  2. The arbitrary union of open sets is open.

  3. Thus, RN is open.

  4. Thus, N is closed.

Theorem 3.9

Arbitrary intersections of closed sets are closed sets.

Proof. We proceed as follows:

  1. Let {Ai}iI be a family of closed subsets of X.

  2. Then {XAi}iI is a family of open subsets of X.

  3. Then, their union iI(XAi) is an open set.

  4. Hence, its complement XiI(XAi) is a closed set.

  5. But, by De Morgan’s law: XiI(XAi)=iIAi.

  6. Thus, iIAi is a closed set.

Theorem 3.10

Finite unions of closed sets are closed sets.

Proof. We proceed as follows:

  1. Let {A1,A2,,An} be a finite collection of closed subsets of X.

  2. Then, {XA1,XA2,,XAn} are open.

  3. Then, their union i=1nXAi is open.

  4. Then, its complement Xi=1nXAi is closed.

  5. By De Morgan’s law: Xi=1nXAi=i=1nAi.

  6. Thus, i=1nAi is a closed set.

3.2.5. Interior#

Definition 3.12 (Interior point)

Let (X,d) be a metric space. A point x is called an interior point of a set AX if there exists an open ball B(x,r) such that B(x,r)A.

By definition an interior point of a set belongs to the set too. xB(x,r)A.

Remark 3.2 (Interior point in terms of closed balls)

Let (X,d) be a metric space. Let AX.

Then, xA is an interior point of A if and only if there exists a closed ball B[x,r]A.

This result shows that interior points of a set A can be identified either through open balls or closed balls around them totally contained in the set A.

We emphasize that this interpretation of interior points is not applicable for all topological spaces but is perfectly valid for metric spaces.

Proof. Let x be an interior point of A.

  1. Then, there exists an open ball B(x,s)A.

  2. By Theorem 3.2, there is a closed ball B[x,r]B(x,s) with some r<s.

  3. Thus, xB[x,r]B(x,s)A.

For the converse, let there be a closed ball B[x,r]A.

  1. Then, B(x,r)B[x,r]A.

  2. Thus, xB(x,r)A.

  3. Thus, x is an interior point of A.

Definition 3.13 (Interior)

Let AX. The largest open set in X that is contained in A is called the interior of A (relative to X) and is denoted by intA or intXA.

Note that intAA.

Proposition 3.3

Let OA be an open set. Then OintA.

In words, every open set that is contained in A, is contained in its interior.

Proof. Since O and intA are open sets, hence their union OintA is open. But OA and intAA implies OintAA. Also, intAOintA. But intA is the largest open subset of A. Thus, intA=OintA. Thus, OintA.

Theorem 3.11 (Interior is set of interior points)

The interior of a set A is the collection of all the interior points of A.

Proof. Let I be the collection of all interior points of A. Thus, IA. Our claim is that I=intA.

We first show that IintA. Let xI. Then, there exists an open ball B(x,rx)A. Then, the union C=xIB(x,rx)A. But C is an arbitrary union of open sets. Hence C is open. But every open subset of A is a subset of intA. Thus, CintA. Also ICIintA.

For the converse, we proceed as follows.

intA is open. Thus, for every xintI, there exists an open ball B(x,rx)IA. Thus, x is an interior point of A. In other words, intAI.

This result is simply a different characterization of the interior of A. We could have started with the definition that the interior is the set of all interior points and then proved that it is the largest open set contained in A.

Proposition 3.4

A is open if and only if A=intA.

Proof. Assume, A is open. Then, A is the largest open set contained in A. Hence, A=intA.

Assume A=intA. Then, since intA is open, hence A is open.

Theorem 3.12

If AB then intAintB.

Proof. We have:

intAAB.

But intA is open. Hence, by Proposition 3.3:

intAintB.

Alternate proof.

  1. Let x be an interior point of A.

  2. Then x is an interior point of B too since AB.

  3. Thus, xintB.

  4. Thus intAintB.

3.2.6. Closure#

Definition 3.14 (Closure point)

A point xX is called a closure point of a subset A of X if every open ball at x contains (at least) one point in A.

In other words:

B(x,r)Ar>0.

Proposition 3.5

Every point in A is a closure point of A.

Proof. Let xA. Then xB(x,r) for every r>0. Thus, xB(x,r)A for every r>0. Thus, B(x,r)A is not empty for every r>0. Thus, x is a closure point of A.

Definition 3.15 (Closure)

Let AX. The smallest closed set in X that contains A is called the closure of A (relative to X) and is denoted by clA or clXA.

Proposition 3.6

Let C be a closed subset of X such that AC. Then clAC.

In words, every closed set that contains A, contains its closure.

Proof. Consider the set D=CclA. Since, intersection of two closed sets is closed, hence D is closed. Since AC and AclA, hence AD. At the same time, DclA. Thus, we have ADclA. But clA is the smallest closed set containing A. Hence, D=clA must be true. But DC. Hence clAC.

Theorem 3.13 (Closure is set of closure points)

The closure of a set A is the collection of all the closure points of A.

Proof. Let D be the set of all closure points of A.

We first show that D is a closed set.

  1. Let xD (i.e. xXD).

  2. Then, there exists an open ball B(x,r) such that B(x,r)A=.

  3. For any yB(x,r), then, there exists an open ball B(y,s) such that B(y,s)B(x,r) (since B(x,r) is an open set).

  4. Thus, B(y,s)A=.

  5. Hence, yD. In other words, the open ball B(x,r) doesn’t contain any closure point of A.

  6. Thus, B(x,r)XD.

  7. Thus, for every point xXD, there exists an open ball B(x,r)XD.

  8. Thus, XD is open.

  9. Consequently, D is closed.

Next, we show that clAD.

  1. Since every point of A is a closure point, hence AD.

  2. Since clA is a subset of any closed set containing A, hence clAD.

Last, we show that DclA. In fact, we show a stronger result that DC for any closed CA.

  1. Let C be a closed set such that AC.

  2. Then, XC is open.

  3. For any xXC, there exists an open ball B(x,r)XC.

  4. Thus, B(x,r)C=.

  5. In particular, B(x,r)A= (since AC).

  6. Thus, x is not a closure point of A (i.e. xD).

  7. Thus, no point in XC is a closure point.

  8. Thus, every closure point belongs to C.

  9. Thus, DC.

  10. We have established that D is a subset of any closed set that contains A.

  11. In particular, DclA since clA is closed and AclA (by definition).

Together, D=clA.

This result is simply a different characterization of the closure of A. We could have started with the definition that the closure is the set of all closure points and then proved that it is the smallest closed set containing A.

Proposition 3.7

A is closed if and only if A=clA.

Proof. Assume, A is closed. Then, A is the smallest closed set containing A. Hence, A=clA.

Assume A=clA. Then, since clA is closed, hence A is closed.

Theorem 3.14

A closed ball is a closed set.

Proof. Let xX. Let r>0. Let

C=B[x,r]={yX|d(x,y)r}

be a closed ball. We proceed by showing that its complement is open:

  1. Let yXC. Then d(x,y)>r.

  2. Let r1=d(x,y)r>0.

  3. Consider the open ball B(y,r1).

  4. For any zB(y,r1):

    d(x,y)d(x,z)+d(z,y)<d(x,z)+r1d(x,z)>d(x,y)r1=r.
  5. Thus, zC, i.e. zXC.

  6. Thus, B(y,r1)XC.

  7. In other words, for every point in XC, there exists an open ball contained in XC.

  8. Thus, XC is an open set.

  9. Thus, C is closed.

Theorem 3.15

Let AX. Then

X(intA)=cl(XA).

In words, complement of interior is the closure of the complement.

Proof. We proceed as follows:

  1. Let xX(intA), i.e., x is not an interior point of A.

  2. Then, for every r>0, B(x,r) is not a subset of A.

  3. Thus, for every r>0, B(x,r)XA is not empty.

  4. Thus, x is a closure point of XA.

  5. Thus, X(intA)cl(XA).

The same logic for the converse:

  1. Let xcl(XA), i.e., x is a closure point of XA.

  2. Thus, for every r>0, B(x,r)(XA) is not empty.

  3. Thus, for every r>0, B(x,r) is not a subset of A.

  4. Thus, x is not an interior point of A.

  5. Thus, cl(XA)X(intA).

Together, the equality is established.

Theorem 3.16

If AB then clAclB.

Proof. We have: AclA and ABclB. But then by Proposition 3.6 every closed set that contains A contains its closure. Thus:

clAclB.

Alternate proof.

  1. Let x be a closure point of A.

  2. Then x is a closure point of B too since AB.

  3. Thus, xclB.

  4. Thus clAclB.

Theorem 3.17

Closure of union contains union of closures.

Proof. Let A be a family of subsets of X. Let

K=AAA.

Let

L=AAclA.
  1. Let xL.

  2. Then xclA for some AA.

  3. Then x is a closure point of some AA.

  4. Then x is a closure point of K.

  5. Thus, xclK.

We arrive at the result:

AAclAcl(AAA).

Theorem 3.18

Closure of a finite union is equal to the union of closures.

i=1n(clAi)=cl(i=1nAi).

Proof. Let {A1,A2,,An} be a finite family of subsets of X.

Let

K=i=1nAi.

Let

L=i=1n(clAi).

From the previous result, we have established that

LclK.

We now seek to prove that clKL.

  1. clAi are closed.

  2. L is a finite union of closed sets.

  3. Hence, L is closed due to Theorem 3.10.

  4. AiclAi.

  5. Thus, AiclAi.

  6. Thus, KL.

  7. Thus, clKL due to Proposition 3.6.

3.2.7. Boundary#

Definition 3.16 (Boundary point)

A point xX is called a boundary point of A if every open ball B(x,r) at x contains points from A as well as XA.

In other words, B(x,r)A and B(x,r)XA for every r>0.

Definition 3.17 (Boundary)

The boundary of a set AX (relative to X), denoted by bdA or bdXA is defined as the set of all boundary points of A.

Theorem 3.19

bdA=clAintA.

Proof. Let x be a boundary point of A.

  1. B(x,r)A for every r>0.

  2. Thus x is a closure point of A.

  3. B(x,r)XA for every r>0.

  4. Thus, there is no r>0 such that B(x,r)A.

  5. Thus, x is not an interior point of A.

  6. Thus, xclAintA.

  7. Thus, bdAclAintA.

Let xclAintA.

  1. x is a closure point.

  2. Thus, B(x,r)A for every r>0.

  3. x is not an interior point.

  4. Thus, there is no r>0 such that B(x,r)A.

  5. Thus, for every r>0, B(x,r)XA.

  6. Combining, x is a boundary point.

  7. Thus, clAintAbdA.

Proposition 3.8

Boundary of a set is closed.

Proof. By definition:

bdA=clAintA=clA(XintA).
  1. intA is open. Hence XintA is closed.

  2. clA is closed.

  3. Thus, bdA is an intersection of two closed sets.

  4. Thus, bdA is closed (Theorem 3.9).

3.2.8. Frontier#

Definition 3.18 (Frontier point)

A frontier point of a set A is a boundary point that belongs to A.

Definition 3.19 (Frontier)

The set of all frontier points of a set A, denoted by frA, is called its frontier.

Note

Some authors treat boundary points as frontier points. They treat boundary and frontier as synonymous. There is no consistent terminology for the set AintA. Our definitions distinguish between boundary points and frontier points.

Proposition 3.9

frA=AintA=bdAA.

Proof. frA=bdAA is by definition as a frontier point is a boundary point which belongs to A.

If xA is an interior point then it’s not a boundary point. All other points in A are boundary points. Hence,

frA=AintA.

Proposition 3.10

For a closed set, the frontier and boundary are same.

Proof. Every boundary point belongs to the closed set.

Proposition 3.11

The frontier of a closed set is closed.

Proof. For a closed set frA=bdA. The boundary of any set is closed.

3.2.9. Accumulation#

Definition 3.20 (Deleted neighborhood)

Let xX. The deleted neighborhood of radius r around x is defined as the open ball of radius r around x excluding x itself.

Bd(x,r)=B(x,r)x.

Definition 3.21 (Accumulation point)

A point xX is called an accumulation point of a set AX, if every open ball B(x,r) contains a point in A distinct from x.

B(x,r)A{x}r>0.

In other words, every deleted neighborhood of x contains a point from A.

Bd(x,r)Ar>0.

Note that an accumulation point need not belong to the set A.

Remark 3.3

Every accumulation point is a closure point.

Although, every closure point need not be an accumulation point.

Definition 3.22 (Derived set)

The set of accumulation points of a set A is called its derived set and is denoted by A.

Definition 3.23 (Isolated point)

A point xA is called isolated if there is an open ball B(x,r) which doesn’t contain any other point of A.

In other words, there exists an r>0 such that:

B(x,r)A{x}=.

Proposition 3.12

A closure point is either an accumulation point or an isolated point.

Proof. Let xclA. Assume that x is not an accumulation point.

We need to show that xA and x is isolated.

  1. Since x is not an accumulation point, there exists r>0 such that B(x,r)A{x}=.

  2. Since x is a closure point, hence B(x,r)A is not empty.

  3. Then, B(x,r)A must be {x}.

  4. Thus, xA.

  5. Finally, since B(x,r)A{x}=, x is an isolated point of A.

Proposition 3.13

clA=AA.

This is a restatement of the previous result.

Theorem 3.20

A set is closed if and only if it contains all its accumulation points.

Proof. AAAA=A. But AA=clA. Thus, AAA=clA.

A=clAA=AAAA.

Recall that d(x,A) denotes the distance between a set A and a point x.

Theorem 3.21

If x is an accumulation point of A then d(x,A)=0.

Proof. Let x be an accumulation point of A.

  1. If xA then d(x,A)=0.

  2. So consider the case where xA.

  3. For every r>0, there exists aA such that d(x,a)<r.

  4. Thus, d(x,A)<r for every r>0.

  5. At the same time, d(x,A)0.

  6. Thus, d(x,A)=0.

3.2.10. Interior II#

Theorem 3.22

Interior of a finite intersection is the intersection of interiors.

inti=1nAi=i=1n(intAi).

Proof. Define:

A=i=1nAi.

From Theorem 3.15:

intA=X(cl(XA))=X(cl(X(i=1nAi)))=X(cl(i=1n(XAi)))=X(i=1n(cl(XAi)))=X(i=1n(X(intAi)))=X(X(i=1n(intAi)))=i=1n(intAi).

In this derivation, we have made use of the facts that:

  1. complement of interior equals closure of complement (Theorem 3.15).

  2. closure of a finite union is a union of the closures (Theorem 3.18).

3.2.11. Dense Sets#

Definition 3.24 (Dense subsets)

A subset A of X is called dense in X if clA=X.

Theorem 3.23

A set A is dense if and only if OA holds for every nonempty open set O in X.

Proof. Assume A is dense and O is open and non-empty.

  1. Let xO.

  2. There exists r>0 such that B(x,r)O.

  3. Since xclA, B(x,r)A.

  4. Thus, OA is not empty.

Assume AO for every open and nonempty O.

  1. Let xX. If xA then x is a closure point of A. So assume xXA.

  2. Let r>0 and consider the open ball B(x,r).

  3. Since B(x,r) is nonempty and open, hence B(x,r)A.

  4. Thus, x is a closure point of A.

  5. Thus, A is dense in X.

Theorem 3.24

Complement of a dense set has an empty interior.

Proof. Recall from Theorem 3.15 that:

X(intA)=cl(XA).

Now, let B be dense and A=XB. Then,

X(intA)=cl(XA)X(intA)=clBX(intA)=XintA=XX=.

Thus, A 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#

Definition 3.25 (Equivalent metrics)

Let da:X×XR and db:X×XR be two different metrics on X. Then, the metrics are said to be equivalent if they determine the same topology on X. In other words, the family of open sets in (X,da) is identical to the family of open sets in (X,db); i.e., O is an open set in (X,da) if and only if O is an open set in (X,db).

Theorem 3.25 (Balls inside balls)

Let da:X×XR and db:X×XR be two different metrics on X which are equivalent. Let Ba(x,r) and Bb(x,r) denote the open balls at xX of radius r in metric spaces (X,da) and (X,db) respectively.

Then, for every xX and for every r>0, there exists r>0 such that

Ba(x,r)Bb(x,r).

Similarly, for every xX and for every r>0, there exists r>0 such that

Bb(x,r)Ba(x,r).

Proof. Let xX and for every r>0.

  1. Consider the open ball O=Bb(x,r).

  2. Then, O is an open set of (X,db).

  3. Since both metrics determine same topology, hence O is also an open set of (X,da).

  4. Since xO, hence x is an interior point of O in (X,da).

  5. Thus, there exists r>0 such that

    Ba(x,r)O=Bb(x,r).

We proceed similarly for the other way round.

  1. Consider the open ball O=Ba(x,r).

  2. Then, O is an open set of (X,da).

  3. Since both metrics determine same topology, hence O is also an open set of (X,db).

  4. Since xO, hence x is an interior point of O in (X,db).

  5. Thus, there exists r>0 such that

    Bb(x,r)O=Ba(x,r).

We can actually prove a stronger result.

Theorem 3.26 (Metric equivalence characterization as balls inside balls)

Let da:X×XR and db:X×XR be two different metrics on X. Let Ba(x,r) and Bb(x,r) denote the open balls at xX of radius r in metric spaces (X,da) and (X,db) respectively.

The two metrics da and db are equivalent if and only if

for every xX and for every r>0, there exists r>0 such that Ba(x,r)Bb(x,r)

and

for every xX and for every r>0, there exists r>0 such that Bb(x,r)Ba(x,r).

Proof. One part of this result was proved in Theorem 3.25.

We now assume that for every xX and for every r>0, there exists r>0 such that Ba(x,r)Bb(x,r) and for every xX and for every r>0, there exists r>0 such that Bb(x,r)Ba(x,r).

  1. Let O be an open set of (X,da).

  2. Let xO.

  3. Then, x is an interior point of O in (X,da).

  4. There exists an open ball Ba(x,r)O.

  5. Then, there exists an open ball Bb(x,r)Ba(x,r)O.

  6. Thus, x is an interior point of O in (X,db) too.

  7. Thus, O is an open set in (X,db) too.

  8. Similarly, we can show that if O is an open set in (X,db) then it is an open set in (X,da) too.

Thus, both metric spaces determine same topology. Hence, they are equivalent.

Theorem 3.27 (Equivalent metrics as equivalence relation)

Let X be an arbitrary set. Consider the set of metrics on X denoted as D:

D{d:X×XR|d is a metric }.

Let da,db by any two metrics in D. Let dadb if the two metrics are equivalent. Then, is an equivalence relation on the set of metrics D.

Proof. [Reflexivity]

  1. Let d be an arbitrary metric on X.

  2. Then, dd since it determines same topology.

[Symmetry]

  1. Let da and db be two metrics on X.

  2. If they are equivalent, then they introduce same topology.

  3. Thus, dadb implies dbda.

[Transitivity]

  1. Let dadb and dbdc.

  2. Then, all three da,db,dc determine the same topology on X.

  3. Thus, dadc holds too.

Definition 3.26 (Strongly equivalent metrics)

Let da:X×XR and db:X×XR be two different metrics on X. Then, the metrics are said to be strongly equivalent if there exist constants M,M>0 such that

da(x,y)Mdb(x,y) and db(x,y)Mda(x,y)x,yX.

Theorem 3.28

Two strongly equivalent metrics are equivalent.

Proof. Let da:X×XR and db:X×XR be two metrics on X which are equivalent.

  1. Let O be an open set of (X,da).

  2. Let xO be an interior point.

  3. Then, there exists an open ball Ba(x,r)O.

  4. Consider the open ball Bb(x,rM).

  5. For any point yBb(x,rM)

    db(x,y)<rMMdb(x,y)<rda(x,y)Mdb(x,y)<r.
  6. Thus, yBa(x,r).

  7. Thus, Bb(x,rM)Ba(x,r)O.

  8. Thus, x is an interior point of O in (X,db).

  9. Thus, O is an open set in (X,db).

  10. Similarly, if O is an open set in (X,db) then it is an open set in (X,da).

  11. Thus, both metric spaces determine the same topology of open sets.

  12. Thus, they are equivalent.

3.2.13. Connectedness#

Definition 3.27 (Connectedness)

A metric space (X,d) is called connected if it cannot be expressed as a union of two non-empty disjoint open sets.

Example 3.9 (Connected vs non-connected spaces)

  1. R is connected.

  2. (0,1)(1,2) is not connected.

Definition 3.28 (Connected subsets)

Let (X,d) be a metric space. A subset AX is called connected if the metric subspace (A,d) is connected.

3.2.14. Counter Examples#

This subsection is a collection of some examples which illustrate some salient aspects of topology.

Example 3.10 (Empty Interior)

Consider the set R with the usual metric. The subset Q of rational numbers has an empty interior. But clQ=R. Thus, the closure of Q doesn’t have an empty interior.