1 Set theory notation 1

1 Set theory notation

The (Cartesian) product of two sets X and Y is defined by

X × Y := {(x, y) : x ∈ X, y ∈ Y }

where (x, y) denotes the pair formed by its first and second coordinates. Arule of assignment from X to Y is a subset r ⊂ X × Y such that for eachx ∈ X there exists a unique y ∈ Y such that (x, y) ∈ r. A function isa triple formed by a nonempty set X, called the domain, a set Y , calledthe codomain, and a rule of assignment from X to Y . The notation f :X → Y refers to the triple (X, Y, r) for some definite, but unspecified rule ofassignment r.1

The collection of all subsets of a set X is called the power set of X,denoted by P(X). A subset of P(X) is a collection A of subsets of X. Toindex the elements of a nonempty collection A is to specify a surjective map,called the indexing function, from some set J , called the index set, ontothe collection A. Formally speaking, an indexed family of subsets ofX may be considered a pair (A, ι) formed by a subset A ⊂ P(X) and afunction ι : J → P(X) whose image is A. Usually, the indexing function isnot mentioned explicitly, as in the notation

{Aα}α∈J .

The union and intersection of the elements of an indexed family are de-noted respectively by ⋃

α∈J

Aα and⋂α∈J

Aα.

Sometimes, the index set may also be suppressed, e.g.⋃Aα and

⋂Aα.

Exercise 1.1. Show that A ∩B = ∅ iff A ⊂ Bc iff B ⊂ Ac.

Exercise 1.2. Show that A ⊂ B ∪ C and A ∩ C = ∅ implies A ⊂ B.

Exercise 1.3. Let f : X → Y , A ⊂ X and B ⊂ Y . Show that f(A) ⊂ B iffA ⊂ f−1(B).

1 Injectivity makes sense as a property of a rule of assignment, but surjectivity does notbecause the codomain cannot be inferred from the collection of pairs. The grammaticaldistinction between a function and its rule of assignment is justified by the duality that ismade possible between the two notions.

2 Boundary of a subset of Rn 2

Exercise 1.4. Show that f(A ∪B) = f(A) ∪ f(B).

Exercise 1.5. Show that f(A∩B) ⊂ f(A)∩ f(B) and that equality can fail.

Exercise 1.6. Show that f−1(⋃α∈J Uα) =

⋃α∈J f

−1(Uα).

Exercise 1.7. Show that f−1(⋂α∈J Uα) =

⋂α∈J f

−1(Uα).

Exercise 1.8. Show that (g ◦ f)−1 = f−1 ◦ g−1.

2 Boundary of a subset of Rn

The distance between two points in Rn is given by the Euclidean metric

d(x,y) = |x− y| =√

(x− y) · (x− y) =

(n∑i=1

(xi − yi)2)1/2

.

The open (Euclidean) ball centered at x of radius r > 0 is denoted by

B(x, r) = {y ∈ Rn : d(x, y) < r}.

The complement of a subset A ⊂ Rn is denoted by

Ac := Rn \ A.

Definition 1. Let A ⊂ Rn and x ∈ Rn. We say x is an interior point ofA if there exists a δ > 0 such that B(x, δ) ⊂ A. We say x is an exteriorpoint of A if it is an interior point of Ac. It is a boundary point if it isneither an interior point nor an exterior point; that is,

BdA := (IntA ∪ ExtA)c .

Equivalently, x is a boundary point of A if for every δ > 0, the ball B(x, δ)contains points of A as well as points of Ac:

B(x, δ) ∩ A 6= ∅ and B(x, δ) ∩ Ac 6= ∅.

Lemma 1. IntA ⊂ A and A ∩ ExtA = ∅.

Proof. For any x ∈ IntA there is a ball B containing x such that B ⊂ A.Hence x ∈ A, establishing the first assertion. For the second assertion,ExtA = IntAc ⊂ Ac, which means ExtA is disjoint from A.

2 Boundary of a subset of Rn 3

Consequently, Rn admits a partition into disjoint sets:

Rn = IntA ∪ BdA ∪ ExtA.

Definition 2. Let A ⊂ Rn. We say A is open if it consists entirely ofinterior points; in other words,

A ⊂ IntA.

It is closed if it contains all its boundary points, in other words,

BdA ⊂ A.

Theorem 1. A is open if and only if A is disjoint from BdA.

Proof. If A is open, then A ⊂ IntA. Since IntA is disjoint from BdA, itfollows that A is disjoint for BdA. For the converse, note that A∩ExtA = ∅because ExtA ⊂ Ac. Hence, if A ∩ BdA = ∅, then A ⊂ IntA.

Theorem 2. A is open if and only if Ac is closed.

Proof. Since BdAc = BdA, the statement that Ac is closed is equivalent toBdA ⊂ Ac; in other words, BdA is disjoint from A. Since A is disjoint fromExtA, this in turn is equivalent to A ⊂ IntA.

Note that the sets Rn and ∅ are both open as well as closed.

Theorem 3. The union of any collection of open sets is open.

Proof. Let {Aα} be a collection of open subsets. Let x be a point in ∪Aα.Then x ∈ Aα0 for some fixed α0. Since Aα0 is open, x is an interior point.Hence, there is some δ > 0 such that B(x, δ) ⊂ Aα0 ⊂ ∪Aα. It follows thatx is an interior point of ∪Aα. And since x ∈ ∪Aα was arbitrary, it followsthat ∪Aα is open.

Definition 3. The closure of a subset A ⊂ Rn is defined by

C`A := A ∪ BdA = IntA ∪ BdA

where the second equality holds by the second assertion of Lemma 1.

In the next section, we shall show that an open Euclidean ball is indeedopen as a subset of Rn. Using this fact, we obtain

2 Boundary of a subset of Rn 4

Theorem 4. For any A ⊂ Rn, IntA is open and C`A is closed.

Proof. Given x ∈ IntA, there is an open ball B containing x such thatB ⊂ A. By Theorem 6, B is open as a subset of Rn. Hence, for any y ∈ Bthere is a δ > 0 such that B(y, δ) ⊂ B ⊂ A. Thus B ⊂ IntA, which showsthat x is an interior point of IntA. Since x is an arbitrary point in IntA,it follows that IntA is open. Applying this result to the set Ac we concludethat ExtA = IntAc is open. Hence, C`A = (ExtA)c is closed.

Theorem 5. For any A ⊂ Rn, IntA is the largest open set contained in Aand C`A is the smallest closed set containing A.

Proof. If U is an open set contained in A, then it consists entirely of interiorpoints. Hence, U ⊂ IntA. By Theorem 4 IntA is open and by Lemma 1IntA ⊂ A. Hence, IntA is the largest among all open sets contained in A.

It is clear from its definition that C`A contains A. Theorem 4 impliesthat it is closed. To show that C`A is the smallest among all closed setscontaining A, it remains to prove the following.

Claim: If F is a closed set that contains A then C`A ⊂ F . Indeed, sinceF c is open, for any x ∈ F c there is a ball B about x such that B ⊂ F c ⊂ Ac.Thus B is disjoint from A so that x ∈ ExtA. It follows that F c ⊂ ExtA, orequivalently, C`A = (ExtA)c ⊂ F , proving the claim.

Exercise 2.1. What is the set BdB(x, r) ? (See Exercise 3.4 for answer.)

Exercise 2.2. Show that Bd I consists of two points for any interval I byfirst making a suitable definition of the concept of an interval.

Exercise 2.3. Determine the boundary of the rectangle [a, b]× [c, d].

Exercise 2.4. For each condition below, give an example of a nonemptysubset A ⊂ R2 satisfying it.

1. A ∩ BdA = ∅ (disjoint)

2. BdA = A

3. BdA ( A (proper containment)

4. A ( BdA

5. A ∩ BdA 6= ∅ and A 6⊂ BdA and BdA 6⊂ A (nontrivial intersection)

3 Metric properties of Euclidean space 5

Exercise 2.5. Show that IntAc = ExtA and BdAc = BdA.

Exercise 2.6. Show that A is open if and only if A = IntA.

Exercise 2.7. Show that A is closed if and only if C`A = A.

Exercise 2.8. A set is bounded if it is contained in some Euclidean ball. Abox in Rn is a product of intervals. Show that a set is bounded if and onlyif it is contained in some box.

Exercise 2.9. Show that a bounded subset of R2 has finite area. Give anexample of an unbounded set with infinite area.

Exercise 2.10. Show that BdA is a closed set such that Int BdA = ∅.( Remark: It can nevertheless be uncountable and even have positive volume!)

Exercise 2.11. Show that Int IntA = IntA and C`C`A = C`A. Does theidentity Bd BdA = BdA necessarily hold?

Exercise 2.12. (*) If A is both open and closed, is it necessarily Rn or ∅?

3 Metric properties of Euclidean space

The Euclidean metric satisfies the triangle inequality (see Exercise 3.7)

d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z ∈ Rn.

Theorem 6. For any x ∈ Rn and any r > 0, the ball B(x, r) is open.

Proof. Given y ∈ B(x, r), let δ = r − d(x, y). Then for any z ∈ B(y, δ)

d(x, z) ≤ d(x, y) + d(y, z) < d(x, y) + δ = r

so that z ∈ B(x, r). Hence, B(y, δ) ⊂ B(x, r). It follows that y is an interiorpoint of B(x, r).

Lemma 2. The intersection of any two open balls is open.

Proof. We shall show that given any two balls B(x, r) and B(y, s) with apoint z in common there exists a δ > 0 such that B(z, δ) ⊂ B(x, r)∩B(y, s).

Suppose 0 < δ ≤ r − d(x, z) and w ∈ B(z, δ). Then

d(w, x) ≤ d(w, z) + d(z, x) < δ + d(z, x) ≤ r.

Hence, w ∈ B(x, r) so that B(z, δ) ⊂ B(x, r) for all δ ∈ (0, r − d(x, z)].Similarly, B(z, δ) ⊂ B(y, s) for all δ ∈ (0, s−d(z, y)]. It follows that B(z, δ) ⊂B(x, r) ∩B(y, s) for all 0 < δ ≤ min(r − d(z, x), s− d(z, y)).

3 Metric properties of Euclidean space 6

Theorem 7. The intersection of a finite collection of open sets is open.

Proof. First, we consider the intersection of two open sets, say U and V .Suppose z ∈ U ∩ V . Since U is open, z is an interior point of U . Hence,there exists an open ball B1 centered at z such that B1 ⊂ U . Similarly, sinceV is open, there exists an open ball B2 centered at z such that B2 ⊂ V . ByLemma 2, there exists a δ > 0 such that B(z, δ) ⊂ B1 ∩B2 ⊂ U ∩ V . Hence,z is an interior point of U ∩ V and since z ∈ U ∩ V was arbitrary, it followsthat U ∩ V is open. The general case follows by induction since

U1 ∩ · · · ∩ Un = (U1 ∩ . . . Un−1) ∩ Un

for any n > 2. 2

Since {x} =⋂r>0B(x, r), the next theorem shows that the intersection

of an infinite collection of open sets need not be open.

Theorem 8. For any x ∈ Rn, the singleton {x} is closed and not open.

Proof. For any y 6= x, note that x 6∈ B(y, δ) if we set δ = d(x, y). Hence,B(y, δ) ⊂ {x}c so that y is an interior point of {x}c. Since y is an arbitrarypoint in {x}c, this shows that {x}c is open. Hence, {x} is closed. Since thecontainment B(x, δ) ⊂ {x} never holds, x is not an interior point of {x}.Hence, {x} is not open.

Let f : X → R be a function defined on some subset X ⊂ Rn. Recallthat f is continuous if it is continuous at each point in its domain. Pointwisecontinuity is given by the infamous (ε, δ)-definition: f is continuous at a pointx ∈ X if for every ε > 0 there exists a δ > 0 such that for all y ∈ X

|x− y| < δ implies |f(x)− f(y)| < ε. (1)

The implication (1) may be rewritten in terms of metric balls as

f(BX(x, δ)) ⊂ B(f(x), ε) where BX(x, δ) := B(x, δ) ∩X. (2)

Pointwise continuity is most naturally formulated in the context of a mapbetween metric spaces.

2 Best practice is to avoid duplicate variable names, unless there is no risk of confusion.

3 Metric properties of Euclidean space 7

Definition 4. Let f : X → Y be a map between two metric spaces andx ∈ X. Then f is continuous at x if for every ε > 0 there exists a δ > 0such that

f(BX(x, δ)) ⊂ BY (f(x), ε).

The map f is continuous if it is continuous at each point in its domain.

A more general definition in terms of open sets will be given later.

Exercises

Exercise 3.1. Show that d(x, y) ≥ |d(x, z)− d(y, z)| for all x, y, z ∈ Rn.(The distance between two points is at least the absolute difference betweentheir distances to a third point.)

Exercise 3.2. Verify directly from definitions that (1, 1), (2, 2) and (√

2,√

2)are respectively interior, exterior and boundary points of the disk x2+y2 < 4.

Exercise 3.3. Prove that every closed ball in Rn is a closed set; that is, forevery x ∈ Rn and any r > 0, the set B(x, r) = {y ∈ Rn : d(x, y) ≤ r} isclosed. (Hint: Show that the complement is open.)

Exercise 3.4. Prove that BdB(x, r) = {y ∈ Rn : d(x, y) = r}.

Exercise 3.5. Prove that the set

A = {x ∈ Rn : r1 < |x| < r2}

is open for any 0 < r1 < r2.

Exercise 3.6. Prove that |u + v| ≤ |u|+ |v| for any two vectors u,v ∈ Rn

by deriving it from the Cauchy-Schwarz inequality u · v ≤ |u||v| where thedot product is given by u · v = u1v1 + · · ·+ unvn.

Exercise 3.7. Prove that the Euclidean metric satisfies the triangle inequal-ity by applying the result of Exercise 3.6 to the pair u = x−y and v = y−z.

Exercise 3.8. Show that B(0, r) is convex, i.e. for any x, y ∈ B(0, r)the line segment joining x to y is contained in B(0, r). (Hint: Use theparametrization (1−t)x+ty, 0 ≤ t ≤ 1, then apply the result of Exercise 3.6.)

4 Topological spaces 8

Exercise 3.9. Let ϕ : Rn → Rn be the map given by ϕ(x) = x+ a for somea ∈ Rn. Show that ϕ(A) is convex if and only if A is convex. Then use theresult of the previous exercise to show that B(x, r) is convex.

Exercise 3.10. Prove the Cauchy-Schwarz inequality by explicitly computingthe minimum value of the function

f(t) = |u + tv|2 = (u + tv) · (u + tv)

and observing that it is necessarily ≥ 0.

Exercise 3.11. Show that given any two points in Rn there is an open setabout one that misses the other. That is, for any x, y ∈ Rn there is aneighborhood U of x such that y 6∈ U .

Exercise 3.12. Show that given any two points in Rn there is an open setabout each that is disjoint from from another. That is, for any x, y ∈ Rn thereare neighborhoods U and V about x and y, respectively, such that U ∩V = ∅.

4 Topological spaces

Definition 5. A topology on a set X is a collection T of subsets of Xsatisfying

1. ∅ and X belong to T .

2. The union of an arbitrary collection of elements of T belongs to T .

3. The intersection of a finite collection of elements of T belongs to T .

A topological space is a pair (X, T ) formed by a nonempty set X togetherwith a topology T on X. An open set in X refers to an element of T .

The pair (X, T ) is often referred to as the “space X” without explicitmention of the topology T , which must then be inferred from the context.For example, the space Rn refers to the topological space (Rn, T ) where T isthe standard topology on Rn, i.e. the collection of open sets in Rn. Thespace Rn is also known as Euclidean space and is the main ingredient inthe formulation of the all-important concept of a differentiable n-manifold,3 abasic tool used in the analysis involving a finite collection of real parameters.

3 A 1-manifold is a curve, a 2-manifold is a surface, and so on.

4 Topological spaces 9

Definition 6. If two topologies are defined on the same set X and one iscontained in the other as subsets of P(X) then the larger collection is said tobe finer than the smaller collection and the smaller coarser than the larger.

Definition 7. Let X be a topological space. A sequence (xn)n∈Z+ of pointsin X converges to a point x ∈ X if every neighborhood of x contains allpoints of the sequence from some point onwards, i.e. for any neighborhood Uof x, there exists an N such that n > N implies xn ∈ U . In this case, x iscalled the limit of the sequence.4

Example 4.1. On any set X the trivial topology is given by {∅, X} whilethe discrete topology is given by P(X). It is easy to see that the axioms fora topology are satisfied. They are respectively the coarsest and finest amongall topologies that can be put on a set X. Every sequence in X converges inthe trivial topology; in the discrete topology the only sequences that convergeare those that are eventually constant. The limit of a convergent sequence isalways unique in the discrete topology but not in the trivial topology if X hasmore than one point.

Example 4.2. The finite complement topology on a set X is given by

{F c : F is a finite set in X } ∪ {∅}.

It is closed under the operation of taking arbitrary unions because

⋃α∈J

F cα =

(⋂α∈J

Fα

)c

and any subset of a finite set is finite. It is closed under the operation oftaking finite intersections because

n⋂i=1

F ci =

(n⋃i=1

Fi

)c

and the union of a finite collection of finite sets is a finite set.

Definition 8. Let X be a topological space. A closed set in X refers to aset A ⊂ X whose complement Ac := X \ A is open. A clopen set is onethat is both open and closed as subsets of X.

4 The uniqueness of the limit is only guaranteed under an additional hypothesis. SeeExercise 4.6.

4 Topological spaces 10

By De Morgan’s laws, we have

Theorem 9. Let collection of closed sets in a topological space X satisfies

1. ∅ and X are closed.

2. The intersection of any collection of closed sets is closed.

3. The union of a finite collection of closed sets is closed.

Example 4.3. In the finite complement topology, every finite set is closedand these are the only closed sets apart from the whole space itself.

4.1 Subspaces

One rarely specifies a topology on a set explicitly as in the preceding exam-ples. Instead, there are many techniques for constructing new topologicalspaces out of old ones. The following is an example of such a technique.

Theorem 10. Given a space (Y, TY ) and a subset X ⊂ Y , the collection

TX := {A ⊂ X : A = U ∩X for some U ∈ TY }.

defines a topology on X, a.k.a. subspace topology on X inherited from Y .

Proof. The first condition follows from ∅ = ∅∩X and X = Y ∩X. For thesecond condition, consider a collection {Aα}α∈J of elements in TX . Then foreach α ∈ J there is an open set Uα in Y such that Aα = Uα ∩X. Hence

⋃α∈J

Aα =⋃α∈J

Uα ∩X =

(⋃α∈J

Uα

)∩X.

Since⋃α∈J Uα is an open set in Y , it follows that

⋃α∈J Aα belongs to TX .

Similarly, if {Ai}ni=1 is a collection of elements in TX , then for each i = 1, . . . , nthere is an open set Ui in Y such that Ai = Ui ∩X. Hence

n⋂i=1

Ai =n⋂i=1

Ui ∩X =

(n⋃i=1

Ui

)∩X

and since⋂ni=1 Ui is an open set in Y , it follows that

⋂ni=1Ai ∈ TX .

4 Topological spaces 11

In the context where A ⊂ X where X ⊂ Y the term relatively open maybe used to refer to A being open in X as opposed to being open as a subsetof Y . Similarly, A being relatively closed means its complement X \ A isrelatively open, as opposed to being a closed subset of Y ; equivalently, thereexists a closed set B in Y such that A = B ∩X.

Example 4.4. Let I = [0, 1) and X = (−1, 1). Then I is closed in Xbecause (−1, 0) is open in X and I = X \ (−1, 0). However, I is not open inR because (−δ, δ) 6⊂ I for any δ > 0 so that 0 is not an interior point of I.

Example 4.5. Let A = {(x, 0) : 0 < x < 1} be an interval on the x-axisX = {(x, 0) : x ∈ R}. Then A is open as a subset of the x-axis becauseA = B((1/2, 0), 1/2) ∩ X, but it is not open as a subset of R2, because itsinterior IntA = ∅ since I does not contain any open disks.

Example 4.6. The closed unit square [0, 1]2 = {(x, y) : 0 ≤ x ≤ 1, 0 ≤y ≤ 1} admits a partition into 9 open cells: a 2-cell (0, 1)2, four 1-cells ofwhich the set A in the previous example is one, and four 0-cells which arethe singletons determined by the corners. These cells are open relative to thesmallest affine subspace containing the cell. In the case of A in the previousexample, this subspace is X. In the case of the 1-cell on the right side of thesquare, the subspace is the line x = 1. In the case of a 0-cell, the subspace isthe singleton formed by the corresponding corner of the square.

Example 4.7. A subset of a space X is said to be discrete if the topologythat it inherits as a subspace is the discrete topology on the subset. Z is adiscrete subset of R, but Q is not, for singletons are relatively open in Z butnot in Q.

4.2 Interior, closure, boundary and exterior operators

Now we generalize the set operations introduced earlier for Euclidean space.

Definition 9. Let X be a topological space. The interior of a subset A ⊂ Xis the largest open subset contained in A; in other words, it is an open set

IntA ⊂ A

that is not contained in any other open set contained in A. It is well-definedbecause the union of all open sets contained in A is open and cannot beproperly contained in any other open subset inside A.

4 Topological spaces 12

Definition 10. The closure of a subset A ⊂ X is the smallest closed setcontaining A; in other words, it is a closed set

C`A ⊃ A

that does not properly contain any closed subset of X that contains A. It iswell-defined because the intersection of all closed sets containing A is closedand cannot properly contain any other closed set that contains A.

Definition 11. The boundary and exterior of a subset A ⊂ X are definedby

BdA = C`A \ IntA and ExtA = X \ C`A.

An interior point (resp. exterior point and boundary point) refers toan element of IntA (resp. ExtA and BdA). A neighborhood of a pointrefers to an open set containing the point.5

That these notions reduce to the familiar ones in the case of Euclideanspace is a straightforward exercise starting with the following observation.

Lemma 3. x ∈ IntA if and only if it has a neighborhood U such that U ⊂ A.

Proof. For necessity, note that IntA is open and contained in A, by definition.Hence, if x ∈ IntA, then U = IntA is a neighborhood of x that is containedin A. For sufficiency, suppose U is a neighborhood of x such that U ⊂ A.Then U ⊂ IntA, since IntA is the union of all open sets contained in A.Since x ∈ U , it follows that x ∈ IntA.

A subset of a space X is said to dense if its closure is all of X.

Lemma 4. A set is dense if and only if it meets every nonempty open set.

Proof. Suppose A is dense. If U is an open set that is disjoint from A, thenU c is a closed set containing A. Hence, U c ⊃ C`A = X, i.e. U is the emptyset. Hence, if U is nonempty, it must contain points of A. Conversely, if Ameets every nonempty open set, then the only closed set that contains A isX itself, for if F is any proper subset of X that is closed and contains A,then F c is a non empty open set that misses A. It follows that the closureof A is X, hence A is dense.

5 Some authors use the term in a broader sense to mean any set that contains x as aninterior point; for example, this can be inferred by the expression compact neighborhood.

4 Topological spaces 13

4.3 Topology generated by a basis

One can think of the topology of Rn as being determined by the collection ofopen balls by taking arbitrary unions of open balls. Abstracting the essentialproperties of the collection of open balls that guarantees this constructionleads to a topology on X leads us to following concept.

Definition 12. A collection B ⊂ P(X) is a basis for a topology on X if

(i) For each x ∈ X there exists a B ∈ B such that x ∈ B.

(ii) For each two elements B1, B2 ∈ B with a common point x ∈ B1 ∩ B2

there exists a B3 ∈ B such that x ∈ B3 and B3 ⊂ B1 ∩B2.

Remark. A collection B that satisfies (i) is said to be a covering of X.

Often a topology is determined by specifying a basis for it. We shall seelater how different bases can sometimes generate the same topology.

Theorem 11. If B is a basis for a topology on X, then the collection T ofall possible unions of collections of elements in B is a topology on X.

Proof. ∅ ∈ T because we can take the empty collection whereas X ∈ Tby (i). An arbitrary union of elements in T is a still a union of elementsof B, since each element is a union of elements of T . To show that theintersection of a finite collection of elements in B is in T , it is enough toconsider two elements B1, B2 ∈ B. By (ii), for each point x ∈ B1 ∩ B2 thereis an element Bx

3 ∈ B such that x ∈ B3 and B3 ⊂ B1 ∩ B2. It follows thatB1∩B2 =

⋃x∈B1∩B2

Bx3 is a union of elements in B. Hence, B1∩B2 ∈ T .

The topology T in Theorem 11 is said to be generated by the basis B.The following is a more convenient characterization of T in terms of B.

Lemma 5. Let B be a basis generating the topology of a space X. ThenA ⊂ X is open if and only if for any x ∈ A there exists a basis elementB ∈ B such that x ∈ B and B ⊂ A.

Proof. If A is open, then it is expressible as a union of elements in B. Hence,every x ∈ A is contained in some B ∈ B such that B ⊂ A. This provesnecessity. For sufficiency, we note that since basis elements are open, everypoint x ∈ A is an interior point, by Lemma 3. Hence, A is open.

4 Topological spaces 14

Often a notion defined in terms of open sets does not change meaningwhen we restrict to basis elements. For example, Lemma 4 says a set isdense if and only if it meets every nonempty open set.

Lemma 6. Let B be a basis generating the topology of a space X and A ⊂ X.Then A is dense if and only if every (nonempty) basis element intersects A.

Proof. By Lemma 4, A is dense if and only if it meets every nonempty openset. In particular, it meets every nonempty basis element. Conversely, ifA meets every nonempty basis element, then given any nonempty open setU , the previous lemma ensures the existence of a nonempty basis elementB ⊂ U . Since A meets B, it also meets U . It follows that A is dense.

A space is said to be separable if it contains a countable dense subset.

Example 4.8. Rn is separable because Qn is a countable dense subset.

Example 4.9. The set X of all functions from [0, 1] to the set {0, 1} is ametric space in which the distance between any two points is one. It inducesthe discrete topology on X, and provide an example of a metric space that isnot separable because the only subset that is dense is the whole space itself.

Manifolds are usually assumed to have a countable basis.

Theorem 12. Let X be a metric space. Then X has a countable basis if andonly if it has a countably dense subset.

Proof. If X has a countable basis B, let Y be the subset of X obtained bychoosing an element from each (nonempty) basis element in B. Then Y iscountable, and it is dense, by Lemma 6. Conversely, given a countably densesubset Y ⊂ X, it is readily seen (using Lemma 12 that the collection

{B(x, 1/n) : x ∈ Y, n ∈ Z+}

is a basis for the topology of X.

We remark that a basis for a topology can be restricted to any subset togive a basis for the subspace topology. That is, if B is a basis for a topologyon Y and X ⊂ Y , then

BX = {B ∩X : B ∈ B}

is a basis for the subspace topology on X, by virtue of the distributive law:⋃α∈J

(Bα ∩X) =

(⋃α∈J

Bα

)∩X

4 Topological spaces 15

4.4 Limit points (bonus material)

There is a simple characterization of points in the closure of a set.

Theorem 13. x ∈ C`A if and only if every neighborhood of x intersects A.

Proof. If x has a neighborhood U that is disjoint from A, then U c is closedand contains A. Hence, C`A ⊂ U c so that x 6∈ C`A. This proves (thecontrapositive of) necessity. For sufficiency, suppose every neighborhood ofx meets A and let F be a closed set containing A. If x 6∈ F then F c would aneighborhood of x that misses A, which does not exist by hypothesis. Hence,x ∈ F and since the closed set F is an arbitrary closed set containing A, itfollows that x ∈ C`A.

Definition 13. A deleted neighborhood of a point x refers to a set of theform U \ {x} where U is a neighborhood of x. A limit point of A is a pointsuch that every deleted neighborhood of x intersects A, or equivalently, everyneighborhood of x meets A in a point different from x.

Theorem 14. C`A = A ∪ A′ where A′ denotes the set of limit points of A.Hence, a set is closed if and only if it contains all its limit points.

Proof. If x is a limit point that is not in A, then every neighborhood of xmeets A as well as Ac. Hence x is a boundary point. Conversely, if x is aboundary point that is not in A, then every deleted neighborhood of x meetsA so that x is a limit point of A.

Remark. A limit point of a subset A ⊂ Rn can also be characterized as apoint x (not necessarily in A) for which there exists a sequence in A \ {x}that converges to x. For subsets of a general topological space, limit pointsare similarly characterized in terms of convergent nets.

Definition 14. An isolated point of A is a point x ∈ A that has a neigh-borhood U containing no other points of A, i.e. U ∩ A = {x}. A set isperfect if it is closed and has no isolated points; or equivalently, if A′ = A.

Exercises

Exercise 4.1. Show that the finite complement topology on an infinite set Xhas no clopen sets other than ∅ and X itself.

4 Topological spaces 16

Exercise 4.2. The T1-axiom for a space X is the condition that given anytwo points in X, each has a neighborhood that does not contain the other.Prove that T1-axiom is equivalent to the condition that singletons are closed.

Exercise 4.3. Show that finite complement topology is coarser than anytopology satisfying the T1-axiom.

Exercise 4.4. Show that any sequence of distinct points in X converges withrespect to the finite complement topology. The hypothesis means the sequence{xn}n∈Z+ is injective as a function from Z+ → X.

Exercise 4.5. The Hausdorff axiom for a space X is the condition thatany two points in X has a pair of disjoint neighborhoods. Show that a set Xwith the finite complement topology does not satisfy the Hausdorff axiom ifit has more than one point.

Exercise 4.6. Show that the limit of a convergent sequence is uniquely de-termined in a Hausdorff space, i.e. one that satisfies the Hausdorff axiom.

Exercise 4.7. Verify that (IntA) ∪ (BdA) ∪ (ExtA) is a partition of X.

Exercise 4.8. Prove that

(i) A is open if and only if IntA = A, and

(ii) A is closed if and only if C`A = A if and only if BdA ⊂ A.

Exercise 4.9. Show that Int IntA = IntA and that C`C`A = C`A.

Exercise 4.10. Prove the identity Int(A ∩B) = (IntA) ∩ (IntB). Does theidentity Int(A ∪B) = (IntA) ∪ (IntB) hold?

Exercise 4.11. Prove the identity C`(A ∪ B) = (C`A) ∪ (C`B). Does theidentity C`(A ∩B) = (C`A) ∩ (C`B) hold?

Exercise 4.12. Let A = {(x, y) ∈ X : x + y > 1} where X = [0, 1]2. Showthat A is neither open nor closed nor relatively closed, but it is relativelyopen.

Exercise 4.13. Let A = {(x, y) ∈ X : y > 0} where X = B(0, 1). Show thatA is neither open nor closed nor relatively open, but it is relatively closed.

Exercise 4.14. Verify that IntA defined in this section generalizes the oneintroduced earlier for subsets of Euclidean space by letting Int′A denote, saythe Euclidean version, and checking that x ∈ Int′A if and only if x ∈ IntA.

Exercise 4.15. Repeat the previous exercise for C`A, BdA and ExtA.

5 Continuous functions 17

5 Continuous functions

Definition 15. A function between topological spaces is continuous if theinverse image of every open set in the codomain is an open set in the domain.

Note that continuity of a function is a condition on the set map f−1.Specifically, f : (X, TX) → (Y, TY ) is continuous if the restriction of theset-map f−1 : P(Y )→ P(X) to the topology TY has image contained in TX .

Theorem 15. The composition of two continuous functions is continuous.Specifically, if f : X → Y and g : Y → Z are continuous functions, theng ◦ f : X → Z is continuous.

Proof. If U is open in Z, then g−1(U) is open in Y so that (g ◦ f)−1(U) =f−1(g−1(U)) is open in X.

If A is a subset of a space X, the subspace topology on A is preciselywhat makes the inclusion map A ↪−→ X continuous, for if U is open in X,then its inverse image under the inclusion map is U ∩ A, which belongs tothe subspace topology, by definition. Composing a map f : X → Y with theinclusion maps A ↪−→ X and Y ↪−→ Z, one obtains

Corollary 1. The restriction of a continuous function f : X → Y to anysubset A ⊂ X is continuous as is any extension of the codomain to a spaceZ having Y as a subspace.

Continuous functions can also be obtained by restricting the codomain.

Lemma 7. Let f : X → Y be continuous such that f(X) ⊂ Z ⊂ Y . Thenthe restriction of f in the codomain to the subspace Z is also continuous.

Proof. Let f∣∣Z

: X → Z denote the restriction of f in the codomain to Z so

that f = ιZ ◦ f∣∣Z

where ιZ : Z ↪−→ Y is the inclusion map of Z into Y . Givenan open set W in Z, we have W = V ∩ Z for some open set V in Y . Then

W = ι−1Z (V ) so that f∣∣−1Z

(W ) = f∣∣−1Z

(ι−1Z (V )) = f−1(W ) is open in X.

Lemma 8. Continuity between topological spaces generalizes (ε, δ)-definition.

Proof. If a continuous function f : X → Y happens to be between metricspaces then it satisfies Definition 4 since for any x ∈ X and any ε > 0 theball BY (f(x), ε) is open in Y so that f−1BY (f(x), ε) is open in X, and since

5 Continuous functions 18

it contains x as an interior point, there exists δ > 0 such that BX(x, δ) ⊂f−1BY (f(x), ε), or equivalently, f(BX(x, δ)) ⊂ BY (f(x), ε).

Conversely, if f satisfies Definition 4, then given any open set V in Y , weclaim that f−1(V ) is open in X. Indeed, for any x ∈ f−1(V ) we have f(x) ∈V so that since V is open, there exists ε > 0 such that BY (f(x), ε) ⊂ V .Choose δ > 0 so that f(BX(x, δ)) ⊂ BY (f(x), eps) ⊂ V . Hence, BX(x, δ) ⊂f−1(V ), which shows that x, which was an arbitrary point in f−1(V ), isinterior to it. Hence, f−1(V ) is open.

Continuity of a function depends explicitly on the choice of topology onthe domain as well as the codomain.

Example 5.1. Let T1 ⊂ T2 be topologies on a set X such that T2 is strictlyfiner than T1, i.e. T1 ( T2. Then the identity map 1X : (X, T1)→ (X, T2) isnot continuous. Note, however, that 1X maps open sets to open sets.

To prove continuity of a map f : X → Y where the codomain is generatedby a basis it is enough to check that the inverse image of every basis elementis open, because (see Exercise 1.6)

f−1

(⋃α∈J

Bα

)=⋃α∈J

f−1(Bα)

expresses the pre-image of any open set in Y as a union of open sets in X.As an illustration, we prove

Lemma 9. The Euclidean norm ‖ · ‖ : Rn → R is continuous.

Proof. It is enough to show the inverse image of every open interval (a, b)is open. There are 3 cases. If 0 ∈ (a, b), the inverse image is an open ball,which is open. If 0 ≤ a, the inverse image is the intersection of an open ballof radius b with the complement of a closed ball of radius a, i.e. an openannulus. If b ≤ 0, the inverse image is empty.

The next result is called pasting lemma.

Lemma 10. Let X and Y be closed subsets of a space W such that W =X ∪ Y . If f : X → Z and g : Y → Z are continuous maps into some spaceZ such that f(x) = g(x) for all x ∈ X ∩ Y , then h(w) = f(w) if w ∈ X andh(w) = g(w) if w ∈ Y defines a continuous map h : W → Z.

5 Continuous functions 19

Proof. Let C be a closed subset of Z. Then f−1(C) is a relatively closedsubset of X, hence closed in W . Similarly, g−1(C) is a relatively closedsubset of Y , hence also closed in W . Therefore, h−1(V ) = f−1(V ) ∪ g−1(V )is closed in W . Hence, h is continuous.

5.1 Product topology (finite version)

Given two spaces X and Y , it is straightforward to verify that the collection

B = {U × V : U open in X, V open in Y }

is a basis for a topology on X × Y . Indeed, the main observation is that theintersection of two elements in B is again in B:

(U1 × V1) ∩ (U2 × V2) = (U1 ∩ U2)× (V1 × V2)

Hence, (ii) in Definition 12 is satisfied by taking B3 = B1 ∩ B2, while (i) issatisfied by taking B = X × Y .

Definition 16. The product topology on the Cartesian product X × Y isthe topology generated by the basis consisting of products of an open set inX with an open set in Y . The set X × Y endowed with the product topologydefines the product space of two topological spaces X and Y .

Let π1 : X × Y → X and π2 : X × Y → Y be the projections to thefirst and second coordinates. From π−11 (U) = U × Y and π−12 (V ) = X × Vit follows that both projections are continuous. They are also open maps,i.e. they map open sets to open sets, for given an open set in X × Y , we canexpress it as a union of basis elements. Considering then the first projection,we have (see Exercise 1.4)

π1

(⋃α∈J

Uα × Vα

)=⋃α∈J

π1(Uα × Vα) =⋃α∈J

Uα,

which is open in X. The argument for the second projection is similar.

Lemma 11. f : Z → X × Y is continuous if and only if both componentfunctions f1 := π1 ◦ f and f2 := π2 ◦ f are continuous.

Proof. Necessity is clear since the composition of two continuous functionsis continuous. For sufficiency, observe that f−1(U × V ) = f−1(U) ∩ f−1(V ),which shows that the inverse image of any basis element of is open.

5 Continuous functions 20

The product of a finite collection of spaces is the Cartesian productn∏i=1

Xi = X1 × · · · ×Xn

endowed with the topology generated by the basis

B =

{n∏i=1

Ui : Ui open in Xi

}Next, let us show that the product topology on Rn coincides with the

standard topology defined using Euclidean balls. In both cases, the topologiesare constructed using bases. The basis elements for the product topology areopen boxes while that for the standard topology are open Euclidean balls.

Lemma 12. Let Bi be a basis generating a topology Ti on a set X for i = 1, 2.If for each B1 ∈ B1 and for each x ∈ B1 there exists a B2 ∈ B2 such thatx ∈ B2 and B2 ⊂ B1 then T1 ⊂ T2.Proof. The assumption on B1 shows that each x ∈ B1 is in the interior ofB1 relative to T2. Hence, B1 is open in T2, i.e. B1 belongs to T2. Sinceeach element of T1 is a union of basis elements in B1 and T2 is closed underarbitrary unions, it follows that T1 ⊂ T2.

The fact that the product topology on Rn coincides with the standardtopology can now be readily verified.

5.2 Quotient topology

The circle S1 can intuitively be thought of as an interval, say [0, 1] with itsendpoints identified. Likewise, a torus can be represented as the unit square[0, 1]2 by gluing its parallel edges together, which is often more convenientthan representing it as the boundary of some open subset of R3.

Definition 17. Let X be a space with an equivalence relation. Let X∗ be theset of equivalence classes and p : X → X∗ the maps that sends an elementto its equivalence class. Note that each U∗ ⊂ X∗ is a (disjoint) collectionof equivalence classes and p−1(U∗) ⊂ X is the union of these equivalenceclasses. The collection

T ∗ = {U∗ ⊂ X∗ : p−1(U∗) is open in X}

defines a topology on X∗, called the quotient topology.

5 Continuous functions 21

Note that the map p : X → X∗ is continuous, by design, and the topologyThese operations are formalized by endowing the set X∗ of equivalence

classes with a suitable topology.In each case, we start with a topological space X together with an equiva-

lence relation on X, which represents the gluing instructions, then using thisdata form a new space (X∗, T ∗) where X∗ is the set of equivalence classesand T ∗ is a topology on X∗ that we shall define in this section. the mapp : X → X∗ that sends an element of X to its equivalence class in X∗

Example 5.2.

Given an equivalence relation on a set X, let p : X → X∗ be the map thatsends an element to its equivalence class, where X∗ denotes the partition ofX into equivalence classes. Each subset A∗ ⊂ X∗ is a collection of disjointsubsets of X so it makes sense to take the union, giving a map Σ : P(X∗)→P(X) that preserves the Boolean structure. The T ∗ be the inverse imageunder Σ of some topology TX on X. In other words, A∗ ∈ T ∗ if and only ifΣ(A∗) ∈ TX , i.e. the union of the equivalence classes in A∗ is an open subsetof X. Since Σ preserves Boolean operations, we have

Σ(⋃α∈J

A∗α) =⋃α∈J

Σ(A∗α) ∈ TX

if each A∗α ∈ T ∗, which would imply⋃α∈J A

∗α ∈ T ∗. Similarly,

Σ(n⋂i=1

A∗i ) =n⋂i=1

Σ(A∗i ) ∈ TX

if each A∗i ∈ T ∗, which would imply⋂ni=1A

∗α ∈ T ∗. Hence, T ∗ is a topology

on X∗, called the quotient topology.surfaces, connected sum, classification of surfaces

Exercises

Exercise 5.1. Show that a constant function is continuous.

Exercise 5.2. Show that a continuous map need not be open.

Exercise 5.3. Let π1 : X × Y → X and π2 : X × Y → Y be the projectionsto the first and second coordinates. Show that π1 and π2 are open maps.

6 Connected subsets of Rn 22

6 Connected subsets of Rn

Connectivity is a central theme that the subject of topology deals with.Methods of algebraic topology provides various ways to assign invariants totopological spaces to describe the multitude of ways a space can be connected.In this section we investigate the most basic of these connectivity properties,which concerns whether a space can be divided up into smaller constituents.

Definition 18. A separation of a set X is a partition of X into a pair ofdisjoint nonempty subsets, i.e. X = A ∪ B such that A 6= ∅, B 6= ∅ andA ∩B = ∅. This relation between X and the subsets A and B is denoted by

X = A tB.

A topological separation of a space X is a separation with the additionalrequirement that A and B be (non empty) open sets. A space X is connectedif it does not admit any topological separations; equivalently, this means theonly clopen subsets of the set X are the empty set ∅ and the set X itself. Aspace that is not connected is also said to be disconnected.

Example 6.1. Let X be any set. In the trivial topology, X is connected,whereas in the discrete topology, every separation of X is topological; hence,X is disconnected as soon as it has at least 2 points.

To say that a subset X of a space Y is connected means that it is con-nected in the subspace topology. Hence, it makes sense to talk about theconnected subsets of a topological space. In the case of R we shall seethat a classification of all connected subsets is achievable. In higher dimen-sions, no such classification is known nor expected.

If X ⊂ Rn admits a separation X = A t B, the additional requirementthat A and B be relatively open subsets is equivalent to the conditions

A ∩B = ∅ and A ∩ B = ∅

(where A denotes the closure of A in Rn.) Indeed, the first condition isequivalent to A being relatively closed, since it implies that A = A ∩ X isthe intersection of a closed set with X and conversely, if A = F ∩X for someclosed set F , then A ⊂ F and F ∩B = ∅ so that A ∩B = ∅. Likewise, thesecond condition is equivalent to B being relatively closed. Since A and Bare complements of each other, they are both relatively open.

6 Connected subsets of Rn 23

Example 6.2. Let o denote the origin in Rn and e any point a distance 2from o. Then the space X = B(o, 1) t B(e, 1) is obviously not connected,since the open balls B(o, 1) and B(e, 1) are disjoint and open in X and theirunion is all of X. Hence, they form a topological separation of X.

On the other hand, the separation of the space Y = B(o, 1)tB(e, 1) is nottopological because B(o, 1) is not relatively open, as can be seen by observingthat B(o, 1) ∩ B(e, 1) 6= ∅, so that B(e, 1) is not relatively closed.

While the separation of the space Y in Example 6.2 is not topological,this does not imply that Y is disconnected, for there are many other ways toseparate the set Y and to prove that the space Y is connected one needs toargue that none of these separations can be topological. We shall establishlater that the space Y is indeed connected. To prove this, the well knownfact that any interval of real numbers is connected as well as some generalfacts about connected sets will be used.

Theorem 16. Let f : X → Y be a continuous function on a connecteddomain X. Then the image f(X) is a connected subset of the codomain Y .

Proof. If f(X) = AtB is a separation into relatively open subsets of f(X),then X = f−1(A) t f−1(B) is a separation of X. Moreover, since A =U ∩ f(X) for some open set U in Y that is disjoint from B, it follows thatf−1(A) = f−1(U) is open in X. Similarly, B = V ∩ f(X) for some open setV in Y that is disjoint from A, so that f−1(B) = f−1(V ) is open in X.

6.1 Linear continua and order topology

The main ingredient needed to prove that R is connected is the following.

Definition 19. Recall that a totally ordered set (X,<) has the least upperbound property if every nonempty A ⊂ X that is bounded above has aleast upper bound. It has the in-betweenness property if for every pairx, y ∈ X such that x < y there exists z ∈ X such that x < z < y. A totallyordered set that has both these properties is called a linear continuum.

We shall assume the standard fact that the real numbers with its usualordering < is a linear continuum. Given this, it is not hard to see that aninterval, whether open, closed or half-open, is also a linear continuum withrespect to the order topology induced by <. Note that the rationals Q (withthe usual ordering) has the in-betweenness property but not the least upper

6 Connected subsets of Rn 24

bound property whereas the integers Z has the least upper bound property,but not the in-betweenness property.

Recall that the standard topology on Rn is generated by a basis consistingof open balls. In dimension one, the basis consists of open intervals and theconstruction generalizes naturally to the setting of totally ordered sets.

Definition 20. An open interval in a totally ordered set (X,<) is anonempty subset of the form {x ∈ X : a < x < b}. An open ray is anonempty subset of the form {x ∈ X : a < x} or of the form {x ∈ X : x < b}for some elements a, b ∈ X. The order topology on X is generated by thebasis formed by the collection of all open intervals and open rays.

One of the main goals of this section is to establish that every linearcontinuum is connected in the order topology. It will follow that all intervalsare connected with respect to the order topology. The fact that the orderand subspace topologies coincide is a due to a special property of intervals.

Definition 21. A subset A of a totally ordered set (X,<) is convex if forany two points a, b ∈ A such that a < b the closed interval [a, b] ⊂ A.

Lemma 13. Let (Y,<) be a totally ordered set and X a subset of Y . Let TXbe the order topology on X induced by the restriction of the linear order toX. Let T∗ be the subspace topology on X inherited from the order topologyon Y . Then TX is coarser than T∗. If X is convex, then TX = T∗.

Proof. The topology TX is generated by the basis BX formed by the collectionof open intervals and open rays in X. Let BY be the basis for the space Yformed by collection of open intervals and open rays in Y . Then each B ∈ BXhas the form B = B′∩X for some B′ ∈ BY that is open in Y . Hence, B ∈ T∗,from which it follows that TX ⊂ T∗. Conversely, if B′ ∈ BY open interval oropen ray in Y and X is convex, then B′ ∩ X is either ∅ or X or an openinterval or open ray in X, depending on the number of endpoints of B′ in X.In any case, B′ ∩X is open with respect to TX . Hence, T∗ ⊂ TX .

Example 6.3. The dictionary order on R2 is the linear order given by(x1, y1) ≺ (x2, y2) if x1 < x2, or, x1 = x2 and y1 < y2. The order topologyon X = [0, 1]2 is strictly coarser than the subspace topology it inherits fromthe dictionary order topology on R2, for the latter contains the set A ={1/2} × [0, 1) = ({1/2} × (−1, 1)) ∩ X, but the former does not, because(1/2, 0) is not an interior point of A with respect to the order topology on X.

6 Connected subsets of Rn 25

Lemma 14. Let (X,<) totally ordered set and suppose x ∈ X is the leastupper bound of A ⊂ X. Then x ∈ C`A. Likewise, inf A ∈ C`A if it exists.

Proof. Suppose x 6∈ A. Then for any y < x there exists a ∈ A such thaty < a < x for otherwise y would be an upper bound of A that is strictly lessthan x. That means every neighborhood of x contains points of A. Sincex 6∈ A, it follows that x ∈ BdA. Hence, x ∈ C`A. A similar argument canbe given for the greatest lower bound.6

Theorem 17. Every linear continuum (in particular, R) is connected.

Proof. Suppose X = AtB is a separation into disjoint open sets. Let a ∈ Aand b ∈ B and without loss of generality, assume a < b. The subspaceX0 = [a, b] = {x ∈ X : a < x < b} inherits a separation X0 = A0 t B0

where A0 = A ∩ X0 and B0 = B ∩ X0. Note that A0 is nonempty sincea ∈ A0 while B0 is nonempty since b ∈ B0. Furthermore, A0 and B0 arerelatively closed. Let x be the least upper bound of A0, whose existence isensured by the fact that b is an upper bound for A0. Since A0 is relativelyclosed, Lemma 14 implies x ∈ A0. Let y be the greatest lower bound ofB1 = {y ∈ B0 : y > x}, which is nonempty since b ∈ B1 and bounded belowby x. Again, Lemma 14 implies y ∈ C`X0 B1 ⊂ C`X0 B0 = B0, since B0 isrelatively closed. Thus, y > x. By the in-betweenness property, there existsa z ∈ X such that x < z < y. Since x is an upper bound for A0, z 6∈ A0.Hence, z ∈ B0. In fact, z ∈ B1 since z > x, but this now contradicts the factthat y is a lower bound for B1. This contradiction shows that the separationof X into disjoint open sets cannot exist.

Theorem 18. A subset of R is connected if and only if it is convex.

Proof. For necessity, let X is a connected subset of R, a, b ∈ X with a < b,and suppose there exists a z ∈ (a, b) that is not in X. Then X = A t Bwhere A = (−∞, z)∩X and B = (z,∞)∩X is a separation of X into disjointrelatively open subsets of X, which contradicts the fact that X is connected.Hence, [a, b] ⊂ X, showing that X is convex.

For sufficiency, suppose X be a convex subset of R. Since the subspacetopology coincides with the order topology, it is enough to check that X is alinear continuum. If A is a nonempty subset of X that has an upper bound

6 Alternatively, the identity map (X,<) → (X,>) being order-reversing implies thatinf A = sup> A ∈ Bd> A = BdA, since a monotone bijection is a homeomorphism.

6 Connected subsets of Rn 26

in X. Let x = supA be the least upper bound in R. Since x lies on the linesegment joining any point in A to any upper bound of A in X, convexityimplies x ∈ X. Hence, X has the least upper bound property. Now letx, y ∈ X with x < y be given. Then there exists z ∈ R such that x < z < y.Convexity implies z ∈ X. Hence, X has the in-betweenness property. Itfollows that X is a linear continuum and is thus connected.

We now obtain a general form of the Intermediate Value Theorem.

Theorem 19. Let f : X → R be a continuous, real-valued function definedon a connected domain. Then f(X) is a convex subset of R.

6.2 Path-connectedness

Using Theorems 16 and 17 we have the following application

Example 6.4. The unit circle S1 is connected, because it is the image of Runder the map t→ (cos t, sin t), which is continuous since cos t and sin t areboth differentiable, hence continuous functions.

Definition 22. A path in X is a continuous map α : [0, 1]→ X. It is saidto join a point x ∈ X to a point y ∈ X if α(0) = x and α(1) = y. A spaceis path-connected if every two points can be joined by a path.

Theorem 20. Every path connected space is connected.

Proof. Let X be a path connected space and suppose it admits a separationX = A t B into disjoint nonempty open sets. Let α be a path that joinssome point a ∈ A to some points b ∈ B. Then [0, 1] = α−1(A) t α−1(B)is a separation of [0, 1] into disjoint open subsets of [0, 1]. However, such aseparation does not exist since [0, 1] is connected. Hence, X is connected.

Definition 23. A subset A ⊂ Rn is convex if for any two points in A, theline segment joining them is also contained in A.

Example 6.5. Every convex subset of Rn is path-connected, hence connected.In particular, Rn is connected and so is every ball in Rn.

If α is a path from x to y and β is a path β from y to z, then theconcatenation of α and β, denoted

α ? β,

6 Connected subsets of Rn 27

is defined as the path following α from x to y at twice the speed until timet = 1/2, then continues from y to z following the path of β at twice thespeed. More formally, for any t ∈ [0, 1] set

α ? β(t) =

{α(2t) t ≤ 1/2

β(2t− 1) t ≥ 1/2

and observe that the maps being pasted have a common point t = 1/2 wherethey agree; hence, α ? β is continuous by the pasting lemma.

As a consequence we have

Lemma 15. If A,B ⊂ X are path-connected subsets with a point in common,then A ∪B is also path-connected.

Proof. Let z ∈ A ∩ B. Then any two points in A ∪ B can be joined to z,hence to each other via a concatenation of their paths to z.

Using Lemma 15, it is not hard to see that Example 6.2 is path-connected.

Example 6.6. Rn \ {0} is connected for any n ≥ 2, because it is the unionthe half-spaces {x ∈ Rn : ±xi > 0}, any two of which share a common point.

Example 6.7. Sn is connected for any n ≥ 1, for it is the image of Rn+1\{0}under the map x 7→ x

|x| .

Theorem 21. The closure of a connected subset is also connected.

Proof. Let X be a connected subset of a space Y . Suppose there exists aseparation C`X = A t B into open subsets of C`X. Then X = A0 t B0

where A0 = A ∩ X and B0 = B ∩ X is a separation of X. (Note: A0 = ∅implies A ⊂ BdX, which contradicts the fact that BdX has no interior. SeeExercise 2.10.) Since A and B are open in C`X, A0 and B0 are open relativeto X. This contradicts the hypothesis that X is connected. Hence, C`X isconnected.

Example 6.8. The topologist sine curve S is the graph of the functiont ∈ (0, 1]→ sin(1/t). Its closure is S = S ∪ ({0} × [−1, 1]) is connected, butnot path-connected. (Remark: To see that it is not path-connected, supposeα(t) = (x(t), y(t)) is a path joining (0, 0) to (1, sin(1)). Let t0 = sup{t ∈[0, 1] : x(t) = 0}. Then y fails to be continuous at t0, because using theIntermediate Value Theorem applied to x(t) we can construct two sequencestn → t+0 where one satisfies y(tn)→ 1 while the other satisfies y(tn)→ −1.)

6 Connected subsets of Rn 28

6.3 Remarks on path-components

An equivalence relation on a space X is obtained by defining two points tobe related if they can be joined by a path in X. The equivalence classes ofthis relation are called the path-components of X. A weaker equivalencerelation is obtained by defining two points to be related if there is a connectedset containing both of them. Its equivalence classes are called the connectedcomponents of X. A space is connected (resp. path-connected) if and onlyif it has exactly one connected (resp. path-connected) component. A space issaid to be locally path-connected if every point has a neighborhood thatis path-connected.

Theorem 22. A locally path-connected space is connected if and only if it ispath-connected.

Proof. Let A be a path-component. Then each point x ∈ A has a path-connected neighborhood U , which is contained inA since U is path-connected.It follows that x is an interior point of A, so that A is open. That is, everypath-component is open. Each path-component is also closed because itscomplement is a union of path-components. Hence, a nontrivial clopen setwould exist if there was more than one path-component.

Every open subset of Rn is locally path-connected. A manifold is atopological space satisfying certain assumptions, the most important of whichis that it is locally Euclidean, which means every point has a neighborhoodthat is homeomorphic to an open subset of Rn. In particular, for manifolds,the two notions of connectedness coincide. The closed topologist sine curveprovides an example of a connected space (single connected component) thathas two path-components.

A space is said to be totally disconnected if its only nonempty con-nected subsets are singletons. For example, any set with the discrete topologyis totally disconnected. However, a totally disconnected set need not be dis-crete, as the example of Q shows. Indeed, any subset of Q with at least twopoints can be separated into two nonempty relatively open subsets, so theonly nonempty connected sets are singletons; but no singleton can be ex-pressed as the intersection of Q with an open subset of R. Hence, singletonsare not closed in Q.

Exercise 6.1. Let X be a set with finite complement topology. Show that asubset of X is connected if and only if it is infinite.

7 Compact subsets of Rn 29

7 Compact subsets of Rn

Again, unlimited possibilities. Heine-Borel, other formulations, Tychonofftheorem.