The University of Sydney

MATH 3901

Metric Spaces 2004

Tutorial 12

PROBLEM SET 12

1. Let N be the set of all nonnegative integers, i.e.,

N = {0, 1, 2, 3, 4, . . . }.

For each k ∈ N, let Uk = {k, k + 1, k + 2, . . . }. and let τ = {∅, Uk (k ∈ N)}.Show that τ is a topology for N.

Solution.

(a) Clearly ∅ ∈ τ . Since U0 = {0, 1, 2, . . . } = N it follows that N ∈ τ.

(b) We see thatU0 ⊃ U1 ⊃ U2 ⊃ . . . .

Thus, for any subfamily {Un1 , Un2 , Un3 , . . . } of τ , we can assume, with-out loss of generality, that n1 = min{nk | k = 1, 2, . . . }. Then Un1 ⊇ Unk

for all k, so that ⋃k

Unk= Un1 ∈ τ.

(c) For any finite subfamily {Un1 , . . . , Unm} of τ , we can assume, without

loss of generality, that n1 ≤ · · · ≤ nm. Then

Un1 ∩ · · · ∩ Unm = Unm ∈ τ.

Hence τ satisfies all the required axioms and so τ is a topology for N.

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2. Let X be an infinite set and let τ be the family of all subsets U of X such thatX \ U is finite, together with ∅. Prove that τ is a topology for X.

Solution.

(a) Clearly ∅ ∈ τ and X ∈ τ .(b) Let {Ui} ⊆ τ . Then either Ui = ∅ or X \ Ui is finite. Then either Ui = ∅

for all i and⋃i

Ui = ∅, or X \ Uj is finite for some j and

X \( ⋃

i

Ui

)=

⋂i

(X \ Ui)

is finite. Thus⋃i

Ui is in τ.

(c) If U and V are in τ , then X \ U and X \ V are finite so

X \ (U ∩ V ) = (X \ U) ∪ (X \ V )

is finite and thus U ∩ V is in τ.

Hence τ satisfies all the required axioms and so τ is a topology for X.

3. Let X = {a, b, c, d, e}. Determine whether or not each of the following familiesof subsets of X is a topology on X. For each of those which is not a topology,find a smallest topology that contains it.

(i) τ1 = {∅, {a}, {b}, {a, b}, {b, c}, X}(ii) τ2 = {∅, {a, b, c}, {a, b, e}, {a, b, c, e}, X}(iii) τ3 = {∅, {a}, {b}, {a, b}, {a, c, d}, {a, b, c, d}, X}

Solution.

(i) Since {a, b}, {b, c} ∈ τ1, but {a, b} ∪ {b, c} = {a, b, c} is not in τ1, itfollows that τ1 is not a topology on X. The smallest topology containingτ1 is

τ = {∅, {a}, {b}, {a, b}, {b, c}, {a, b, c}, X}.

(ii) Since {a, b, c}, {a, b, e} ∈ τ2, but {a, b, c} ∩ {a, b, e} = {a, b} is notin τ2, it follows that τ2 is not a topology on X. The smallest topologycontaining τ1 is

τ = {∅, {a, b}, {a, b, c}, {a, b, e}, {a, b, c, e}, X}.

(iii) τ3 is a topology since it satisfies all the required axioms.

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4. Let τ∞ be the family of subsets of R consisting of R, ∅, and all open infiniteintervals Ua = (a, ∞) with a ∈ R. Show that τ∞ is a topology on R. Then

(i) Determine the closed subsets of the topological space (R, τ∞).(ii) Determine the interiors, closures and boundaries of the sets A = (4, 9],

B = {−2, 0, 6, 57}, C = {−4, 4, 8, 12, . . . } and E = [5, ∞) in (R, τ∞).

Solution.

(a) Clearly ∅ ∈ τ∞ and R ∈ τ∞.(b) Let A ⊆ τ∞ \ {∅, R}; that is,

A = {Ui | i ∈ I},

where I is some subset of real numbers. If I is not bounded from below,then

⋃i∈I

Ui = R. If I is bounded from below, say inf I = i0, then⋃i∈I

Ui =

(i0, ∞) = Ui0 . In either case,⋃i∈I

Ui ∈ τ∞.

(c) For any real numbers a and b with a < b, we see that Ub ⊂ Ua. Thus forUa, Ub ∈ τ∞, with a < b, it follows that Ua ∩ Ub = Ub ∈ τ∞.

Hence τ∞ satisfies all the required axioms and so τ∞ is a topology for X

(i) A set is closed if and only if its complement is open. Hence the closedsubsets of (R, τ∞) are ∅, R and all closed infinite intervals R \ Ua =(−∞, a], where a ∈ R.

(ii) Let G be any subset in (R, τ∞). An interior of G is the largest opensubset in (R, τ∞), contained in G and the closure of G is the smallestclosed subset in (R, τ∞), containing G.

Hence

Int A = ∅, Int B = ∅, Int C = ∅, Int E = (5, ∞),

and

A = (−∞, 9], B = (−∞, 57], C = (−∞, ∞) = R, E = R.

Now X \ A = (−∞, 4] ∪ (9, ∞), we see that X \A = R. Next, it is easyto see that

X \B = X \ C = R and X \ E = (−∞, 5].

Therefore

Fr A = (−∞, 9], Fr B = (−∞, 57], Fr C = R, Fr E = (−∞, 5].

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5. Draw the Hasse diagrams for all the 33 different topologies for a set of fourelements X = {a, b, c, d}.

Solution.

The Hasse diagrams of the 33 different topologies are:

X

∅τ1 τ2 τ3 τ4 τ5 τ6 τ7 τ8 τ9 τ10 τ11 τ12

τ13 τ14 τ15 τ16 τ17 τ18 τ19 τ20 τ21 τ22 τ23

τ24τ25 τ26 τ27 τ28 τ29 τ30 τ31

τ32 τ33

The Hasse diagrams of τ24 and τ33 may be drawn as follows:

τ24 τ33

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6. Let X = {a, b, c, d} and let τ be the topology given by

τ ={∅, {a}, {a, b}, {a, c}, {a, b, c}, X

}.

Let A = {a, b, d}, B = {a, d} and C = {b, c, d}. For each of the sets A, B andC, find the interior, the derived set, the closure, the boundary and the inducedtopology of the set.

Solution.

We haveInt A = {a, b}, Int B = {a}, Int C = ∅.

Consider the set A. Then the point a ∈ X is not a limit point of A since{a} is an open set containing no other points of A. The open sets of b are{a, b}, {a, b, c} and X, each of which contains the point a in A with a 6= b sothat b is a limit point of A. Similarly c and d are limit points of A. Thus thederived set A′ of A is

A′ = {b, c, d}.

In the similar way, one can show that

B′ = {b, c, d}.

Consider the set C. Then a is not a limit point of C since {a} is an open setcontaining no other points of C. Next, b is not a limit point of C since {a, b}is an open set containing no other points of C. Similarly c is not a limit pointof C. The only limit point of C is d. Thus the derived set C ′ of C is

C ′ = {d}.

The closed sets of X are

∅, {d}, {b, d}, {c, d}, {b, c d}, X.

HenceA = X, B = X, C = C.

The complements of A, B and C are

X \A = {c}, X \B = {b, c}, X \ C = {a}.

ThereforeX \A = {c, d}, X \B = {b, c, d}, X \ C = X,

and soFr A = {c, d}, Fr B = {b, c, d}, Fr C = C.

Finally the induced topologies are given by

τA = {∅, {a}, {a, b}, A}, τB = {∅, {a}, B}, τC = {∅, {b}, {c}, {b, c}, C}.

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7. Let X = {a, b, c, d} , and τ the topology on X given by

τ ={∅, {a}, {a, b}, {a, c}, {a, b, c}, {a, b, d}, X

}.

Write down all the closed sets in (X, τ).Let A = {a, c, d} . Is A open? Is A closed? Find Int A, A′, A and Fr A of A .

Solution.

The closed sets of X are

∅, {c}, {d}, {b, d}, {c, d}, {b, c, d}, X.

Since A = {a, c, d} is not in τ , it follows that A is not open. Since X \A = {b}is not in τ , we conclude that A is not closed.

NowInt A = {a, c}, A′ = {b, c, d}, A = X, Fr A = {b, d}.

8. Let A be any subset of a discrete topological space (X, D). Show that thederived set A′ of A is empty.

Solution.

Let x be any point in X. Recall that every subset of a discrete topologicalspace X is open. Hence, in particular, the singleton set G = {x} is an opensubset of X. Since

(G \ {x}) ∩A = ∅ ∩A = ∅,

it follows that the derived set A′ of A is empty.

9. Let (N, τ) be the topological space as given in Question 1.

(i) List all the open sets containing the integer 6.(ii) Let A = {3, 10, 26, 47}. Find A′, A and Fr A.(iii) Determine those subsets A such that A′ = N.

Solution.

(i) Since the open sets are of the form

Uk = {k, k + 1, k + 2, . . . },

with k ∈ N, the open sets containing the positive integer 6 are the follow-ing:

U1 = {0, 1, 2, . . . } = N,

U2 = {1, 2, 3, . . . },U3 = {2, 3, 4, . . . },U4 = {3, 4, 5, . . . },U5 = {4, 5, 6, . . . },U6 = {5, 6, 7, . . . },U7 = {6, 7, 8, . . . }.

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(ii) Note that the open sets containing any point n ∈ N are the sets Uk, wherek ≤ n. If n0 ≤ 46, then every open set containing n0 also contains 47 ∈ Awhich is different from n0. Thus any n0 ≤ 46 is a limit point 0f A. Onthe other hand, if n0 > 46, then the open set

Un0 = {n0, n0 + 1, n0 + 2, . . . }

contains no point of A different from n0. So any n0 > 46 is not a limitpoint of A. Hence the derived set A′ OF A is

A′ = {0, 1, 2, 3, . . . , 46}.

ThenA = A ∪A′ = {0, 1, 2, 3, . . . , 47}.

Next, X \A = N so that the boundary Fr A of A is

Fr A = A ∩ (X \A) = {0, 1, 2, 3, . . . , 47}.

(iii) If A is an infinite subset of N, then A is not bounded from above. Thusany open set containing any point x ∈ N will contain points of A otherthan x. Hence A′ = N.

On the other hand, if A is finite, then A is bounded from above, say, byn0 ∈ N. Then the open set Un0+1 contains no point of A. Hence n0 + 1is not a limit point of A, and so A′ 6= N.

Hence the subsets A such that A′ = N are all the infinite subsets of N.

10. Let (N, τ) be the topological space as given in Question 1.

(i) Determine the closed subsets of (N, τ)

(ii) Determine the closure of the sets A = {5, 34, 79} and B = {4, 8, 12, . . . }.(iii) Determine those subsets of N which are dense in N.

Solution.

(i) The closed sets of N are as follows:

N, ∅, {1}, {1, 2}, {1, 2, 3}, . . . , {1, 2, . . . , m}, . . . .

(ii) The closure of a set is the smallest closed set containing that set. Hence

A = {1, 2, . . . , 78, 79}, B = N.

(iii) It follows from the previous question that if a subset A os N is infinite,then A = N; that is, A is dense in N; and if A is finite, then its closure isnot N, that is, A is not dense in N.

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11. Show that in the indiscrete topological space (X, I), any sequence (xn) in Xconverges to every point x in X.

Solution.

Let (xn) be any sequence in X. Since X is indiscrete, X is the only open setcontaining any point x ∈ X. Since X contains every term of the sequence (xn),it follows that the sequence (xn) converges to every point x ∈ X.

12. Show that in the discrete topological space (X, D), a sequence (xn) in Xconverges to a point x in X if and only if the sequence (xn) is of the form(x1, x2, . . . , xN , x, x, x, . . . ).

Solution.

Let (xn) be any sequence in X. Since X is indiscrete, for any x ∈ X, thesingleton set {x} is an open set containing x. If (xn) converges to a pointx ∈ X, then the open set {x} must contain almost all xn. Hence a sequence(xn) in X converges to a point x in X if and only if the sequence (xn) is of theform (x1, x2, . . . , xN , x, x, x, . . . ).

13. Let (X, τX) and (Y, τY ) be two topological spaces and f : X → Y a mapping.Let BY be a base for τY of Y . Prove that f is continuous on X if and only iffor any B ∈ BY , f−1(B) is open in X, i.e., it is in τX .

Solution.

Suppose that f is continuous on X. Since any B ∈ BY is in τY , it follows thatf−1(B) is open in X, i.e., it is in τX .

Suppose that for any B ∈ BY , f−1(B) is open in X. Let WτY . Then W =⋃i Bi for some Bi ∈ B, and so

f−1(W ) = f−1( ⋃

Bi

)=

⋃i

f−1(Bi).

Since each f−1(Bi) is open in X, it follows that f−1(W ) is open in X andhence f is continuous on X.

14. Let X and Y be two nonempty sets. Show that for any subsets U1, U2 of Xand V1, V2 of Y ,

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

Solution.

We have(a, b) ∈ (U1 × V1) ∩ (U2 × V2)

⇐⇒ (a, b) ∈ U1 × V1 and (a, b) ∈ U2 × V2

⇐⇒ a ∈ U1, b ∈ V1, a ∈ U2 and b ∈ V2

⇐⇒ a ∈ U1 ∩ U2 and b ∈ V1 ∩ V2

⇐⇒ (a, b) ∈ (U1 ∩ U2)× (V1 ∩ V2).

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15. Let X and Y be topological spaces with A ⊂ X and B ⊂ Y . In the productspace X × Y , prove that A×B = A×B.

Solution.

Since A ⊆ A and B ⊆ B, we have

A×B ⊆ A×B.

It follows from the previous question that

A×B = (A ∩X)× (Y ∩B) = (A× Y ) ∩ (X ×B).

Since(X × Y ) \ (A× Y ) = (X \A)× Y

and X \A is open in X, (X × Y ) \ (A× Y ) is open in X × Y and so A× Y isclosed in X × Y . Similarly X × B is closed in X × Y . Hence A × B is closedin X × Y so that

A×B ⊆ A×B.

Now let a = (a1, a2) ∈ A×B so that a1 ∈ A1 and a2 ∈ A2. Next, let U1 × U2

be any open set containing a in X × Y . Then U1 is an open set containing a1

in X and U2 is an open set containing a2 in Y . Since a1 ∈ A1 and a2 ∈ A2, itfollows that

U1 ∩A 6= ∅ and U1 ∩B 6= ∅,

and therefore

(U1 × U2) ∩ (A×B) = (U1 ∩A)× (U2 ×B) 6= ∅,

which implies thata = (a1, a2) ∈ A×B.

HenceA×B ⊇ A×B,

and consequently,A×B = A×B.

16. Prove that a topological space (X, τ) is compact if and only if there exists abase B for τ such that every open covering of X by members of B has a finitesubcovering.

Solution.

Suppose that X has such a base B and let C be any open covering of X. Foreach x ∈ X, let Ux be a member of C containing x. Since B is a base, for eachx ∈ X, there is, a member Bx in B such that x ∈ Bx ⊆ Ux. Then {Bx | x ∈ X}is an open covering of X by members of B and therefore has a finite subcovering{Bxi}n

i=1. The corresponding family {Uxi}ni=1 is then a finite subcovering of C

for X. Hence X is compact.

If X is compact, then every open covering of X, and in particular every coveringby basic open sets, must have a finite subcovering. This completes the proof.