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INTRODUCTION TO TOPOLOGY (MA30055)

SEMESTER 2 MATHEMATICS: PROBLEM SHEET 5: SOLUTIONS

Let Xbe a topological space. Then X is said to be:

a T1-space if, for any x=y X, there is an open set U with y U and xU; a T2- or Hausdorff space, if for any x = y X, disjoint open U, V with x U and

y V.

Recall also the topology on N = N{} whose closed subsets are either finite or contain .

1. True or false? (Give reasons.)

(a) X is T1 if and only if x X, the point {x} is closed.

(b) Any T1-space is Hausdorff.

(c) Any countable Hausdorff space is discrete

(d) Let a, b:N Xbe continuous with a(n) =b(n) for all n N. Then a() =b().

(e) As in (d) but assuming also Xis Hausdorff.

Solution:

(a) True. If{x} is closed, then U =X\ {x} is open and contains everyy = x, fulfilling

the requirement forT1. Conversely, if the topology isT1, then for everyy=x there is an

open Uy withy Uy andxUy. Then X\ {x}=y=xUy is a union of open sets, hence

open. Thus{x} is closed.

(b) False. The cofinite topology on an infinite set is T1 by part (a) above, but is not

Hausdorff since two cofinite sets, i.e., two non-empty open sets, are never disjoint.

(c) False. The topology on Q induced from the Euclidean topology on R is Hausdorff,

because it is induced from a metricd (forx, y Q, let = d(x, y)/2 so that U = B(x)

and V = B(y) are disjoint open sets containing x and y rspectively). However, the

topology on Q is not discrete since singletons are not open.

(d) False. TakeXto have more than one point and the trivial topology. Then any functions

a, b:N Xare continuous, and they can be chosen to agree forn Nand disagree at .

(e) True. This is a reformulation of uniqueness of limits in Hausdorff spaces. Here is a

proof of the contrapositive. Supposea, b:N X are continuous. Ifa() = b() thenthey can be separated by disjoint open setsU, V. Nowa(U) and b(V) are open and

contain, so they must be cofinite; in particulara(U) Nandb(V) Nhave nonempty

intersection (the union of their complements is finite), i.e., n N with a(n) U and

b(n) V; in particulara(n)=b(n).

2. Prove thatXis Hausdorff if and only if the diagonal X :={(x1, x2) XX :x1 =x2}

is a closed subspace of the product space X X. [Hint: it is possible to write down the


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definition of what it means for X to be closed using a basicopen set in X X.]

Solution: The diagonal is closed if and only if, for all(x1, x2) (X X) \ , there is

a basic open set W =U V with (x1, x2) W andW = , whereU, V are open in

X. In other words, ifx1=x2 inX, then there are open setsU, V withx1Uandx2V

such that

(U V) ={(x, x) |x U, x V}= .

But this set is empty precisely when U V is empty, i.e. U andV are disjoint. Thusis

closed if and only ifXis Hausdorff.

3. Prove that ifX1 and X2 are Hausdorff, so is the product space X1 X2.

Solution:

IfXandX2 are Hausdorff spaces, consider distinct points(x1, x2)and(y1, y2)in X1 X2.

They are distinct, so eitherx1=y1 orx2 =y2. First suppose thatx1=y1. Then there are

disjoint open setsU1 x1 andU2 y1, becauseX1 is Hausdorff. HenceU1 X2(x1, x2)

andU2 X2 (y1, y2) are disjoint open sets in X X2. If insteadx2 =y2, then we use

a similar argument, relying on the fact that X2 is Hausdorff. ThusX1 X2 is Hausdorff.

This argument can easily be adapted to arbitrary products: two points differ if they differ

at one of their coordinates, and finding disjoint open sets for this coordinate gives disjoint

open sets in the product.

4. Letf1: Y X1 and f2: Y X2 be continuous maps. A space Zwith continuous maps

p1: X1Zand p2: X2Zis called a pushout off1, f2 iffp1f1 =p2f2 and for any other

space W with such maps qi: Xi W there is a unique continuous map h: Z W with

qi =hpi.

(a) Show that the pushout is a quotient of the disjoint union X1

X2. [Hint: it is a

coequalizer.]

(b) SupposeX1, X2are open subspaces of a spaceV andY =X1X2withfithe inclusions

ofY into Xi. Show that the unionZ = X1 X2 (with inclusion maps pi: Xi Z the

inclusion maps) is a pushout. What ifX1 and X2 are not open?

Solution:

(a) Let X1

X2 be the disjoint union with inclusions 1: X1 X1

X2 and 2: X2

X1

X2. The mapsfi: Y Xi determine continuous mapsgi=ifi: Y X1

X2. Let

p: X1

X2Zbe the coequalizer ofg1 andg2, andpi =pi: Xi Z.

We claim that Zwith these two mapspi is a push-out. Indeed,

p1f1 =p1f1 =pg1 =pg2 =p2f2 =p2f2,


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so the first property of pushout is satisfied precisely because the first property of a co-

equalizer is satisfies. Draw a diagram!

For the second (universal) property, supposeqi: Xi W satisfyq1f1 =q2f2. First of all,

by the universal property of the disjoint union, we may patchq1andq2to give a continuous

mapq: X1

X2W (with qi =qi). Add this to your diagram!

Now since qg1 = qg2, there is a unique continuous maph: Z W with q= hp (by the

universal property of the coequalizerp: X1

X2Z). It follows thatqi =hpias required.

Since coequalizers are quotient maps, we are done.

(b) Since all maps are inclusions of subspaces, it is clear thatp1f1 =p2f2(they both include

the intersection into the union). It remains to check that the union has the universal

property of a pushout.

So supposeqi: XiWare continuous maps with q1f1 =q2f2. The latter condition means

that q1|X1X2 =q2|X1X2 . However, X1 andX2 form an open cover ofX1 X2 and thisis the patching condition for the existence of a continuous map h: X1X2 W with

qi =h|Xi =hpi.

IfX1 andX2 are not open, patching may not apply (it does ifX1 andX2 are both closed,

but not in general). By the construction in (a), the pushout is the quotient X1

X2/X1

X2(i.e. by the equivalence relation where points in X1X2 are identified). The underlying

point set of this space is |X2| |X2|, but equipped with the quotient topology (from

X1

X2) rather than the subspace topology (from V).

5. Let X be the set {0, 1} with the discrete topology and let Z = XN be the set of

sequences in X, with the (infinite) product topology.

(a) Define

f: Z R: (xn)nN

nN

xn2n

Show that f is continuous. Is f injective? What is its image?

(b) Define

g: Z R: (xn)nN

nN

2xn

3n

Show that g is continuous. Is g injective? The image ofg is called the Cantor set. Show

that it is a closed subset of [0, 1].

Solution:

(a) The map f takes a sequence x = (x1, x2, . . .) Z to the real number with binary

expansion 0.x1x2x3. . .. Hence the image off is[0, 1] (noting that 1 = 0.111 . . .), but f is

not injective, because, e.g.,1/2 = 0.100 . . .= 0.0111 . . ..


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To show that f is continuous at xZwe must show, for any basic open neighbourhood

V off(x), that f(V) contains a basic open neighbourhood ofx. That is, for any open

ballB(f(x)) aroundf(x) = 0.x1x2. . . R, there is a basic open neighbourhood

B(x1, . . . , xk) ={y = (yn) Z |yi=xi i k}

ofx such that f(B(x1, . . . , xk)) B(f(x)).

More simply still, for any > 0, there is ak such that, for anyy Z with yi = xi for

i k, we have|f(y) f(x)|< . But with this assumption on y we have

|f(x) f(y)|

n=1

|xn yn|

2n

n=k+1

1

2n =

1

2k

and so we succeed by choosingk large enough that 1/2k < .(b) The continuity of g follows by a similar argument to above, with the last estimate

being

|g(x) g(y)|

n=1

2|xn yn|

3n

n=k+1

2

3n

= 1

3k.

The map g takes a sequence (x1, x2, . . .) to the real number with ternary expansion

0, y1y2. . ., where yn = 2xn. As for binary, ternary expansions are not always unique.

However, coincidences only occur when one expansion contains the digit 1, e.g.0.022 . . .=

0.100 . . .or0.122 . . .= 0.200 . . .and hence the two expressions cannot come from applyingg to a sequence(xn). Thusg is injective.

The Cantor setC(the image ofg) consists of all real numbers in [0, 1]which have a ternary

expansion in which all the digits are either0 or2. In other words.

C= [0, 1] \

nN

Tn,

whereTn is the set of real numbers whosenth ternary digit mustbe1. Note, e.g. 1/3 =

0.100 . . . = 0.022 . . ., so 1/3 T1. Thus T1 = (1/3, 2/3). Similarly, T2 = (1/9, 2/9)(4/9, 5/9) (7/9, 8/9). In general, Tn is a union of open intervals of length 1/3

n, and

hence open. HenceCis the complement of a union of open sets, hence closed.

GKS, 10/3/15

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