The University of Sydney

School of Mathematics and Statistics

Solutions to Tutorial 8

MATH3961: Metric Spaces Semester 1, 2012

Lecturer: Laurentiu Paunescu

1. Let (X, d) be a metric space, and let A and K be disjoint subsets of X, with A

closed and K compact.

Show that ∃δ > 0 such that d(a, k) ≥ δ for all a ∈ A and k ∈ K.

Solution: We give three different arguments. Take your pick!

(a) Suppose not. Then there are sequences {an}n≥1 in A and {kn}n≥1 in K suchthat d(an, kn) < 1

nfor all n ≥ 1. Since K is compact {kn}n≥1 has a convergent

subsequence, with limit L, say. Clearly L ∈ A = A and so A ∩K 6= ∅, contrary tohypothesis.

(b) Let f(x) = inf{d(a, x) | a ∈ A}, for all x ∈ X. Then f(x) ≥ 0 for all x ∈ X,and f(x) = 0 if and only if x ∈ A. Moreover |f(x) − f(y)| ≤ d(x, y), for allx, y ∈ X, and so f is continuous. Hence f(K) is a compact subset of (0,∞), andso δ = inf f(K) > 0.

(c) Since A is closed, if x is not in A there is a δx > 0 such that d(a, k) ≥ δk for alla ∈ A. Suppose that 0 < rk ≤ δk, for each k ∈ K. Clearly K ⊆ ∪k∈KB(k, rk) andso K ⊆ ∪f∈FB(f, rf ), for some finite F ⊆ K, by compactness. Let δ = min{rf |f ∈F . (Note that F may depend on the choices we have made.) Then δ > 0. Whenwe check to see whether this works, we find that if k ∈ B(f, rf ) and a ∈ A thend(a, k) ≥ d(a, f) − rf . Thus if we choose rf < 1

2δf we shall have d(a, k) ≥ rf ≥ δ

for all a, k.

2. Let X be a Hausdorff space and let x ∈ V ⊆ C, where V is open and C is compact.For each y 6= x there is a pair of disjoint open sets Ay, By with x ∈ Ay and y ∈ By.Then {V }∪{By | y 6= x} is an open cover of C. Use compactness to conclude thatthere is a smaller open set W with x ∈ W and W ⊆ V . (Hint: W may be takento be the intersection of finitely many of the Ays.)

Solution: (My hint needed a little modification.) Since C − V = C ∩ (X − V )is a closed subset of C it is also compact, and therefore C − V ⊆ ∪y∈FBy for somefinite subset F ⊆ C − V . Let W = V ∩ (∩y∈FAy). Then W is open and x ∈ W .Since W ⊆ C − ∪y∈FBy = C ∩ (X − ∪y∈FBy), which is closed, it follows thatW ⊆ C − ∪y∈FBy ⊆ V .(C \ V ⊆ B → C \B ⊆ V .)

3. Let K0 = [0, 1], and construct a nested descending sequence of closed subsetsKn be “removing middle thirds”. Thus K1 = K0 − (1

3, 23) = [0, 1

3] ∪ [2

3, 1]. K2 =

K1−(19, 29)−(7

9, 89), and so on. The intersection C = ∩n≥1Kn is called the (standard)

Cantor set.

Show that C consists of all real numbers in [0, 1] which have base-3 expansionsinvolving only the digits 0 and 2 (with infinitely recurring 2s allowed).

Copyright c© 2012 The University of Sydney 1

Show that C is compact, uncountable, and has empty interior.

Solution:

The first assertion is almost self-evident.

Note that the endpoints of the intervals removed are 0, 1 and a subset of therational numbers m

3n, with 0 < m < 3n and m not divisible by 3. The left hand

endpoints have numerator m ≡ 2 mod (3), while the right-hand endpoints (1, 1

3, 1

9,

7

9. . . ) have numerator m ≡ 1 mod (3) and are exactly the elements of [0, 1] which

have base-3 expansions involving no 1s and ending with recurring 2s.

Since each Kn is closed C is closed and since each Kn is compact it follows that Cis compact.

There are countably many endpoints. Define a function h : [0, 1] → C as follows.Let x = 0.a1a2a3 . . . be the base-2 expansion of x ∈ [0, 1], in non-recurrent form.(Thus 1

2= 0.1000 . . . , rather than 0.011111 . . . .) Let h(x) be the real number with

base-3 expansion 0.b1b2b3 . . . , where bn = 2an. Then h is 1-1, and maps the interval[0, 1] onto C − R, where R is the (countable) set of right-hand endpoints. HenceC has the cardinality of the continuum: |C| = 2ℵ0 .

We obtain C from [0, 1] by removing a countable union of intervals, of total lengthΣn≥12

n−13−n = 1, which is the length of [0, 1]. Thus C can contain no interval ofpositive length, and so IntC is empty.

If you do not wish to use such a “measure theoretic” approach, note that rationalsof the form 3m+1

3nwith 0 ≤ m < 3n−1 are dense in [0, 1], and no such number is in

C.

The construction of C may be varied to produce closed subsets of [0, 1] with emptyinterior, but with total length λ, for any 0 ≤ λ < 1. (Instead of removing middlethirds of each interval, we remove “middle 1

n’ths.)

It can be shown that C is homeomorphic to the product of countably many copiesof the discrete 2-point space (with the infinite product).

4. Let X be a topological space and ∞ an object not in X. Let X+ = X ∪ {∞} as aset, and declare a subset U ≤ X+ to be open if either U is an open subset of X orK = X+ − U is a compact closed subset of X.

Show that the set of all such subsets defines a topology on X+.

Show that X+ is compact.

Show that X+ is Hausdorff if and only if X is locally compact and Hausdorff.

(This new space is called the 1-point compactification of X.)

Show that the inclusion of X in X+ is an embedding.

Given f : X → Y continuous we can define f+ : X+ → Y + by specifying f(∞) =∞. Show that f+ is continuous if f is proper.

Solution: ∅ and X+ are open in X+, since ∅ is open in X and ∅ = X+ −X+ isis also a compact subset of X.

If U is open in X and K is compact and closed in X then U ∩ (X+ −K) = U −K

is open (in X). Hence intersections of open subsets are again open (in X+).

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If {Ki}i∈I is a family of compact closed subsets of X then ∩i∈IKi is closed, hencecompact, and so the union ∪i∈I(X

+ −Ki) is open (in X+).

It is now straightforward (but slightly tedious) to verify that the above rules definea topology.

Let U be an open cover of X+. Then ∞ ∈ U∞ for some U∞ = X+ −K in U . AsU − {U∞} is an open cover of the compact subset K we see that a finite subset ofU covers X+.

Suppose that X is locally compact and Hausdorff. Since open subsets of X remainopen in X+ distinct points of X have disjoint neighbourhoods in X+. If x ∈ X

has a neighbourhood N with compact closure then IntN and X+ − N are disjointopen subsets of X+ containing x and ∞, respectively. Thus X+ is Hausdorff.

Conversely, if X+ is Hausdorff then so is X. If x ∈ X and ∞ are contained indisjoint open subsets U and X+ −K then U ⊆ K and so U has compact closure.Thus X is locally compact.

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