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Problem Set 2: Topological spaces

Your name:

Due: Thursday, February 11

Problem 1 (7). Let (M, d) be a metric space, and let x be a point in M . Showthat the subset M \ {x} is open in the metric topology τ d.

Solution 1. To check this we have to show that for any y ∈ M \ {x}, there

exist an

> 0 such that B(y,

) ⊂ M \ {x}, i.e. x 6∈ B(y,

). But as y 6= x,hence we can just take = d(x, y) > 0.

Problem 2 (12). Let X be a space.

a. Suppose (τ i)i∈I is a family of topologies on X indexed by I . Prove thatTi∈I τ i is a topology on X .

b. Suppose τ , τ 0 are topologies on X . Is τ ∪τ 0 a topology on X ? Justify yourclaim.

c. Let A be a basis for a topology on X , and let I be the collection of topologies τ on X such that A ⊂ τ . Prove that τ A =

Tτ ∈I τ . In other

words, τ A is the coarsest topology that contains A. Is this true if A isonly a sub-basis for a topology on X ?

Solution 2. a. We have to check the axioms. X, ∅ ∈ T

i∈I τ i as they arein all τ i. Let (U j)j∈J be a family of elements in

Ti∈I τ i. Then it is a

family of elements of any τ i and soS

j∈J U j ∈ τ i for all i. ThusS

j∈J U j ∈Ti∈I τ i.The third axiom is checked similarly; given U, V ∈

Ti∈I τ i, we

have U, V ∈ τ i for all i; hence, U ∩ V ∈ τ i as τ i is a topology. But thisimplies U ∩ V ∈

Ti∈I τ i. Hence,

Ti∈I τ i is a topology.

b. The answer is no. A simple counterexample is as follows: Let X = R,A = R≥0 ⊂ R and B = (−1, 0] ⊂ R. Let τ = {∅, A , X } and τ

0 = {∅, B , X },they can easily seen to be topologies. Then both A and B are in τ ∪ τ 0

but A ∩ B = {0} 6∈ τ ∪ τ 0. Hence, it is not a topology. Union axiom failsas well.

c. Let τ 0 =T

τ ∈I τ . As A ⊂ τ A, τ A ∈ I . Thus, τ 0 ⊂ τ A. On the other hand,given τ ∈ I and U ∈ τ A, we can write U as a union of elements of A ⊂ τ ,hence as a union of open subsets of τ . This implies U itself is in τ . Hence,τ A ⊂ τ for all τ ∈ I , i.e. τ A ⊂

Tτ ∈I τ = τ 0. Thus, τ A = τ . The same

holds for sub-bases A, if we define τ A to be the topology with basis theset of finite intersections of A.

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Problem 3 (18). A totally ordered set is a set X together with a subset R ⊂X × X satisfying the following properties:

• for every x ∈ X, (x, x) does not belong to R;

• for every x, y ∈ X , exactly one of (x, y) and (y, x) belongs to R.

• for every x, y,z ∈ X , if (x, y) and (y, z) belong to R then (x, z) belongs toR.

We often write x < y as an abbreviation for (x, y) ∈ R. The subset R is calledthe total ordering . Given a, b ∈ X , we define the following subsets of X :

(−∞, a) ≡ {x ∈ X : x < a}

(a, ∞) ≡ {x ∈ X : a < x}

(a, b) ≡ (a, ∞) ∩ (−∞, b).

Denote by B the collection of all such subsets, as well as X itself.

a. Show that B is a basis for a topology τ R on X .

b. Recall that R has a total ordering, x < y ⇐⇒ y−x is positive. Prove thatthe order topology on R coincides with the standard (metric) topology onR.

c. Note that R2 = R × R has a total order defined as follows:

(a, b) < (c, d) ⇐⇒ a < c or (a = c and b < d).

This is called the lexicographical order (think of how words are ordered

in the dictionary). Denote this ordering L ⊂ R2 × R2. Observe that thesubset C = [0, 1] × [0, 1] ⊂ R2 inherits a total ordering from R2, which wedenote by T ⊂ C ×C . This means we have three topologies on [0, 1]×[0, 1]:

• (τ L)[0,1]×[0,1], the subspace topology from the ordering on R2;

• τ T , the topology from the induced ordering on [0, 1] × [0, 1];

• the subspace topology on [0, 1] × [0, 1] from the standard topology onR

2.

Compute the closure of the set A = {(x, 0) : x ∈ [0, 1)} in each of thesetopologies.

Solution 3. a. One needs to check that finite intersections of elements of B

can be covered by elements of B . But clearly intersection of two elements,is again in B , for instance (a, b) ∩ (c, d) = (max{a, c},min{b, d}). Hence,B is a basis.

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b. Let τ d and τ o denote the metric and order topologies respectively. Clearly,elements of B are open in τ d: finite intervals are balls centered around their

mid-points and infinite ones are infinite unions of finite intervals. Hence,τ o ⊂ τ d as in the previous question. On the other hand, given U ∈ τ d wecan form U o =

S(a,b)⊂U , finite interval(a, b) ⊂ U and this set is open in order

topology. But given x ∈ U , there exist an > 0 such that (x−, x+) ⊂ U so (x − , x + ) ⊂ U o. Thus U = U o is also open in order topology andτ d = τ o.

c. • For any x ∈ [0, 1), ((x, 0), (x, 2))∩ C is open in the subspace topologyfor the order topology on R2. But this set is just {x} × (0, 1]. Sim-ilarly, {1} × [0, 1] = ((1, −1), (1, 2)) ∩ C , where the interval is takenwith respect to given order on R2, is open. But, these sets cover allthe compliment of A in C . Hence A is closed and its closure is itself.

• Any basic open set, i.e. an open interval, in the complement would besupported in a single x-coordinate, otherwise it would intersect theelements of A with x-coordinate in between. So the basis elementsin the complement are subsets of the intervals ((x, 0), (x, 1)), for 0 ≤x

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hand, U × {0} ∩ V × {1} = ∅, B ̀ satisfies the desired property; hence, itis a basis.

c. The elements of B Q are open in the metric topology: If (x, y) ∈ U × V ,where U and V are open, then there exist , 0 > 0 such that B (x, ) ⊂ U and B(y, 0) ⊂ V . Then it is easy to show that B((x, y),min{, 0}) ⊂U × V . As (x, y) ∈ U × V was arbitrary, this shows that the sets U × V are metric open. Thus, all the basis elements are metric open and thetopology they generate is contained in τ d, the metric topology. On theother hand, as open subsets in metric topology are unions of ball, it isenough to show that the balls are open in the product topology. Given(x, y) = b ∈ B(a, 0) ⊂ R2, we can find > 0 such that B (b, ) ⊂ B(a, 0).Then, clearly (x − /2, x + /2) × (y −/2, y + /2) ⊂ B((x, y), ) = B(b, ).As b,a, 0 was arbitrary, this shows the open balls are product open, hencemetric open subsets are open in the product topology. Thus, they are the

same.

Problem 5 (20). The goal of this problem is to show that there are infinitelymany prime numbers. This result is known as Euclid’s theorem, its first recordedproof having been published by Euclid around 300 B.C. The surprising topo-logical proof that we will see here was discovered in 1955 by H. Furstenberg.

Recall that a natural number p ∈ N is prime iff p 6= 1 and if its only divisorsare 1 and p. Let P ⊂ N be the set of all prime numbers.

a. Show that every natural number n 6= 1 is divisible by a prime number.

b. For x, n ∈ Z letx + nZ = {x + nz : z ∈ Z}.

Call a subset U ⊂ Z open iff for every x ∈ U there exists n ∈ N \ {0} suchthat x + nZ ⊂ U . Show that this defines a topology on Z.

c. Show that for every n ∈ N \ {0}, the set nZ is both open and closed in Z.

d. Using (a), show that

Z \ {1, −1} =[

p∈P

pZ.

e. Conclude that P is infinite.

Solution 5. a. Assume the converse and let n 6= 1 be the smallest naturalnumber not satisfying this condition. Every natural number p 6= 1 dividingn should be less than than or equal to n but if p < n then p and hence n

is divisible by a prime number. This cannot happen by assumption; henceevery such p is equal to n, i.e. n is prime. But then it is divisible by aprime, a contradiction; hence, the statement holds.

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b. First notice that the family {x + nZ : x ∈ Z, n ∈ N \ {0}} is closed underfinite non-empty intersection: Indeed, if a ∈ (x + nZ) ∩ (y + mZ), then

(a + nZ) = (x + nZ) and (a + mZ) = (y + mZ) and (x + nZ) ∩ (y + mZ) =(a + nZ) ∩ (a + mZ) = a + gcd(m, n)Z, which is of the same type. Now, tocheck the given family of opens define a topology becomes easy: Opennessof ∅ and Z is vacuously true. Union axiom is also obvious. Hence we onlyhave to check this family of sets is closed under finite intersections. Butif U and V are open, then given x ∈ U ∩ V , we have n and m such thatx + nZ ⊂ U and x + mZ ⊂ V . Thus their intersection x + gcd(m, n)Z isa subset of U ∩ V . This implies U ∩ V is open as well and given familydefines a topology.

c. As x ∈ y + nZ implies x + nZ = y + nZ, the sets x + nZ are open. Inparticular, nZ and its complement

Si=1,...,n−1 i + nZ are open, so nZ is

also closed.

d. Clearly the union is a subset of Z \ {1, −1} for 1 or -1 are not divisible byany prime. On the other hand, given x ∈ Z \ {1, −1}, part (a) implies itis divisible by some p ∈ P, i.e. x ∈ pZ. So the result follows.

e. If P were finite the unionS p∈P pZ would be closed in the topology above,

by part (c), i.e. as pZ is closed. Thus by part (d), {−1, 1} would be open.Thus it would contain 1 + mZ, for some m 6= 1, yet it is finite and this isa contradiction. Thus, P is infinite.

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