Math 500: Topology

Homework 1
Lawrence Tyler Rush
<me@tylerlogic.com>
January 12, 2013
http://coursework.tylerlogic.com/courses/math500/homework01

Problems

P-1

Let X be a topological space with some subset A such that for all x ∈ A there exists an open set U such that
U ⊂ A and x ∈ U
(P-1.1)

Let U be the set ⋃ x∈AUx where Ux is a particular, yet arbitrary, set for which the property in P-1.1 holds with respect to x. As U is the union of subsets of A, clearly U ⊂ A. Also for each a ∈ A, a ∈ Ua ⊂U, so A ⊂U. Thus A equals U, a union of open sets, i.e. A is open.

P-2

Let X be some set with a collection of subsets, T , such that U ∈T if X \ U is countable or is the set X itself. Since X \∅ is X, than ∅ in T and X is in T since its complement in itself is the empty set, which is countable.

Now let {Uα} be an arbitrary collection of sets in T . The complement of the union of these sets, X \⋃ αUα, is equal to ⋂ α(X \ Uα) by DeMorgan, which has a cardinality that is no larger than that of its largest set. As each X \ Uα is countable, then ⋂ α(X \ Uα), and therefore X \⋃ αUα, is countable. So T is closed under arbitrary union of its elements.

Let {Ui} be a finite collection of sets in T . The complement of the intersection of these sets, X \⋂ iUiA, by DeMorgan, is also given by ⋃ i(X \Ui). This set, however, is a countable union of countable sets and thus has countable cardinality. So T is closed under finite union of its elements.

With the above two properties of T and, as we saw, the fact that it contains ∅ and X, then T is a topology on X.

Exercises

E-1

Let X be a set with A a collection of subset of X. Allow x ∈ X \⋃ A∈AA. Then for all A ∈A, x is not in A. In other words x is in each and every X \ A; so x is in ⋂ A∈A(X \ A). This gives us X \⋃ A∈AA ⊂⋂ A∈A(X \ A).

Now let x ∈⋂ A∈A(X \ A). Thus for all A ∈A, x is in X \ A. Then there is no A for which x ∈ A, so x is in X \⋃ A∈AA. Hence ⋂ A∈A(X \ A) ⊂ X \⋃ A∈AA.
 
With these two results, we have that

⋃ ⋂ X \ A = (X \ A) A ∈A A∈A

which we can use to prove that

⋂ ⋃ X \ A = (X \ A). A∈A A∈A

Using a little complement-trickery, we have that X \⋂ A∈AA is equal to X \⋂ A∈A(X \ (X \ A)), which from our above proof, we know to be equal to X \ (X \⋃ A∈A(X \ A)). However, this is simply ⋃ A∈A(X \ A)). Thus we have

⋂ ⋃ X \ A = (X \ A) A ∈A A∈A

as desired.

E-2

The following matrix has entries with values of ⊂, ⊃, =, and ⊃⊂ standing for strictly finer than, strictly courser than, equal to, and not comparable. The entry at the ith, jth position corresponds to the relationship between the ith and jth subsets of the book’s Figure 12.1. We will number the topologies in the book’s figure from left-to-right, top-to-bottom. Given this, the relationships are as follows, leaving out the redundancies.
= ⊃ ⊃ ⊃ ⊃ ⊃ ⊃ ⊃ ⊃ = ⊃ ⊂ ⊃ ⊂ ⊃ ⊂ ⊃ ⊂ ⊃ ⊂ ⊂ = ⊃ ⊃ ⊂ ⊂ ⊃ ⊃⊂ ⊂ = ⊃ ⊂ ⊂ ⊃⊂ ⊂ ⊂ = ⊃ ⊂ ⊃⊂ ⊃⊂ ⊂ = ⊃ ⊃⊂ ⊂ = ⊂ ⊂ = ⊂ =

E-3

(a)

Certainly ∅ and ℝ are in T since the empty set satisfies the required property as it has no such x, and for ℝ any interval that contains such an x will be contained in ℝ.

So let {Uα} be a subset of elements of T of arbitrary size. Allow x to be an element of ⋃ αUα. Then x is in some Uα and since Uα ∈T , we can find an interval (a,b) such that x ∈ (a,b) ⊂ Uα. Therefore (a,b) is also contained in ⋃ αUα and so it’s an element of T . Thus T is closed under arbitrary union.

Now let {Ui} be a finite collection of elements of T . If ⋂ iUi is the empty set, then its in T as per above, so let it be nonempty. Let x ∈⋂ iUi. Then there is a collection {(ab)i} of intervals such that for each i, x ∈ (a,b)i ⊂ Ui. This then implies that x ∈⋂ i(a,b)i, but ⋂ i(a,b)i ⊂ Ui for each i. Hence x ∈⋂ i(a,b)i ⊂⋂ iUi, and since ⋂ i(a,b)i is an open interval, then ⋂ iUi ∈T ; i.e. it is closed under finite union.

(b)

Let U be in the finite complement topology on ℝ. Then X \ U is finite and denote it by {a1,…,an} where ai < aj for i < j. Let x ∈ U. Then either x < a1, x > an, or a1 < x < an. If x < a1, then x ∈(x- 1,a - |a1--x|) 1 2⊂ U and likewise x ∈(a + |an-x|,x + 1) n 2⊂ U when x > an. Now let a1 < x < an. Here we can find i such that ai is the greatest element in X \ U which is less than x. With this element we have that x ∈( ) ai + |ai-2-x|,ai+1 - |ai+12-x|⊂ U. Thus in all three cases we can find an interval contained in U, so U must also be in T . Hence T is finer than the finite complement topology.