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Topology: Lecture 1
MAT246: Concepts in Abstract Math
July 17th 2024
Open and closed sets
Topology begins by studying the namesake of the open (0 , 1) and closed [0, 1]. For this topic we will exclusively study subsets of the n-dimensional real space Rn. A crucial subset is the open ball Ba,r , where a ∈ Rn is its center and r > 0 ∈ R is its radius. Given a subset A ⊂ Rn we define
Definition 1. A point x is said to be in the interior of A if some ball centered at x is a subset of A.
Definition 2. A point x is said to be a limit point of A if every ball centered at x contains at least one point of A.
Consider the open interval (0, 1). The points 0 < x < 1 are clearly interior points and the points 0, 1 are clearly limit points. The difference between (0 , 1) and [0, 1] is on one hand that the closed interval contains all its limits points, and on the other hand the point 0, 1 in the closed interval are not interior points. This leads to the following definitions.
Definition 3. A set A is called open if each of its points is an interior point.
Definition 4. A set A is called closed if it contains all of its limit points.
Operations on open/closed sets
Lets see what happens if you apply standard set operations to open/closed sets.
Lemma 1. The union of any number of open sets is open.
Proof. Given a collection of open sets Ai we want to prove that i(u)Ai is also an open set. By definition of union a point x belongs to i(u)Ai if it belongs to one of the sets Ai.
Since the set Ai is open some ball around x will be a subset of Ai , which means it will also be a subset of the union i(u)Ai.
Lemma 2. The intersection of a finite number of open sets is open.
Proof. Given a collection of open sets A1,..., An we want to prove that Ai is open. By definition a point x in the intersection UAi belongs to each of the i sets Ai. For each of those sets there is a ball centered at x that is contained in that set. Since there is a finite number of these balls, the smallest of these ball will be contained in each of the sets Ai.
We might now want to prove analogous lemmas for closed sets, but instead of doing this directly we will use the following important lemma and de Morgan’s laws.
Lemma 3. A set A is open if and only if its complement Ac is closed.
Proof. Given an open set A we want to show that its complement Ac is closed. Consider a limit point x of the set Ac , if it belongs to A, then there would be a ball Br centered at x that is a subset of A, in particular this ball would not contain any points of Ac , so it would not be a limit point of Ac. This shows that every limits point of Ac must belong to Ac , so Ac is closed.
Given a set A such that its complement Ac is closed we want to prove that A is open. Let x be an arbitrary point in A. We want to find a ball Br centered at x such that Br is a subset of A. If for all r > 0 we have that Br is not a subset of A this would mean that for all r > 0 the ball Br centered atx contains a point in Ac making x alimit point of Ac , but we assumed that Ac was closed, so it must contain all its limit points, in particular no point x in A is a limit point of Ac.
This implies that there must exist an r > 0 such that the ball Br centered at x is a subset of A.
This lemma can be restated in the following way.
Lemma 4. The complement of an open set is closed, the complement of a closed set is open.
We can now prove the lemmas.
Lemma 5. An arbitrary intersection of closed sets is closed.
Lemma 6. A finite union of closed sets is closed.
By using de Morgan’s laws and the lemmas we prove previously for open sets.
Proof. Given a collection of closed sets Ai. By previous lemma their comple-ments Ai(c) are open. To prove that Ai is closed we can instead prove that its complement ( Ai )c which according to de Morgan’s law is equal to i(U)Ai(c) , which is open since it is union of open sets. A similar argument works for a finite union of closed sets.
Exercise 1. What about infinite unions of closed sets or infinite intersections of open sets?
Basics of Rn
The points in Rn can be viewed as vectors and you now from linear algebra that vectors have a norm. The norm of a vector ⃗x = (x1,..., xn ) is the same as the distance from the origin to its endpoint x. This norm is denoted as ||x|| and can be expressed as √x1(2) + ... + xn(2) . The crucial property we will need to prove basic facts about subsets of Rn is the triangle inequality.
Lemma 7. For any two points x,y ∈ Rn we have Ⅱx + yⅡ ≤ ⅡxⅡ + ⅡyⅡ .
Convergence in Rn
Lets rewrite the definition of a sequence (xn ) converge to a point L in a way that works for both R and Rn.
Definition 5. A sequence of points (xn ) converges to the points L ifevery ball centered at L contains all but finitely many points of the sequence.
We can also write this in a more familiar fashion using the norm on Rn.
Definition 6. A sequence of points (xn ) converges to a point L if for every ε > 0 there exists an N such that for all n > N we have Ⅱxn − LⅡ < ε .
Exercise 2. Check that these two definitions are equivalent.
It is important to compare the notion of limit to that of a limit point.
Exercise 3. Show that if a sequence (xn ) converges to a point L, then L is a limit point of the set (xn ).
The converse is not quite true, even if we make sure the limit point in question is not an element of the sequence.
Exercise 4. Construct a sequence which has two limits points that do not belong to it.
A sequence can have many limits points, but it can only have one limit.
Lemma 8. A sequence of points (xn ) can only have one limit.
Proof. Suppose a sequence of points (xn ) converges to a point L. For any other point L′ we can choose small balls one centered at L the other centered at L′ such that these balls do not intersect. By definition of convergence all but finitely many element of the sequence will belong to the ball centered at L, which implies that only finitely many points can belong to the ball centered at L′ . This means that the sequence does not converge to L′ .