In this brief post, we investigate the topological foundations of analysis. We limit the scope to the topology of a metric space. In particular, we provide the knowledge that is needed for the application to probability and measure theory.
A good introduction for all German-speakers is also provided by Prof. Dr. Edmund Weitz as embedded in the Appendix.
The objective of the next section is to prepare the very basic concept for the definition of a topological space. We start by defining an open and a closed ball, a neighborhood, interior point, an open and a closed set. All these definitions are interlinked to each other as we will see later on.
Parts of this post are based on  and .
Basic Definitions & Properties
Balls, Open & Closed Sets & Neighborhoods
Let denote a metric space.
An (open) ball of radius and center is the set of all points of distance less than from , i.e. .
A closed ball of radius is the set of all points of distance less than or equal to from , i.e. .
A subset of is called a neighborhood of if there is some such that .
An element is called an interior point of if there is a neighborhood of such that .
A set of is called open if every point of is an interior point.
The introduced concepts “interior point” and “open set” (and hence also the concept of “closed sets”) depend on the surrounding metric space . It is sometimes useful to make this explicit by saying “ is an interior point of with respect to ” or “ is open in “.
By definition, every superset of a ball is considered to be a neighborhood. Neighborhoods can be defined without mentioning explicitly the corresponding metric space.
An open ball in is an open interval . However, if we consider embedded in , then is apparently not an open ball in .
For the sake of illustration we sketch a 2-dimensional open ball (i.e. an open interval) as a rectangle even though it would actually be a line without end-points. These end-points are illustrated by the lines crossing the x-axis.
The heuristic reasoning for that is that a one-dimensional interval cannot contain an open two-dimensional ball not matter how small the radius of this ball might be.
Apparently, is an interior point of if and only if there is some real such that . In addition, is open if and only if is a neighborhood of each of its points.
Let us consider open and closed balls in along with the standard Euclidean metric . Every open ball , in is an open set. Let be an arbitrary point. Set and consider with .
For all we have
and so is contained in . This shows that the arbitrary point is an interior point of and thus an open ball is an open set.
That is, a set is open if and only if for every an open ball exists such that .
Definition 1.2 (Boundary Point)
Let and . Then, is called boundary point for if
for all . The set of all boundary points is denoted by .
Boundary points are just points on the boundary between the set and the surrounding basis set (i.e. of the metric space .
can be contained in or in . If all boundary points are outside of , i.e. if , then it is an open set.
Definition 1.3 (Closed Set)
A set of is called closed in if its complement is open in .
Equivalently, if all boundary points of are inside the set , i.e. if , then it is a closed set. This also means that the set is always a closed set no matter what is. The set is then also called the closure of .
(a) The interval is closed in .
Let denote a real number that is not an element of . Then, and the ball with has empty intersection with .
Hence, the complement is open and so the set is closed.
For the sake of illustration we sketch a 2-dimensional open ball as a rectangle even though it would actually be a line without end-points.
Note that all closed intervals in are of the form , , , and .
Let us now consider the set . For all and we have . Hence, the set is open in .
Note that all open intervals in are of the form , , , and .
(b) Now consider the metric space with with the same distance function , as in (a).
The set is an open set since for all we consider the open ball with . That is, we determine the shortest distance between and the two points and and divide it by 2. Then, the corresponding open ball is contained within the set . For the point , we can consider the open ball with , for instance. This open ball equals since all the figures right from until 5 are not contained in . Hence, is open.
The set is also closed since is included in . Note that is not a boundary point since it is not an element of the metric space . The point is not a boundary point since for , for instance, holds true.
(c) Let us consider the metric space of all integers together with the discrete metric for all . It follows that the open ball of radius and center at is either if or if .
(d) Let us consider open and closed balls in along with the standard Euclidean metric . Every closed ball , in is a closed set. That is, we need to show that the complement set is open. Let . Set and consider the open ball , that does not intersect with the closed ball due to the choice of : Let , then the following holds due to the triangular property.
and because of .
Note that this also means that every isolated point in along with the Euclidean metric is a closed set.
In the following section we are going to outline basic properties of open & closed sets, which will provide the heuristic for the definition of a topological space.
Properties of Open & Closed Sets
The following two propositions state basic properties of open and closed sets. It turns out that these properties characterize a topological space.
Proposition 1.1 (Open Sets)
Let be a family of open sets.
(ii) if for an arbitrary index set ;
(iii) if for a finite index set .
(i) The empty set possesses every property since it does not have an element. Hence, the empty set is closed and open. Correspondingly, the set is open.
(ii) If then there must be an index such that with . All the more holds true.
(iii) Let with be open sets. Since there exists a for each with . If we set then and the assertion follows.
Why (iii) holds only true for finite index sets can also be seen if we try to prove that the intersection of two open balls at is open again. Just chose for the open ball of the intersection . This can be iterated provided that the index set is finite.
Please note that the proof above would also be valid if we use neighborhoods instead of balls. This would mean that the proof is not dependent on the used metric space and suggests that we could derive a new mathematical structure from it, where –in general– no algebraic operations, distance functions or orders need to be defined. However, given that we restrict our-self to topology for metric spaces, we will not further outline this idea.
The proof of the corresponding proposition for closed sets is apparent given the definition of a closed set.
Proposition 1.2 (Closed Sets)
Let be a family of closed sets.
(ii) if for an arbitrary index set ;
(iii) if for a finite index set .
Apply the definition and Proposition 1.1 (Open Sets).
Let us continue Example 1.3. The intersection of a finite number of intervals is closed in . Without loss of generality, let . All the sets are closed and so is in .
Let us now consider the family of open sets . For all elements of all open sets , we can find an open ball . Hence, the union of these sets also contain open balls for all elements and is thus an open set itself.
An infinite intersection of open sets need not to be open. Consider in . Apparently, there cannot be an such that .
Correspondingly, infinite unions of closed sets need not to be closed. To this end, consider in .
Accumulation and Limit Points
Closely related to closed sets are so-called accumulation and limit points. In addition, this section will pave the way to one of the below application of a topological space — limits.
Definition 1.2 (Accumulation & Limit Point)
Let and . We call an accumulation point of if every neighborhood of in has non-empty intersection with .
The element is called limit point of if every neighborhood of in contains a point of other than . Finally we set
Every point within an open set but also every limit point of this set is an accumulation point as illustrated in the following example. Here, is the open interval in the standard metric space .
Every point in is an interior, an accumulation and a limit point of . Note that an interior point of requires to have a neighborhood that is contained in . Given that is an open set this holds true for all points within the set by definition.
The ‘boundary’ points and are no interior points but limit and accumulation points of . Why?
Because we can define any open ball around it and the intersection with will always contain a point different from .
Let us now consider a set with a single isolated point in the metric space . Is , that is the only element of the set, an accumulation and/or a limit point of ?
Apparently, every neighborhood of in has non-empty intersection with , i.e. . However, is not a limit point since the intersection does not contain a point other than .
According to our thoughts above, elements of are accumulation points of . This fact is also reflected in the next proposition.
Proposition 1.3: Let be a subset of a metric space .
(ii) if and only if is closed.
Proof: (i) See argumentation above.
(ii) ‘ Let . Since is not an accumulation point of , there is some of all neighborhoods of such that . Thus, , that is, is an interior point of . Consequently is open and is closed in .
‘ Let be closed in . Then is open in according to its definition. For any , there is some such that . This means that and are disjoint, and so is not an accumulation point of , i.e., . Hence, we have proved the inclusion , which is equivalent to . With (i), this implies .
The limit points of a set are the limits of certain sequences in .
Proposition 1.4: An element of is a limit point of if and only if there is a sequence in which converges to .
Proof: Let be a limit point of . For each , choose some element in . Then is a sequence in such that .
Conversely, let be a sequence in such that . Then, for each neighborhood of , there is some such that . This means that . Hence, each neighborhood of contains an element of other than .
Corollary 1.1: An element is an accumulation point of if and only if there is a sequence in such that .
Proof: If is a limit point, then the claim follows from Proposition 1.4. Otherwise, if is an accumulation point, but not a limit point of , then there is a neighborhood of such that . Thus, , and the constant sequence with for all has the desired property.
We can also characterize closed sets using convergent sequences by simply using the fact that a closed set is the complement of an open set.
Proposition 1.5: For , the following are equivalent:
(i) is closed;
(ii) contains all its limit points;
(iii) Every sequence in , which converges in , has its limit in .
Proof: ‘(i) (ii)’ Any limit point of is also an accumulation point and so is contained in . Since is closed and Proposition 1.3, , and so all limit points are in .
‘(ii) (iii)’ Let be a sequence in such that in . Then, by Corollary 1.1, is an accumulation point of . This means that, either is in , or is a limit point of , so, by assumption, .
‘(iii) (i)’ This implication follows from Proposition 1.3 and Corollary 1.1.
Compact sets are so important since they possess important properties, that are known from finite sets:
- Set is bounded;
- Set contains a maximal and minimal element;
- An infinite sequence contains a constant subsequence.
The famous Heine-Borel Theorem will show that compact sets in metric spaces do indeed have these properties. However, we will only partly prove this theorem but will provide links and references to full proofs.
The analogy is also outlined in this really nice video (in German only) by Prof. Dr. Edmund Weitz.
Let us start with very basic but important definitions.
Definition 1.3 ((Open) Cover and Open Sub-Cover)
By an open cover of a set in a metric space , we mean a collection of (open) subsets of such that .
A sub-collection with of an (open) cover is called an open sub-cover of .
An (open) cover of is a collection of possibly overlapping open sets in which, after considering their union globally, contains the set inside.
Example 1.6 (Real Line)
Let us consider the set of the metric space . Then, is an open cover of because . This cover apparently contains infinitely many open sets .
is an open subcover of since but . This subcover, however, still contains infinitely many open sets.
Now we have the ingredients for the central definition of this section.
Definition 1.3 (Compact Set)
A subset of a metric space is said to be compact if every open cover of contains a finite subcover.
The definition of a compact set contains a very powerful requirement: it requires the existence of a finite open subcover within any possibly infinite open cover. Please also refer to the analogy to the properties of finite sets as mentioned at the beginning of this section.
It is clear that every finite set is compact.
Example 1.7 (Finite Set & Compactness)
Let be a finite set in the metric space and . Note that we could chose any for this purpose. Apparently, is a finite and open cover of .
Each point of the finite set is contained in some open set by the cover depending upon that point. This directly implies that there is a finite sub-cover most the size of the set .
Let us generalize the approach taken in Example 1.7 for an arbitrary non-empty set . By setting
with , an open set is assigned to each point . Hence, the set is an open cover for any set .
The Heine-Borel theorem states that every closed and bounded set in , is compact. This theorem is very important for analysis, measure theory and beyond.
Theorem 1.1 (Heine-Borel)
Let be a subset of the standard Euclidean metric space with . Then the following statements are eqivalent:
(i) is compact;
(ii) is closed and bounded;
(iii) Every infinite subset of has an accumulation point in .
Part of the Proof for :
First, we assume that is compact i.e. that every open cover admits a finite sub-cover. Define . Then simply because the index set is itself and with is a open cover and thus there exist a finite sub-cover , , , with and . Note that with . Since this interval is bounded so is .
Now, let us assume that is not closed (i.e. complement is open). Then, there must be a limit point of with since a closed set contains all its limit points [according to Proposition 1.5]. Find such that . Define . Apparently, and thus because of the assumption that is compact there must be a finite sub-cover , , , with and . Set and find a such that (which is possible since ). According to the definition of this implies , which is a contradiction. Hence, is closed.
For a general proof of the theorem for please refer to Theorem 3.31 in  or see there. Also note that a visualization by the University of Hannover of one proof can be found on YouTube. There is also a proof analysis of one direction of the Heine-Borel Theorem.
Topological spaces can be defined in many ways, however, given our focus on metric spaces and open sets the following definition is the most natural one. Please note that we could also use the properties of closed sets to define a topological space.
Definition 2.1 (Topological Space)
A topological space is a pair , where is a set and is a family of subsets that satisfies
(ii) if for an arbitrary index set ;
(iii) if for a finite index set .
The following video provides a rather unorthodox way of thinking about a topology. However, it might help to get a heuristic for topological spaces. It also mentions the connection between metrics and a topology.
Example 2.1 (Topologies):
a) If is a metric space and is the set of all open sets, then is a topology according to Proposition 1.1.
b) Let be a set and then is the so-called trivial or indiscrete topology.
c) The power set of a set is the so-called discrete topology. In this topology every subset it open.
Continuous Functions, Open and Closed Sets
Now, we connect two important concepts via the following theorem.
Theorem 3.3 (Continuous Functions & Closed/Open Sets):
Let be a function between metric spaces and . Then the following are equivalent:
(i) is continuous;
(ii) is open in for each open set ;
(iii) is closed in for each closed set .
Proof: ‘(i) (ii)’ Suppose is continuous on and let be an open set. If , then the claim follows from Proposition 1.1 (i). Thus, we suppose that .
Let . Due to the fact that is open, we can chose at least one such that for every . Given that is continuous on , a corresponding exists such that
and . That is, implies . Thus, , which means that is an open set.
‘(ii) (i)’ Let us suppose that is open whenever is open and we aim to prove that is continuous at each point of . Given and , we know that the ball is open in . By assumption so is . Since and is an open set, there must be a such that . This proves the assertion.
Please also refer to some counterexamples illustrating that the image of an open (closed) set under a continuous function does not need to be open (closed).
For all German-speaker, I recommend to listen to the following quite illustrative video about metric spaces and topologies.
In addition, the next video introduces open and closed sets in an illustrative manner.