1.1 Topological Preliminaries
The following definitions are subject to the assumption where the topological space is defined to be . This is satisfactory to the main purpose of our proceeding passage, but it is noteworthy that it can be generalized to more abstract sets (which will be discussed in later sections, especially in Section 3.2.1).
Definition 1.1.1 (Open Ball).
Let denote the -dimensional ball with radius centered at , defined by
Definition 1.1.2 (Open and Closed Sets).
A set is open if for every , there exists an open ball centered at that is fully contained in . A set is closed if its complement, , is open.
(Note that these definitions hold only for the standard topology on induced by the Euclidean metric. For now, this is unimportant, but will be explained later in Section 3.2.1.)
Definition 1.1.3 (Accumulation Point).
A point is an accumulation point of if for any open set containing ,
Theorem 1.1.1.
A set is closed iff contains all of its accumulation points.
Proof.
- We first prove that closedness implies the inclusion of accumulation points. Let be a closed set, and assume that some accumulation point of satisfies . Since is closed, is open, and there exists an open ball that is fully contained in . However, this contradicts the definition of accumulation point, since does not intersect with .
- Assume is an arbitrary set which includes all its accumulation points. We will show that is open, which implies that is closed. Let . Since is not an accumulation point of (as otherwise ), there exists an open ball such that . Then, , and hence is open.
Definition 1.1.4 (Closure).
For a set , define the closure of , or to be the intersection of all closed sets containing . In other words, it is the union of and its accumulation points.
Because the accumulation points of all lie in , the closure of is a closed set. Moreover, is closed iff .
Definition 1.1.5 (Interior).
For a set , the interior of , denoted , is the union of all open sets contained in , or the set of points such that there exists an open neighborhood of that is fully contained in .
Definition 1.1.6 (Compact Set).
A set is compact iff is closed and bounded.
Definition 1.1.7 (Set Covering).
A cover of a set is a collection of sets such that
A cover is open if every set in the collection is open.
Theorem 1.1.2 (Bolzano–Weierstrass).
Every infinite subset of a compact set has an accumulation point in .
Proof.
Since is bounded, there exists a closed cube such that .
Bisect into congruent sub-cubes. Since is infinite and the sub-cubes are finite in number, at least one of the sub-cubes contains infinitely many points of , and choose one to be .
Bisect into sub-cubes, and choose a sub-cube that contains infinitely many points of . We then obtain the recursive sequence
Because the side lengths shrink to zero and the cubes are nested, the intersection
consists of exactly one point. Call this point .
For each , contains infinitely many points of . Because the side length of tends to zero, for any , such that , .
Then, also contains infinitely many points of . Therefore, is an accumulation point of .
We now show that . Suppose for contradiction that . Since is closed, is open, and such that
But then, for sufficiently large , we have , and hence . This contradicts the construction of , which ensures that contains infinitely many points of .
Theorem 1.1.3 (Heine–Borel).
A set is compact iff every open cover has a finite subcover.
Proof.
We will first show that any set satisfying the condition is compact.
First we will show that is bounded. Suppose that , where . Consider the collection of open sets
forms an open cover of . Then by the assumption, there exists a finite subcover in , namely which covers . Then,
By contradiction, must be bounded.
must also be a closed set. For the sake of contradiction, assume that there exists a point . Since , the following open collection of sets covers :
There then exists a finite subcover . Then,
and that . However, by the definition of the accumulation point, every open neighborhood of the accumulation point must intersect . Therefore, by contradiction, is closed.
We then prove the converse. By the assumption that is bounded, such that is contained within the closed cube
Assume that there exists an infinite open cover of without finite subcovering.
Bisect into sub-cubes (for real and complex parts), choose such that has no finite subcover of .
Under the previous assumptions, this is possible since if every had finite subcovering, then would have finite subcovering. Similarly, choose by bisecting similarly, and recursively obtain a sequence of cubes:
Since the side length of each cube tends to 0, consists of a single point . Since covers , such that . Since is open, such that . such that , . Then taking the intersection with on both sides,
This contradicts the assumption that for every , has no finite subcovering, since clearly covers , as it is a single open set that covers a nonempty subset. Therefore by contradiction, every open cover has finite subcovering.
Definition 1.1.8 (Support of a Function).
For a set and a function , the support, denoted by , is the closure of the set for which is nonzero.
Remark.
A notable classification of functions comes from the compactness of support—more specifically, its boundedness. Compactly supported functions in are commonly referred to as bump functions (see Section 3.2.1).