It turns out that metric spaces are not sufficiently general for some applications. This section is a brief introduction to general topology. In making this generalization, the properties of balls which are the conclusion of Theorem 6.1.4 on Page 402 are stated as axioms for a subset of the power set of a given set which will be known as a basis for the topology. More can be found in [?] and the references listed there.
1.) Whenever p ∈ A ∩ B for A,B ∈ℬ, it follows there exists C ∈ℬ such that p ∈ C ⊆ A ∩ B.
2.) ∪ℬ = X.
Then a subset, U, of X is an open set if for every point, x ∈ U, there exists B ∈ ℬ such that x ∈ B ⊆ U. Thus the open sets are exactly those which can be obtained as a union of sets of ℬ. Denote these subsets of X by the symbol τ and refer to τ as the topology or the set of open sets.
Note that this is simply the analog of saying a set is open exactly when every point is an interior point.
Proposition 6.8.2 Let X be a set and let ℬ be a basis for a topology as defined above and let τ be the set of open sets determined by ℬ. Then
Proof: If p ∈∅ then there exists B ∈ℬ such that p ∈ B ⊆∅ because there are no points in ∅. Therefore, ∅∈ τ. Now if p ∈ X, then by part 2.) of Definition 6.8.1 p ∈ B ⊆ X for some B ∈ℬ and so X ∈ τ.
If C⊆ τ, and if p ∈∪C, then there exists a set, B ∈C such that p ∈ B. However, B is itself a union of sets from ℬ and so there exists C ∈ℬ such that p ∈ C ⊆ B ⊆∪C. This verifies 6.8.12.
Finally, if A,B ∈ τ and p ∈ A ∩ B, then since A and B are themselves unions of sets of ℬ, it follows there exists A1,B1 ∈ℬ such that A1 ⊆ A,B1 ⊆ B, and p ∈ A1 ∩ B1. Therefore, by 1.) of Definition 6.8.1 there exists C ∈ℬ such that p ∈ C ⊆ A1 ∩ B1 ⊆ A ∩ B, showing that A ∩ B ∈ τ as claimed. Of course if A ∩ B = ∅, then A ∩ B ∈ τ. This proves the proposition.
Definition 6.8.4 A topological space is said to be Hausdorff if whenever p and q are distinct points of X, there exist disjoint open sets U,V such that p ∈ U,q ∈ V . In other words points can be separated with open sets.
Definition 6.8.5 A subset of a topological space is said to be closed if its complement is open. Let p be a point of X and let E ⊆ X. Then p is said to be a limit point of E if every open set containing p contains a point of E distinct from p.
Note that if the topological space is Hausdorff, then this definition is equivalent to requiring that every open set containing p contains infinitely many points from E. Why?
Proof: Suppose first that E is closed and let x be a limit point of E. Is x ∈ E? If x
Now suppose E contains all its limit points. Is the complement of E open? If x ∈ EC, then x is not a limit point of E because E has all its limit points and so there exists an open set, U containing x such that U contains no point of E other than x. Since x
Proof: If x≠p, there exist open sets U and V such that x ∈ U,p ∈ V and U ∩V = ∅. Therefore,
Note that the Hausdorff axiom was stronger than needed in order to draw the conclusion of the last theorem. In fact it would have been enough to assume that if x≠y, then there exists an open set containing x which does not intersect y.
Definition 6.8.8 A topological space (X,τ) is said to be regular if whenever C is a closed set and p is a point not in C, there exist disjoint open sets U and V such that p ∈ U,C ⊆ V . Thus a closed set can be separated from a point not in the closed set by two disjoint open sets.
Definition 6.8.9 The topological space, (X,τ) is said to be normal if whenever C and K are disjoint closed sets, there exist disjoint open sets U and V such that C ⊆ U,K ⊆ V . Thus any two disjoint closed sets can be separated with open sets.
Lemma 6.8.11 The above definition is well defined.
Proof: Let C denote all the closed sets which contain E. Then C is nonempty because X ∈C.
an open set which shows that ∩C is a closed set and is the smallest closed set which contains E.
Proof: Let x ∈E and suppose that x
Now E ⊆E so suppose x is a limit point of E. Is x ∈E? If H is a closed set containing E, which does not contain x, then HC is an open set containing x which contains no points of E other than x negating the assumption that x is a limit point of E.
The following is the definition of continuity in terms of general topological spaces. It is really just a generalization of the ε - δ definition of continuity given in calculus.
Definition 6.8.13 Let (X,τ) and (Y,η) be two topological spaces and let f : X → Y . f is continuous at x ∈ X if whenever V is an open set of Y containing f(x), there exists an open set U ∈ τ such that x ∈ U and f(U) ⊆ V . f is continuous if f−1(V ) ∈ τ whenever V ∈ η.
You should prove the following.
Proposition 6.8.14 In the situation of Definition 6.8.13 f is continuous if and only if f is continuous at every point of X.
Definition 6.8.15 Let (Xi,τi) be topological spaces. ∏ i=1nXi is the Cartesian product. Define a product topology as follows. Let ℬ = ∏ i=1nAi where Ai ∈ τi. Then ℬ is a basis for the product topology.
Theorem 6.8.16 The set ℬ of Definition 6.8.15 is a basis for a topology.
Proof: Suppose x ∈∏ i=1nAi∩∏ i=1nBi where Ai and Bi are open sets. Say
Then xi ∈ Ai ∩ Bi for each i. Therefore, x ∈∏ i=1nAi ∩ Bi ∈ℬ and ∏ i=1nAi ∩ Bi ⊆ ∏ i=1nAi.
The definition of compactness is also considered for a general topological space. This is given next.
In general topological spaces there may be no concept of “bounded”. Even if there is, closed and bounded is not necessarily the same as compactness. However, in any Hausdorff space every compact set must be a closed set.
Proof: Suppose p
If K is assumed to be compact, there are finitely many of these sets, Ux1,
Definition 6.8.19 Suppose
The complement is taken with respect to
The reason this is called a compactification is contained in the next lemma.
To see the last claim, suppose U contains ∞ since otherwise there is nothing to show. Notice that if C is a compact set, then X ∖ C is an open set. Therefore, if x ∈ U ∖
Proof: Suppose to the contrary that ∅ = ∩K. Then consider
It follows C is an open cover of K0 where K0 is any particular element of K. But then there are finitely many K ∈K, K1,
Proof: Suppose first that X is compact. Then if C is an open cover consisting of basic open sets, it follows it admits a finite subcover because these are open sets in C.
Next suppose that every basic open cover admits a finite subcover and let C be an open cover of X. Then define
In fact, much more can be said than Lemma 6.8.23. However, this is all which I will present here.