The Smirnov- and Bing-Nagata-Smirnov Metrization Theorems
Andreas Granath
Bachelor Thesis, 15hp Bachelor in Mathematics, 180hp
Spring 2019
Department of Mathematics and Mathematical Statistics
Abstract
In this essay we give a sufficient background to, and prove, the classical Bing-Nagata-Smirnov metrization theorem. Thus we show that a topological space is
metrizable if, and only if, it is regular and has a countably locally finite basis. As a corollary of this we also prove Smirnovs metrization theorem.
Sammanfattning
I denna uppsats ger vi en bakgrund till, och bevisar, Bing-Nagata-Smirnovs klassiska metriserbarhetssats. Vi visar således att ett topologiskt rum är metriserbart om, och endast om, det är reguljärt samt har en uppräknelig och lokalt ändlig bas. Som en
följdsats från detta bevisar vi även Smirnovs metriserbarhetssats.
Contents
1. Introduction 1
2. Basic topology 3
2.1. The topology and topological spaces 3
2.2. Separation axioms 6
2.3. Locally finite collections 9
2.4. Continuous functions 12
3. Metrics and metric spaces 19
3.1. The metric and the metric topology 19
3.2. Metrizability 21
4. Metrization theorems 25
4.1. The Bing-Nagata-Smirnov metrization theorem 25
4.2. The Smirnov metrization theorem 28
5. Acknowledgements 31
References 33
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1. Introduction
In the seminal paper [4] by Frechét in 1906, he introduced the concept of a metric and a metric space. Furthermore he related his concept of open sets in metric spaces to neigbourhoods, earlier defined by Georg Cantor in the 1870’s. This clear relation resulted in quite a fruitful program where it was possible to study the properties of semi-abstract spaces using metrics. But, as we shall show in this thesis, the requirement of being a metric space is quite strong. Therefore, the hungarian mathematician Frigyes Riesz proposed in [11], published 1909, that the topology could be defined without the metric.
The definition simply used the concept of limit points which were the building blocks of Cantor’s theory.
Building upon this more abstract framework, in 1914 Felix Hausdorff gives the first definition of a topological space in the seminal textbook [7]. From this point it was known that metrics could generate topologies but the focus were on the new more abstract spaces.
As the modern analysis progressed in parallel with the new subject of abstract topo- logical spaces, the importance of metric spaces was understood and applied to other fields. Since such a breadth of tools to work with was developed for specifically metric spaces, it became attractive to see if one could use these on topological spaces.
Inspired by this, the young prodigy mathematician Pavel Urysohn proved the first of what would be known as metrization theorems. Where a topological space being metriz- able means that there exists a metric defined on the space which generates the given topology on the space. The following result is known as the modern version of Urysohn’s metrization theorem proved in [14] by Urysohn’s friend and collaborator Andrey Ty- chonoff.
Urysohn’s metrization theorem Every regular space X with a countable basis is metrizable.
Although this result is strong, it only provides a sufficient requirement for a space to be metrizable. About twenty-five years later, in what can be described as a miracle year for the subject, three independent proofs were given of the same metrization theorem [1, 10, 12]. This time the theorem provided necessary and sufficient conditions for a space to be metrizable. We will refer to the following theorem as the Bing-Nagata-Smirnov metrization theorem and it will be proved in Section 4.1.
Theorem 4.1.4. A topological space X is metrizable if, and only if, X is regular and has a countably locally finite basis.
The Bing-Nagata-Smirnov metrization theorem is very versatile, but not very handy when working with the spaces known as manifolds. These spaces have a very rich his- tory and had up to the 1950’s been extensively investigated in subjects like differential geometry. In particular, in 1944 the french mathematician Jean Dieudonné published a paper [2], in which he defined a property of a topological space called paracompac- ntess. This property is a generalization of the well-known property of a space being compact, which is highly important in studies on spaces like Rn. Therefore, building upon partial results on paracompact spaces already proven, Smirnov also published the paper [13] in the same year as [12] was published. In this paper he proved what we will
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call the Smirnov metrization theorem, which is a corollary of The Bing-Nagata-Smirnov metrization theorem and is proven in Section 4.2.
Theorem 4.2.4. A topological space X is metrizable if and only if it is a paracompact Hausdorff space that is locally metrizable.
An example of where these metrization theorems are vital is in the quite new subject of analysis on polyfolds. In this subject the Bing-Nagata-Smirnov metrization theorem is crucial in order to verify the existence of metrics on certain sub-structures on manifold- like polyfolds, see for example Corollary 7.2 in [8].
An overview of the essay is as follows. In Section 2 we provide an introduction to general topology, structures on topological spaces and their connections to continuous functions. In Section 3 we introduce metric spaces and the concept of metrizability. In Section 4 we prove the Bing-Nagata-Smirnov metrization theorem, introduce the concept of paracompactness and prove the Smirnov metrization theorem. The essay follows [9].
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2. Basic topology
2.1. The topology and topological spaces. In this section we shall acquaint the reader with the fundamental definitions and lemmas needed to work with what is called point-set topology, aswell as give some concrete examples of topologies. First we shall define the structure giving its name to the subject at hand.
Definition 2.1.1. Let X be a non-empty set, a collection T of subsets of X is called a topology if T satisfies the following:
(1) X ∈ T and ∅ ∈ T , (2) Tn
i=1Xi∈ T for all Xi∈ T , (3) S
α∈JXα∈ T for all Xα∈ T .
A set U is said to be open if it belongs to the topology T of X.
Remark 2.1.2. Note that the third property above holds for general index sets J and not just countable or finite sets.
Definition 2.1.3. Let X be a non-empty set and let T be a topology on X, then the pair (X, T ) is called a topological space.
Remark 2.1.4. We will often supress the bracket notation and simply call X a topological space, or even just a space, letting the topology T be as stated or understood from the context.
Having this definition of a space, we can intuitively from the definition of an open set define a closed set as follows.
Definition 2.1.5. Let X be a topological space and U a subset of X. Then we say that U is closed in X if the set X\U is open in X.
Furthermore, we shall need the concept of the closure of a set.
Definition 2.1.6. Let X be a topological space and U a subset of X. Then we define the closure of U , denoted as ¯U as the intersection of all closed sets containing U .
In general, the topology cannot be explicitly written down in the sense of writing each element since the topology considered may be infinite and sometimes even uncountable.
Thus it is convenient to be able to describe the topology by a collection of elements, to do this we define what we call a basis of the topology.
Definition 2.1.7. Let X be a set, we define a basis for a topology on X to be a collection B of subsets of X, which we shall call basis elements, such that:
(1) For each x ∈ X, there is at least one basis element B containing x.
(2) If x belongs to the intersection of two basis elements B1and B2 in B, then there is a basis element B3 in B containing x such that B3⊂ B1∩ B2.
Proposition 2.1.8. Let X be a non-empty set and let B be a basis of X as defined in Definition 2.1.7. Then B defines a topology T on X and we shall say that B generates the topology T on X.
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Proof. Let X be a non-empty set, let B be a basis on X as defined in Definition 2.1.7 and let T be the collection of sets U such that for each point x ∈ U there is an element B of B such that x ∈ B and B ⊂ U . We shall now show that this collection T is a topology on X and thus that the sets U are open in X.
Now let {Uα}α∈J be an indexed family of elements in T , we shall show that the set U = [
α∈J
Uα
also belongs to T . Let x be a point of X, then by definition there must be an index β such that x ∈ Uβ. Now since Uβ belongs to T there must exist a basis element B such that x ∈ B ⊂ Uβ. Thus we have x ∈ B ⊂ U , and hence U ∈ T by definition.
Finally we turn to intersections. We start by showing that given U1, U2 ∈ T , their intersection also must be an element of T . Let U1, U2 be elements of T such that their intersection is non-empty, otherwise we are vacuously done. Now let x be a point belonging to U1∩ U2. Choose a basis element B1 such that x ∈ B1 and B1 ⊂ U1, also choose a basis element B2 such that x ∈ B2 and B2⊂ U2. Then by the second property of a basis there must be a third basis element B3such that x ∈ B3⊂ B1∩ B2. Thus we have that x ∈ B3⊂ U1∩ U2and hence the intersection must belong, by definition, to T . Now we show by induction that the intersection of a countable collection, say {Uk}nk=1 of elements in T belongs to T . The base case k = 1 follows immediately. Now suppose that it holds for U1∩ U2∩ · · · ∩ Un−1, then by our former result we know that
(U1∩ U2∩ · · · ∩ Un−1) ∩ Un= U1∩ U2∩ · · · ∩ Un−1∩ Un
also must belong to the set T , and the proposition follows. Proposition 2.1.9. Let X = R and let B = {(a, b) : a < b, a, b ∈ R}. Then B is a basis for a topology on X. We shall call the topology generated the standard topology on R.
Proof. Let B = {(a, b) : a < b, a, b ∈ R}, then we need to show that B is a basis for R, then it follows from Proposition 2.1.8 that it generates a topology on R. To show that the first condition of a basis holds we note that for each element x ∈ R we can construct an infinitesimaly small open interval about the point x which belongs to B.
To verify the second criteria, let x belong to the intersection of two open intervals B1
and B2. Now if B1and B2 are the same we are done. Otherwise by the definition of an interval, there must be a smaller sub-interval, B3, such that B3⊂ B1∩ B2. Now since we know that x belongs to their intersection we must have that x ∈ B3⊂ B1∩ B2. It can sometimes be useful to only consider what we shall define as the subbasis for the topology. This is since the topology generated by the subbasis will be the smallest one containing the subbasis itself.
Definition 2.1.10. Let X be a topological space. A collection S is said to be a subbasis for a topology if it is a collection of subsets of X whose union equals X.
Lemma 2.1.11. Let X be a non-empty set, S be a subbasis for a topology on X and let T be the collection of all unions of finite intersections of elements of S. Then T is a topology. We shall say that T is the topology generated by the subbasis S.
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Proof. Let X be a non-empty set and let S be a subbasis for a topology on X. Further- more let T be the collection of all unions of finite intersections of elements in S. Now let B be the collection of all finite intersections of elements in S. Then it is sufficient to show that B is a basis for a topology on X and the proposition will follow from Proposition 2.1.8.
To prove the first criterion for a basis we note that given x ∈ X there is by definition of S an element containing x. Hence it belongs to some intersection of elements and thus it belongs to B.
To verify to second criterion, let
B1= S1∩ S2∩ · · · ∩ Sm and B2= S10 ∩ S20 ∩ · · · ∩ Sn0
be two elements of B where each Si and Sj0 are elements in S. Then their intersection B1∩ B2= (S1∩ S2∩ · · · ∩ Sm) ∩ (S10∩ S02∩ · · · ∩ Sn0) = S1∩ S2∩ · · · ∩ Sm∩ S10∩ S20∩ · · · ∩ Sn0, is itself a finite intersection of elements in S and thus belongs to B. Thus by Proposition
2.1.8 B is a basis for the topology T on X.
Example 2.1.12. Let X = R and define S to be the subbasis S = {(−∞, x), (y, +∞) : x, y ∈ R}.
We first note that the union of all elements in S clearly yields all of R so that it is a proper subbasis. Then we note that the intersection of elements in S yields generic open intervals in R. Hence the topology generated by the subbasis is equivalent with the
standard topology defined in Proposition 2.1.9.
We shall now extend this notion of generating the space in a way which is more intuitive. A collection shall be called a covering of a space if the space itself can be generated by taking the union of the elements in this covering.
Definition 2.1.13. A collection A of subsets of a space X is said to cover X, or to be a covering of X, if the union of the elements of A is equal to X. It is called an open covering of X if its elements are open subsets of X.
Finally in this section we shall define the subspace topology which will be of interest since it allows us to talk about subspaces of topological spaces. Furthermore this topology is useful since it makes certain properties of larger spaces hereditary to the subspaces.
Proposition 2.1.14. Let X be a non-empty set with the topology T and let Y be a subset of X. Then the collection
TY = {Y ∩ U : U ∈ T }
forms a topology on Y which we call the subspace topology. Furthermore we say that Y is a subspace of X.
Proof. The first criterion for a topology is immediate since Y ∩ ∅ = ∅ and Y ∩ X = Y . To show closure under finite intersections we recall the standard set-theoretic law
(U1∩ Y ) ∩ (U2∩ Y ) ∩ · · ·(UN ∩ Y ) = (U1∩ U2∩ · · · ∩ UN) ∩ Y.
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Similarily for arbitrary unions we recall that the following holds [
α∈J
(Uα∩ Y ) = [
α∈J
Uα) ∩ Y.
Hence TY must be closed under arbitrary unions and finite intersections and is therefore
a topology on Y .
Example 2.1.15. Let X = R be a topological space equipped with the standard topol- ogy T as in Proposition 2.1.9. Now let Y = [0, 1] ⊂ X, then the subspace topology is generated by a basis of the form B = {Y ∩ U : U ∈ T }. We know that the elements of the basis for the topology are open intervals U = (a, b) with a < b, thus the elements of the basis are of the form
Y ∩ (a, b) =
(a, b), a, b ∈ Y [0, b), a /∈ Y, b ∈ Y (a, 1], a ∈ Y, b /∈ Y Y or ∅, a /∈ Y, b /∈ Y.
From above we note that the sets [0, b) and (a, 1] are, by definition, open in Y , but they
are not open in X.
2.2. Separation axioms. Starting from Definition 2.1.1 of the topology it is not clear what one can say about the structure of it. Since we want to be able to construct spaces with particular properties it should be natural to have some way of determining its structure via its basic building blocks.
From the definition of the topology it is clear that the most basic things we can work with are the open sets themselves and their elements. Therefore one defines what are called separation axioms, which characterizes how open sets and points interact in the topology.
In this subsection we shall give the definition of the four most fundamental separation axioms, all which are illustrated in Figure 1. But first we shall give a concrete example of why these results are needed.
Example 2.2.1. Let X be a non-empty topological space. We want to know when we can say that each singleton set is closed in X.
Assume that x, y ∈ X, x 6= y are two points. Then if we assume that {x} is closed in X it follows that y ∈ X\{x}, which is an open set, and by definition x /∈ X\{x}. Hence by applying the same argument to the point y we can conclude that the singleton sets are closed in X whenever there exists open sets U1 and U2 such that x ∈ U1, y /∈ U1
y ∈ U2 and x /∈ U2.
In the definitions of the separation axioms we assume that points in the set X are distinguishable in the following sense.
Definition 2.2.2. Let X be a non-empty topological space. We say that the space is Kolmogorov if for each pair of points x, y ∈ X there exists an open set U such that x ∈ U but y /∈ U .
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Definition 2.2.3. A topological space X is called Frechét, if for every x1, x2∈ X, where x16= x2, there are neighbourhoods U1of x1and U2of x2such that x1∈ U/ 2and x2∈ U/ 1. Remark 2.2.4. It is common to say that a set has closed one-point sets instead of writing that it is a Frechét space. For us this will merely be an aestethic choice of what we call it.
Remark 2.2.5. It follows from the above definition and Example 2.2.1 that if all singleton sets in a space X are closed it must be Frechét. It can be shown that it is in fact an equivalence.
Definition 2.2.6. A topological space X is called Hausdorff if for every x1, x2 ∈ X, x16= x2, there are neighbourhoods U1of x1 and U2 of x2 such that U1∩ U2= ∅.
Definition 2.2.7. A Frechét space X is called regular if for every x ∈ X, and closed set V , x /∈ V there exists open sets U1, U2 such that
V ⊆ U1, x ∈ U2, U1∩ U2= ∅.
Definition 2.2.8. A Frechét space X is called normal if for all closed and disjoint sets V1and V2 there are open sets U1 and U2 such that:
V1⊆ U1, V2⊆ U2 and U1∩ U2= ∅.
x1 U1
x2 U2 X1
x1 U1
x2
U2 X2
X3
x U2
V U1
V2
U2
V1
U1 X4
Figure 1. An illustration of a space being: Fréchet (X1), Hausdorff (X2), regular (X3) and normal (X4).
The reason why we enforce the spaces to be Frechét in the definition of regular and normal is since we want to be able to have the hierachy of spaces shown in Figure 2.
Thus we know that a space is e.g Frechét if it is shown to be normal.
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Sometimes it can be quite difficult to verify that a given space has a certain separation axiom, in particular being regular or normal. On the other hand it can be very easy to see that a set is Frechét or Hausdorff. To show this we give the following example.
Example 2.2.9. Let the set X = R be endowed with the topology generated by the basis B = {(a, b), (a, b)\K : a < b a, b ∈ R}, where the set K is given by K = {1/n : n ∈ Z+}.
Then the space X is Hausdorff but not regular. To see that it is Hausdorff, note that for two given points we can find two disjoint and open intervals containing them.
To show that it is not regular, first note that 0 /∈ K and that K is closed. Now suppose that X is regular and that there are two open sets U and V containing 0 and K respectively. Then choose a basis element in U containing 0, it must be of the form (a, b)/K since any interval of the form (a, b) intersects K. Now choose a positive integer k large enough that 1/k ∈ (a, b). Now we choose a basis element about the number 1/k in K, which must be of the form (c, d). Finally we choose a number x such that x ∈ (max{c, 1/(n + 1)}, 1/n) , then x belongs to both U and V such that they are not disjoint, a contradiction.
Kolmogorov Hausdorff
Frechét Regular
Normal
Figure 2. The hierarchy of the defined separation axioms, including the Kolmogorov spaces.
To see a more convoluted example we refer to Example 3 in §31 of [9], where it is shown that the Sorgenfrey-plane, a generalization of the topology mentioned in Proposition 2.1.9, is not normal. The following result will be of use when proving Lemma 2.4.15 in Section 2.4.
Lemma 2.2.10. Let X be a Fréchet-space. Then X is normal if, and only if, given a closed set T and an open set U , such that T ⊆ U , there is an open set V such that T ⊆ V ⊆ ¯V ⊆ U .
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Proof.
⇒: Assume that X is normal, let T and U be sets that are closed and open respec- tively in X such that T ⊆ U . Now let S = X\U , then S is closed and S and T are disjoint. Since X is normal there are now open and disjoint sets V and W such that T ⊆ V and S ⊆ W . Now we note that if a point y ∈ S, then y /∈ V , hence V and S are disjoint and thus we have V ⊆ U .
⇐: Let S and T be two disjoint and closed sets. Now define U = X\S. Then U is open and T ⊆ U . By assumption there is an open set V such that T ⊆ V and ¯V ⊆ U . Thus V and X\ ¯V are open sets which are disjoint and contains the sets respectively.
Hence X is normal.
Furthermore, a property of a given topology which will be of interest is in a sense how large, in the sense of inclusion, the topology is. This can be characterized in a comparative way by talking about finer and coarser topologies.
Definition 2.2.11. Let X be a nonempty set with topologies T1and T2. If it holds that T1⊂ T2we say that the topology T2is finer than the topology T1and that T1is coarser than T2.
Having this definition of finer and coarser topologies we define a way of comparing collections of sets in a space analogously.
Definition 2.2.12. Let A be a collection of subsets of the space X. A collection B of subsets of X is said to be a refinement of of A if for each element B of B, there is an element A of A containing B. We then say that B refines A.
2.3. Locally finite collections. It is not evident from the construction of the topo- logical space that there is any limitation of the amount of overlapping in it. Thus one can construct spaces where sets are infinitely intersecting with each other. In order to prevent this one can enforce that some collection of subsets in the space only intersect any neighbourhoods in the space at most finitely many times.
A collection having this property will be called locally finite in the space and this property turns out to be of great use in the area of metrizability. In particular when proving Theorem 4.1.4.
Definition 2.3.1. Let X be a topological space. A collection A of subsets of X is called locally finite in X if every point of X has a neighbourhood that intersects only finitely many elements of A.
A special case of this which will be of interest is when the collection can be constructed from countably many locally finite subcollections.
Definition 2.3.2. A collection A of subsets of X is called countably locally finite if it can be written as a countable union of collections {An}n∈J each being locally finite.
The following lemma characterizes some useful properties of locally finite collections.
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Lemma 2.3.3. Let A be a locally finite collection of subsets of X, then the following holds.
(1) Any subcollection of A is locally finite.
(2) The collection B = { ¯A}A∈Aof the closures of the elements of A is locally finite.
(3) S
A∈AA =S
A∈AA.¯
Proof. The proof of (1) is immediate. In order to show (2) we let x ∈ A for some A ∈ A, and let U be a neighbourhood of x such that U intersects a finite number of elements in A. Then since A is locally finite U intersects ¯A in up to as many points as it intersects A. Thus the neighbourhood can intersect the collection B up to as many times as A which by definition is finite and hence B = { ¯A}A∈A is locally finite. To show (3) it is now sufficient to prove the inclusion
C ⊂¯ [
A∈A
A¯
where C =S
A∈AA since the reverse inclusion always holds. Let x ∈ ¯C and let U be a neighbourhood of x that intersects a finite number, say k, elements in A. Then we asert that x belongs to one of ¯A1, ¯A2, · · ·, ¯Ak, and thus in the union of closures. If this is not the case we know that the set
U \( ¯A1∪ ¯A2∪ · · · ∪ ¯Ak)
is a neighbourhood of x and doesn’t belong to ¯C, which violates our asumption that
x ∈ ¯C.
Finally we shall prove a lemma which will be of great use when characterizing para- compact spaces in Section 4.2.
Lemma 2.3.4. Let X be a regular space, then the following are equivalent. Every open covering of X has a refinement that is:
(1) An open covering of X and countably locally finite.
(2) A covering of X and locally finite.
(3) A closed covering of X and locally finite.
(4) An open covering of X and locally finite.
Proof. The implication (4) ⇒ (1) is immediate. It is thus sufficient to show the implica- tions (1) ⇒ (2) ⇒ (3) ⇒ (4). This will be done in several steps as follows.
(1) ⇒ (2): Let A be an open covering of X and let B be an open refinement of A which covers X and is countably locally finite. Now write
B =[ Bn,
where every Bn is locally finite. Now given an integer k define Ck = [
B∈Bk
B.
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Then for each positive integer n and each element B in Bn define Dn(B) = B\ [
k<n
Ck
and let En be the collection of the sets Dn(B) for each element B in Bn. From this we note that En is a refinement of Bn since Dn(B) ⊂ B for every B ∈ Bn. Moreover we assert that the collection E = S En is the desired locally finite refinement of A which also covers X.
To show this, given a point x in X, we shall show that x belongs to an element in E and has a neighbourhood which only intersects finitely many elements of E . Let B be given as above, and let N be the smallest integer such that x belongs to an element of BN. Now let B be an element of BN containing x. Now note that since x does not belong to any Bk for k < N we can conclude that x belongs to DN(B), which is an element of E. Furthermore, since the collections Bk are locally finite, we can for each k = 1, 2, ..., N choose a neighbourhood Uk of x that intersects only a finite number of elements of Bk. Now if Uk intersects the element Dk(V ) of Ek, it must intersect the element V in Bk
since Dk(V ) ⊂ V . Therefore Uk only intersects finitely many elements of En. Because B is in BN, B intersects no elements in Ek for k > N . Hence, the neighbourhood
U1∩ U2∩ · · · ∩ UN ∩ B of x intersects only finitely many elements of E .
(2) ⇒ (3): Let A be an open covering of X. Let B be the collection of all open sets U of X such that ¯U is contained in an element of A. Since X is regular, B covers X. Now using our assumption we find a refinement C of B which covers X and is locally finite.
Now let
D = { ¯C : C ∈ C}.
Then D also covers X, refines A and is locally finite by Lemma 2.3.3.
(3) ⇒ (4): Let A be an open covering of X and choose a refinement B of A which is closed, locally finite and covers X. To obtain the desired result we shall expand the elements B in B in a way such that they still are locally finite and refines A.
To do this we shall introduce an auxiliary covering C of X such that C is locally finite and closed. It is clear that each of the elements in C can intersect at most finitely many elements in B. For each x in X we can find a neighbourhood that only intersects finitely many elements of B. Thus the collection of all open sets which intersect only finitely many elements of B is an open covering of X.
Now for each element B in B we let
D(B) = {C : C ∈ C and C ⊂ X\B}.
Furthermore we define
E(B) = X\ [
C∈D(B)
C.
Now because C is a locally finite collection of closed sets, we know by Lemma 2.3.3 that the union of any subcollection of C is closed. Therefore we know that E(B) is an open
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set and B ⊂ E(B) by definition.
Now we must make sure that we have not expanded each B too much so that the collection {E(B)} still is a refinement of A. To remedy this, for each B ∈ B, choose an element F (B) of A containing B. Then define
G = {E(B) ∩ F (B) : B ∈ B}.
The collection G is now a refinement of A. To see this we note that B ⊂ (E(B) ∩ F (B)) and since B covers X the collection G must also cover X.
What now remains to show is that G is locally finite. Given a point x of X, let W be a neighbourhood of x which only intersects finitely many elements of C, say C1, C2,..,Ck. We shall show that W only intersects finitely many elements of G. Since C covers X, the set W is covered by C1,C2,..,Ck. Thus it is sufficient to show that each element C of C only intersects finitely many elements of G. Now if C intersects E(B) ∩ F (B) then it must intersect E(B), hence E(B) is not contained in X\B by definition and thus C must intersect B. Since C only intersects finitely many elements of B, it can intersect
at most the same amount of elements in G.
2.4. Continuous functions. In this section we shall turn to the topological viewpoint of analysis. These definitions and results are together with metric spaces and metrization theorems at the core of the advancement of general topology.
Since we want to be able to talk about functions in more general spaces than R, Rn or even Rω we must give a broader definition of what continuity means. This is since the standard limit-based definitions in real- and complex analysis are dependent on the existence of metrics on the space we are working on.
Definition 2.4.1. Let f : A → B be a function. We say that f is continuous if for every open subset U in B, f−1(U ) is open in A.
Remark 2.4.2. Note that in the above definition, f−1(U ) denotes the preimage of f , and not the standard inverse.
Although the topological definition of continuity is more abstract, we propose that it is equivalent with the standard real-analytic definition of continuity on R. The classical definition is given as follows.
Definition 2.4.3. Let f : R → R be a function and x0 ∈ R be a point. Then given
> 0, we say that f is continuous at x0 if there exists a δ > 0 such that for all x ∈ R we have
|x − x0| < δ ⇒ |f (x) − f (x0)| < .
Proposition 2.4.4. Definition 2.4.1 and Definition 2.4.3 are equivalent on R.
Proof. ⇒: Let f : R → R be a continuous function and let x0 ∈ R and let > 0 be given. Now the set V = (f (x0) − , f (x0) + ) is open in R. Therefore, f−1(V ) is an open set in R. Now since f−1(V ) contains the point x0, there must be some basis ele- ment (a, b) about x0 in the inverse image. Now we let δ = min(x0− a, x0− b), then if
|x − x0| < δ it implies that x belongs to the basis element (a, b) such that x ∈ V and hence |f (x) − f (x0)| < , as desired.
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⇐: Let f : R → R be continuous in the sense of Definition 2.4.3. Furthermore let (a, b) be a basis element of the standard topology on R as given in Proposition 2.1.9.
Now let x ∈ f−1((a, b)), hence f (x) ∈ (a, b). Also define = min{f (x) − a, b − f (x)}, and by the − δ condition there exists a δ such that |x − y| < δ implies that |f (x) − f (y)| < , which by our definition of means
f (x − δ, x + δ) ⊆ (f (x) − , f (x) + ) ⊆ (a, b).
Since (x − δ, x + δ) is open and contains x we know that f−1((a, b)) is open, which implies
our open-set definition of continuity.
Using the definition of continuity in Definition 2.4.1 we can discuss how spaces behaves under mappings. For example, we could ask when the image under a function inherits the same topological structures as the domain itself. This particular question is of interest since a we want to be able to characterize when two topological spaces are, in some sense, equivalent with each other.
To characterize these equivalent spaces we shall introduce the concept of spaces being homeomorphic, which generalizes continuity between spaces. This concept will be our topological way of saying that two spaces are, from a topological viewpoint, equivalent.
Definition 2.4.5. Let X and Y be topological spaces and f : X → Y a function . Then f is called a homeomorphism if it is bijective, continuous and has a continuous inverse f−1. If such a function f exists we say that X and Y are homeomorphic.
Remark 2.4.6. In this definition we denote the inverse as f−1, and not the preimage as in Definition 2.4.1.
Example 2.4.7. We shall show that there always exists a homeomorphism between any two open intervals in R. Let (a, b) and (c, d) be two open intervals in the real line. Let the function f (x) be defined as
f (x) = d − c
b − ax + cb − da b − a .
Then we note that f (x) is a polynomial, hence it is a continuous from R to R and in particular from (a, b) to (c, d). It has an inverse of the form
f−1(x) = b − a
d − cx − cb − da d − c ,
which is continuous between the open intervals by the same argument. Bijectivity follows from f being a linear polynomial. Hence f is a homeomorphism between (a, b) and
(c, d).
Now if we map a topological space onto another with a homeomorphism, we can be certain that the image will inherit any topological structure of the codomain since it will map open sets to open sets. Although the homeomorphism is very handy to work with, it can be made even more useful by enforcing homeomorphicity between the domain and the image itself, in which case we call the map an imbedding.
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Definition 2.4.8. Let X and Y be topological spaces and let f : X → Y be an injective continuous map. If the restriction obtained by restricting the codomain of f to its image set f (X), is a homeomorphism of X with f (X) we call f an imbedding of X in Y .
Before continuing we shall need to define what we mean by products over arbitrary index sets. Since we want to work with spaces like Rn or even Rω we cannot use the traditional cartesian product over finite elements. Instead we define the following generalization.
Definition 2.4.9. Let J be an index set. Given a set X, a J-tuple is defined to be a function x : J → X. The function x evaluated at a particular index α ∈ J is called the α-coordinate of x and we often denote the J -tuple as
(xα)α∈J.
Definition 2.4.10. Let {Aα}α∈J be an indexed collection of sets and let X =S
α∈JAα. Then we define the cartesian product of this indexed family, denoted by
Y
α∈J
Aα,
to be the set of all J -tuples (xα)α∈J of elements in X such that xα∈ Aα for all α in J . That is, the set of all functions
x : J → [
α∈J
Aα, such that x(α) ∈ Aαfor all α in J .
Now since we can take arbitrary products we want to be able to pick out particular elements. Therefore we shall introduce the concept of a projection mapping which will allow us to work with specific coordinates.
Definition 2.4.11. Let {Xα}α∈J be an indexed collection of sets. Then given an index β ∈ J let
πβ: Y
α∈J
Xα→ Xβ
be a function assigning to each product the β’th coordinate. Then the function is called the projection mapping associated with the index β and we write this more compact as
πβ((xα)α∈J) = xβ.
Using this concept of projection mappings we can construct a new kind of general topology called the product-topology, which shall be of great use when studying spaces like Rn and its generalizations.
Proposition 2.4.12. Let {Xα}α∈J be an indexed collection of topological spaces and let Sβ= {πβ−1(Uβ) : Uβ open in Xβ}.
Then the union of each Sβ forms a subbasis for the space X =Q
α∈JXα and we call the topology generated by it the product- topology on X. The set X is called a product space.
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Proof. It is clear that the collection S forms a subbasis for the product space X from the definition. Hence by Lemma 2.1.11 we know that there exists a topology T generated
by S.
When we are performing analysis on product spaces we must ensure that the mappings we are working with maintain their continuity. To do so we shall use the following lemma.
Lemma 2.4.13. Let f : A →Q
α∈JXα be given by the equation f (a) = (fα(a))α∈J
where fα : A → Xα for each α. Let Q
α∈JXα have the product topology. Then the function f is continuous if and only if each function fα is continuous.
Proof.
⇒: Let πβ denote the projection of the product onto its β’th factor. Then we know that the function πβ is continuous, for if Uβ is open in Xβ, the set πβ−1(Uβ) is an element of the subbasis for the product topology on Xα. Now if the function f : A →Q
α∈JXα is continuous, the function fβ = πβ◦ f is a composition of two continuous functions, and is therefore continuous.
⇐: Suppose that each of the coordinate functions fα are continuous. Then to show that f is continuous it is sufficient to show that the inverse image of each of the subbasis elements are open under f in A. A subbasis element of the product topology onQ Xα
can be written as π−1β (Uβ), for some index β, where the set Uβ is open in Xβ. Thus, using the fact that fβ= πβ◦ f we have
f−1(πβ−1(Uβ)) = fβ−1(Uβ).
And since fβ is continuous, this set is open in A.
The following theorem shall be of great use when proving Theorem 4.1.4 and is a good example of the usefulness of homeomorphisms.
Theorem 2.4.14. Let X be a Frechét space. Suppose that {fα}α∈J is an indexed family of continuous functions fα: X → [0, 1] satisfying the requirement that for each point x0
of X and each neighbourhood U of x0, there is an index α such that fαis positive at x0
and vanishes outside U . Then the function F : X → [0, 1]Jdefined by F (x) = (fα(x))α∈J
is an imbedding of X in [0, 1]J.
Proof. Let (fα)α∈J, fα : X → R, be a collection of continuous functions such that for each point x0 in the set X and each neighbourhood U of x0, there is an index α such that fα(x0) > 0 and fα(x) = 0 for x /∈ U . Furthermore let RJ be endowed with the product topology. Then we define the mapping F : X → RJ as
F (x) = (fα(x))α∈J.
We shall show that F is an imbedding. First we argue that F is continuous, this follows from Lemma 2.4.13 since each fα is continuous and X has the product topology. Fur- thermore we argue that F (x) is injective. Let x, y ∈ X be two non-identical points, then
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we can find an index α ∈ J such that fα(x) = 0 and fα(y) > 0 and hence F (x) 6= F (y).
Finally we shall show that F is a homeomorphism of X onto its image F (X), which we shall denote as Y . It is clear that F forms a continuous bijection with Y so it only remains to show that for every open set U in X, the image F (U ) is open in Y . To do this, we shall find an open set Z of Y such that for a point y0∈ Y ,
y0∈ Z ⊂ Y.
Let x0 be the point in U such that F (x0) = y0. Now let β be an index such that fβ(x0) > 0 and fβ(X\U ) = {0}. Now take the open set (0, 1) and let V be the open set
V = π−1β ((0, 1))
of [0, 1]J. Now let W = V ∩ Y , then by the definition of the subspace topology W is open in Y . We now assert that y0∈ W ⊂ F (U ). To see this we first note that
πβ(y0) = πβ(F (x0)) = fβ(x0) > 0.
Then to see that W ⊂ F (U ), first note that if y ∈ W , then y = F (x) for some x ∈ X, and πβ(y) ∈ (0, 1). But we also know that πβ(y) = πβ(F (x)) = fβ(x) and fβ(x) vanishes outside U , hence the point x must belong to U and thus y = F (x) ∈ F (U ) as desired.
Thus F is an imbedding of X in [0, 1]J.
Finally we shall prove a lemma which will be used in proving Lemma 4.1.3 and is due to P. Urysohn and was first proven in [15].
Lemma 2.4.15. Let X be a normal space, U and V be disjoint closed subsets of X and let [u, v] be a closed interval in the real line. Then there exists a continuous map f : X → [u, v] such that
f (x) =
(u, if x ∈ U.
v, if x ∈ V.
Proof. We first show this lemma for a special case and then use homeomorphicity to prove the general lemma. Let u = 0 and v = 1. We know that X\V is an open neighbourhood of the closed set U . Therefore by Lemma 2.2.10 there is an open set W1/2 such that
U ⊆ W1/2⊆ ¯W1/2⊆ X\V.
Then by using Lemma 2.2.10 again we obtain two open sets W1/4 and W3/4 such that U ⊆ W1/4⊆ ¯W1/4⊆ W1/2⊆ ¯W1/2⊆ W3/4⊆ ¯W3/4⊆ X\V.
By continuing in a similar manner, for each rational number on the form t = m
2n, n = 1, 2, ... m = 1, 2, 3..., 2n− 1 we get a collection of open sets Wtsuch that
t1< t2⇒ U ⊆ Wt1⊆ ¯Wt1⊆ Wt2 ⊆ ¯Wt2 ⊆ X\V.
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Now let g : X → [0, 1] be defined by g(x) =
(0, if x ∈ Wtfor all t.
sup{t : x /∈ Wt}, otherwise.
This construction yields that 0 ≤ g(x) ≤ 1, with g(x) = 0 if x ∈ U and g(x) = 1 if x ∈ V . We now need to show that the function g is continuous. Recall that it is sufficient to only consider open subbase elements. Since we are considering the interval [0, 1] we know that [0, a) and (a, 1], 0 < a < 1, are open subbasis elements of [0, 1].
By our construction we have that if g(x) < a, then x ∈ Wtfor some t < a, hence g−1([0, a)) = {x : g(x) < a} = [
t<a
Wt.
Thus f−1([0, a)) ⊆ X is open. Now if instead g(x) > a we have x ∈ X\ ¯Wt for some t > a. Hence we have
f−1([a, 1)) = {x : g(x) > a} = [
t>a
X\ ¯Wt
and f−1((a, 1]) ⊆ X is open. From this we can now conclude that g is continuous. To show the lemma in general, let g : X → [0, 1] be given such that
g(x) =
(0, if x ∈ U , 1, if x ∈ V ,
and it is continuous. Then set f (x) = (v − u)g(x) + u, since f (x) is a composition of continuous functions it is continuous. If x ∈ U , then f (x) = (v − u) · 0 + u = u. Likwise if x ∈ V , then f (x) = (v − u) · 1 + u = v. Hence f (x) is our desired function.
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3. Metrics and metric spaces
3.1. The metric and the metric topology. The most fundamental and intuitive concept in geometry is the concept of distances. Although this concept is easy to under- stand, it is more complicated to describe in mathematics. Thus we define what is called a metric, which will be a scalar function whose value represents the intuitive concept of a distance.
Definition 3.1.1. Let X be a non-empty set. A function d : X × X → R is called a metric if it satisfies:
(1) d(x, y) ≥ 0 for all x, y ∈ X, with equality if and only if x = y, (2) d(x, y) = d(y, x) for all x, y ∈ X,
(3) d(x, y) ≤ d(x, z) + d(z, y) for all x, y, z ∈ X.
We shall call a set X together with a metric d defined on X a metric space, (X, d).
In Example 3.1.2, Example 3.1.3 and Example 3.1.4 we provide a few examples of specific metric spaces.
Example 3.1.2. Let x = (xj)j∈N, y = (yj)j∈N and z = (zj)j∈N be three points in Rω = Q∞
j=1R, then the function d(x, y) = |x − y| = qP∞
j=1(xj− yj)2 is a metric.
Condition (1) follows from the positivity of the square root function and condition (2) follows from the parity of the square function. To show (3) we can write
d(x, z) + d(z, y) = v u u t
∞
X
j=1
(xj− zj)2+ v u u t
∞
X
j=1
(zj− yj)2.
Then by using Minkowski’s inequality (see e.g Theorem 28 in [6]) we have:
d(x, z) + d(z, y) ≥ v u u t
∞
X
j=1
(xj− zj+ zj− yj)2= d(x, y)
as desired. This is known as the Euclidean metric and in particular in R2it is equivalent
with Pythagoras’ theorem.
Example 3.1.3. Let (X, d) be a metric space, then let d(x, y) = min{d(x, y), 1}.¯
We shall show that ¯d is a metric under the standard Euclidean metric d(x, y) = |x − y|, but it can be easily shown for general metrics d. To see that (1) holds, note that since d(x, y) ≥ 0, we can have a minimum of ¯d(x, y) = min{0, 1} = 0. To see (2), note that d(x, y) = d(y, x), hence the minimum will be equal aswell. And to see (3) we note that
d(x, y) = min{|x − y|, 1)} ≤ min{|x − z| + |y − z|, 1}¯
≤ min{|x − z|, 1} + min{|y − z|, 1},
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and hence we have ¯d(x, y) ≤ ¯d(x, z) + ¯d(y, z). Thus ¯d is a metric and under the assump- tions that d(x, y) is the standard Euclidean metric we shall call it the standard bounded
metric.
Example 3.1.4. Let J be an arbitrary index set, ¯d be as defined in Example 3.1.3, together with Example 3.1.2, on RJ and let x = (xα)α∈J, y = (yα)α∈J and z = (zα)α∈J
be three points in RJ. Then the function defined by
¯
ρ(x, y) = sup{ ¯d(xα, yα) : α ∈ J }
is a metric. It follows immediately from the former example that (1) and (2) holds. Now note that for an arbitrary point z = {zα}α∈J in RJ we have by (1)
¯
ρ(x, y) ≤ sup{ ¯d(xα, zα) : α ∈ J } + ¯d(zα, yα) ≤ sup{ ¯d(xα, zα) : α ∈ J }
+ sup{ ¯d(zα, yα) : α ∈ J } = ¯ρ(x, z) + ¯ρ(z, y), and hence (3) follows. In proving Theorem 4.1.4 it shall be of particular interest to consider the subspace [0, 1]J of RJ together with the standard bounded metric. In this subspace we note that for any two points xα and yα in [0, 1], ¯d(xα, yα) = d(xα, yα) =
|xα− yα| since the metric at most can be equal to one. Hence in this subspace we have
¯
ρ(x, y) = sup{|xα− yα| : α ∈ J }.
We shall call this metric the uniform metric.
Using this definition of a metric, we can consider subsets in our metric space which are often referred to as open balls. These balls can in turn be used to define a basis for a topology which we intuitively shall call the metric topology.
Proposition 3.1.5. Let (X, d) be a metric space. Then collection B = {B(x, r) : x ∈ X and r ∈ R≥0}, where
B(x, r) = {z : d(x, z) < r}
denotes the open ball of radius r around the point x, generates a topology. We shall call this topology the metric topology on X.
Proof. Let (X, d) be a metric space and let B denote the collection of all open balls in X. We shall prove the proposition by showing that B is a basis on X.
The first condition of a basis is immediate since by definition of the collection B we have that there exists open balls centered in each point in the set X. In order to show the second condition we shall first show that for a given point y belonging to an open ball B(x, ) there exists an open ball B(y, δ) contained in B(x, ).
Define the number δ = − d(x, y) which is positive since by definition y ∈ B(x, ).
Now let δ be the radius of the open ball about y, then since 0 < − d(x, y) ≤ we have B(y, δ) ⊆ B(x, ).
Now to prove the second condition we let y belong to the two open balls B(x1, 1) and B(x2, 2). Then by our former result we can define an open ball B(y, δ), where δ is given by δ = min{1, 2}. Now y ∈ B(y, δ) which also is included in the intersection
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of B(x, 1) ∩ B(x, 2). Hence the collection B forms a basis and by Proposition 2.1.8 it
generates a topology on X.
We provide an example of this topology.
Example 3.1.6. Let ¯d(x, y) = min{|x − y|, 1} be the standard bounded Euclidean metric defined on Rω and D(x, y) = supd(x¯ k,yk)
k : k ∈ Z+\{0} . Then we assert that D(x, y) is a metric defined on Rω.
The first two requirements on a metric follows immediately from d(x, y) being a metric.
To prove the triangle inequality holds we note that for all integers k the following holds d(x¯ k, yk)
k ≤ d(x¯ k, zk)
k +
d(y¯ k, zk)
k ≤ D(x, z) + D(z, y).
Hence in particular it must hold that
sup
d(x¯ k, yk)
k : k ∈ Z+\{0}
= D(x, y) ≤ D(x, z) + D(y, z).
Thus by Proposition 3.1.5 it follows that the metric D(x, y) generates a topology on
Rω.
3.2. Metrizability. In this section we shall give the definition of metrizability and prove some usefull properties of those spaces which are metrizable.
Definition 3.2.1. Let (X, T ) be a topological space. Then the set X is called metrizable if there exists a metric d defined on X which generates the topology T .
Lemma 3.2.2. Every metrizable space is regular.
Proof. Let X be a metrizable space endowed with a metric d. We shall prove the theorem by showing that X is normal, which by the definition of regularity implies that it is regular. Furthermore let U and V be subsets of X which are closed and disjoint. For every element u in U choose a ρu such that the ball B(u, ρu) does not intersect V . Similarily for an element v in V we choose a number ρv> 0 such that the ball B(v, ρv) does not intersect U . Then we shall define the sets
A = [
u∈U
B(u, ρu/2) and C = [
v∈V
B(v, ρv/2).
Then by constructions A and C are open sets containing U and V respectively, further- more we assert that they are disjoint. Let z ∈ A ∩ C, then there are u ∈ U and v ∈ V such that:
z ∈ B(u, ρu/2) ∩ B(v, ρv/2).
Now the triangle inequality implies that d(u, v) ≤ (ρu+ ρv)/2. Hence we get two cases (z ∈ B(v, ρv) if ρu< ρv
z ∈ B(u, ρu) if ρu≥ ρv
But by construction this is not possible and we have shown that X is normal. Now by letting one of the sets being a singleton set we have our lemma.
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In order to prove the final lemma on metrizable spaces we shall need the following definition and classical theorem.
Definition 3.2.3. A relation ≺ on a set A is called an order relation if it has the following properties:
(1) For every x and y in A such that x 6= y, either x ≺ y or y ≺ x.
(2) For no x in A does the relation x ≺ x hold.
(3) If x ≺ y and y ≺ z, then x ≺ z.
In particular we say that ≺ is a well-ordering on the set X if every non-empty subset of X has a smallest element.
Theorem 3.2.4. If X is a set, then there exists an order relation ≺ on X that is a well-ordering.
Proof. See for example page 68 in [5].
Lemma 3.2.5. Let X be a metrizable space. If A is an open covering of X, then there is an open covering E of X refining A that is countably locally finite.
Proof. Let ≺ be a well-ordering on the collection A, and let us denote three elements of A by U , V and W . Now choose a metric d defined on X and let n be a positive integer.
Given an element U of A, we define a set Cn(U ) to be the set obtained by shrinking the set U . Let
Cn(U ) = {x : B(x, 1/n) ⊆ U }.
Now by using the well-ordering ≺ on A we can create a still smaller set. For each element U in A, let
Dn(U ) = Cn(U )\ [
V ≺U
V.
The sets we have formed are now separated by a distance of at least 1/n. Thus for two distinct elements, say V and W , if x ∈ Dn(V ) and y ∈ Dn(W ) then we have d(x, y) ≥ 1/n. To show that this is the case, if we assume that V ≺ W , since x ∈ Dn(V ), the 1/n-neighbourhood of x lies in V . But since V ≺ W , by the definition of the set Dn(W ), y does not belong to V , thus y does not belong to the 1/n neighbourhood of x.
Since we do not know that these sets necessarily are open, we shall expand them slightly to obtain open sets, which we call En(U ). More specifically we let
En(U ) = [
x∈Dn(U )
B(x, 1/3n).
We now assert that if V and W are unique elements of A, then we have d(x, y) ≥ 1/3n whenever x ∈ En(V ) and y ∈ En(W ), this follows immediately from the triangle inequality. Thus, from our former remark it follows that the sets En of unique elements in A are disjoint.
We shall now define the collection
En= {En(U ) : U ∈ A},
which we asert is a locally finite collection of open sets that refines A. From the definition of En(U ) it is clear that En(U ) ⊂ U for every U ∈ A, thus the collection En is a
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refinement. Furthermore we note that for any point x in X, the 1/6n-neighbourhood can only intersect at most one element of En, therefore it is locally finite.
It is clear that the collection En doesn’t cover X, but we shall show that the collection E = [
n∈Z+
En
does. Let x be a point in X. We know that our initial collection A covers X, so let U be the first element in A, under the well ordering ≺, that contains x. Since U is open, we can choose an integer n such that B(x, 1/n) ⊂ U , then by definition x belongs to Cn(U ). Since U is the smallest element under ≺ we also have that x belongs to Dn(U ) and hence also belongs to the element En(U ) of En, as desired. The property of being metrizable can in some sense be seen as global, since it affects the entire space. We can now consider a more local formulation of this where we only consider metrizability of neighbourhoods around points in the space. Spaces with this property are of high importance since local metrizability is one of the key properties when studying topological manifolds. We shall call these spaces locally metrizable.
Definition 3.2.6. A space X is locally metrizable if every point x of X has a neigh- bourhood U that is metrizable in the subspace topology.
It is readily seen from Definition 3.2.1 that all metrizable spaces are locally metrizable.
The reversed implication is not true in general and is the main reason for the Smirnov metrization theorem.
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