# Topological Spaces and Groups : Part 1: Fast Review

Reference:

Topological Transformation Groups by Deane Montgomery and Leo Zippin, Dover Publications, Available in Amazon India.

My motivation:

We present preliminary and somewhat elementary facts of general spaces and groups. Proofs are given in considerable detail and there are examples which may be of help to a reader for whom the subject is new.

(The purpose of this blog is basically improvement in my understanding of the subject. If, by sharing, it helps some student/readers, that would be great. ) (Also, I found that reading/solving just one main text like Topology by James Munkres left me craving for more topological food: I feel I must also know how the founding fathers discovered topology)…

1.0 Introduction:

We use standard set-theoretic symbols : capitals A, B etc. for sets, $A \bigcup B$ for the union of sets (elements in one or both), and $A \bigcap B$ for the intersection (elements in both) etc.

1.1 Spaces (Topological Spaces);

The term space is sometimes used in mathematical literature in a very general sense to denote any collection whose individual objects are called points, but in topology the term space is used only when some further structure is specified for the collection. As the term will be used in this book it has a meaning which is convenient in studying topological groups. The definition is as follows:

DEFINITION: A topological space (or more simply space) is a non-empty set of points certain subsets of which are designated as open and where, moreover, these open sets are subject to the following conditions:

1. The intersection of any finite number of open sets is open.
2. The union of any number of open sets is open.
3. The empty set and the whole space are open.
4. To each pair of distinct points of space there is associated at least one open set which contains one of the points and does not contain the other.

A space is called discrete if each point is an open set.

Condition (4) is known as the $T_{0}$ separation axiomi in the terminology of Paul Alexandroff and Herr Heinz Hopf. The first three conditions define a topological space in their terminology. The designated system of open sets is the essential part of the topology, and the same set of points can become a topological space in many ways by choosing different systems of subsets designated as open.

1.2. Homeomorphisms

DEFINITION. A homeomorphism is a one-to-one relation between all points of one topological space and all points of a second which puts the open sets of the two spaces in one-one correspondence; the spaces are topologically equivalent.

The notion of homeomorphism is reflexive, symmetric, and transitive so that it is an equivalence relation in a given set of topological spaces.

EXAMPLES of topological spaces:

Let $E_{1}$ denote the set of all real numbers in its customary topology: the open intervals are the sets $\{ y: x < y < z\}$ for every $x < z$. The open sets are those which are unions of open intervals together with the empty set and the whole space.

Let $R_{1} \subset E_{1}$ denote the set of numbers in the closed interval $0 \leq y \leq 1$, where for the moment we take this subset without a topology. This set gives distinct topological spaces as follows:

1. Topologize $R_{1}$ as customarily : the open sets are the intersections of $R_{1}$ with the open sets of $E_{1}$.
2. Topologize $R_{1}$ discretely, that is, let every subset be open.
3. Topologize $R_{1}$ by the choice : open sets are the null set, the whole space and for each x of $R_{1}$ the set $\{ z, x < z \leq 1\}$
4. Topologize $R_{1}$ by the choice: the complement of any finite set is open and the empty set and whole space are open

In the sequel $R_{1}$ will denote the closed unit interval and $E_{1}$ the set of all reals in the customary topology. A set homeomorphic to $R_{1}$ is called an arc. A set homeomorphic to a circle is called a simply closed curve.

1.3 Basis

DEFINITION. A collection $\{ Q_{a}\}$ of open sets of a space is called a basis for open sets if every open set (except possibly the null set) in the space can be represented as a union of sets in $\{ Q_{a}\}$. It is called a sub-basis if every open set can be represented as a union of finite intersection of sets in $\{ Q_{a}\}$ (except possibly the null set).

A collection $\{ Q_{a}\}$ of open sets of a space S is a basis if and only for every open set Q in S and $x \in Q$, there is a $Q_{a} \in \{ Q_{a} \}$ such that $x \in Q_{a} \subset Q$.

If a collection has this property at a particular point x then the collection is called a basis at x.

If a set together with certain subsets are called a sub-basis, then another family of subsets is determined from the sub-basis by taking arbitrary unions and finite intersections. This new family (with the null set added if necessary) then satisfies conditions 1, 2, 3 for open sets of a topological space. Whether 4 will also be satisfied depends on the original family of sets.

EXAMPLE. Let $E_{1}$ denote the space of real numbers in its usual topology. For each pair of rationals $r_{1} < r_{2}$ let $\{ r_{1}, r_{2}$ denote the set of reals $r_{1}. This countable collection of open sets is a basis.

A space is said to be separable or to satisfy the second countability axiom if it has a countable basis. A space is said to satisfy the “first countability axiom” if it has a countable basis at each point.

EXAMPLE. Let S denote a topological space and let F denote a collection of real valued functions $f(x)$, where $x \in S$. If $f_{0}$ is a particular element of F then for each positive integer n let $Q(f_{0},n)=\{ f \in F: |f(x)-f_{0}(x)| < \frac{1}{n}\}$ for all x. We may topologize F by choosing the sets $Q(f_{0},n)$ for all $f_{0}$ and n as a sub-basis. The topological space so obtained has a countable basis at each point in many important cases.

1.4 Topology of subsets

Let S be a topological space, T a subset. Let $Q \bigcap T$ be called open in T or open relative to T if Q is open in S. With open sets defined in this way T becomes a topological space and the topology so defined in T is called the induced or relative topology. If S has a countable basis and $T \subset S$, then T has a countable basis in the induced topology.

DEFINITION. A subset $X \subset S$ is called closed if the complement $S - X$ is open. If $X \subset T \subset S$ then X is called closed in T when $T - X$ is open in T.

Notice that T closed in S and X closed in T implies that X is closed in S. The corresponding assertion for relatively open sets is also true.

It can be seen that finite unions and arbitrary intersections of closed sets are again closed.

DEFINITION. If $K \subset S$, the intersection of all closed subsets of S which contain K is called the closure of K and is denoted by $\overline{K}$. If K is closed, $K = \overline{K}$.

15 Continuous maps

Let S and T be topological spaces and f a map of S into T $f: S \rightarrow T$, that is, for each x in S, $y=f(x)$ is a point of T. If the inverse of each relatively open set in $f(S)$ is an open set in S, then f is called continuous. In case $f(S)=T$ then f is continuous and V open in T imply $f^{-1}(V)$ is open in S. The map is called an open map if it carries open sets to open sets.

If f is a continuous map of S onto T (that is, $f(S)=T$) and if $f^{-1}$ is also single valued and continuous, then f and $f^{-1}$ is also single valued and continuous, then f and $f^{-1}$ are homeomorphisms and S and T are homeomorphic or topologically equivalent.

EXAMPLE. The map $f(t) = \exp{2\pi it}$ is a continuous and open map of $E_{1}$ onto a circle (circumference) in the complex plane.

EXAMPLE. Let K denote the cylindrical surface, described in x ,y, z coordinates in three-space by $x^{2}+y^{2}=1$. Let $f_{1}$ denote the map of K onto $E_{1}$ given by $(x,y,z) \rightarrow (0,0,z)$, let $f_{2}$ the map $(x,y,z) \rightarrow (x,y, |z|)$ of K into K. All three maps are continuous, the first two are open and $f_{1}$ and $f_{2}$ are also closed, that is, they map closed sets into closed sets.

1.6 Topological products

The space of n real variables $(x_{1}, x_{2}, \ldots, x_{n})$ from $-\infty < x_{i} < \infty$, where $i=1,2,\ldots, n$ and the cylinder K of the preceding example are instances of topological products.

Let A denote any non-empty set of indices and suppose that to each $a \in A$ there is associated a topological space $S_{a}$. The totality of functions f defined on A such that $f(a) \in S_{a}$ for each $a \in A$ is called the product of the spaces $S_{a}$. When topologized as below it will be denoted by PROD $S_{n}$; we also use the standard symbol $\times$ thus $E \times B$ is the set of ordered pairs $(a,b)$ $e \in E$, $b \in B$.

The standard topology for this product space is defined as follows: For each positive integer n, for each choice of n indices $a_{1}, a_{2}, \ldots, a_{n}$ and for each choice of a non empty open set in $S_{a_{i}}$

$U_{a_{i}} \subset S_{a_{i}}$ for $i=1,2, \ldots, n$

consider the set of functions $f \in PROD \hspace{0.1in} S_{n}$ for which $f(a_{i}) \in U_{a_{i}}$ for $i=1,2, \ldots, n$

Let the totality of these sets be a sub-basis for the product. The resulting family of open sets satisfies the definition of space in 1.1

EXAMPLE 1.

The space $E_{n}= E_{1} \times E_{1} \times \ldots \times E_{1}$, n copies, is the space of n real variables; here $A=\{ 1, 2, 3, \ldots, n\}$ and each $S_{i}$ is homeomorphic to $E_{1}$ (1.2). Let $x_{i} \in S_{i}$. Then, $(x_{1}, x_{2}, \ldots, x_{n})$ are the co-ordinates of a point of $E_{n}$. It can be verified that the sets $U_{m}(x)$ where $m \in \mathcal{N}$, of points of $E_{n}$, whose Euclidean distance from $x = (x_{1}, x_{2}, \ldots, x_{n})$ is less than $\frac{1}{m}$ form a basis at x. The subset $R_{1} \times R_{1}\times \ldots \times R_{1}$ is an n-cell.

EXAMPLE 2.

Let A be of arbitrary cardinal power and let each $S_{n}$, $a \in A$, be homemorphic to $C_{1}$, the circumference of a circle. Then PROD $S_{n}$ is a generalized torus. If A consists of n objects, the product space is the n-dimensional torus. For n=2, we get the torus.

EXAMPLE 3.

Let $D=S_{1} \times S_{2} \times \ldots \times S_{n} \times \ldots$, where $n \in \mathcal{N}$ where each $S_{i}$ is a pair of points — conveniently regarded as the “same” pair, and designated 0 and 2. This is the Cantor Discontinuum, of Cantor Middle Third set. It is homeomorphic to the subset of the unit interval defined by the convergent series: D: $\sum { a_{n} / 3^{n}}$, where $a_{n}=0, 2$. This example will be described in another way in the next section.

THEOREM :

Let $F_{a}$ be a closed subset of the topological space $S_{a}$, and $a \in A$. Then PROD $F_{n}$ is a closed subset of PROD $S_{a}$.

Proof: The reader is requested to try. It is quite elementary.

1.7 Compactness:

DEFINITION: A topological space S is compact if every collection of open sets whose union covers S contains a finite subcollection whose union covers S.

EXAMPLE 1.

The unit interval $R_{1}$ Thus let $\{ Q\}$ denote a collection of open sets covering $R_{1}$. Let F denote the set of points $x \in R$ such that the interval $0 \leq y \leq x$ can be covered by a finite subcollection of $\{ Q \}$. Then F is not empty and is both open and closed. Hence, by the Dedekind cut postulate, or the existence of least upper bounds, or the connectedness of $R_{1}$ it follows that $F = R_{1}$ To illustrate the concept of compactness, consider the open sets $W_{n} \subset R_{1}$, $W_{n}: \frac{1}{3n}< x < \frac{1}{n}$, $n \in \mathcal{N}$. This collection does not cover $R_{1}$. Let $W_{n}$ be the union of two sets: $0 \leq x < a$ and $1-a < x \leq 1$ for some a, $0 < a < 1$. No matter how $a > 0$ is chosen, there is always some finite number of the $W_{n}$ which together with $W_{a}$ covers $R_{1}$. Of course, $R_{1}$ minus endpoints is not compact and no finite subcollection of the $W_{n}$ in this example will cover it.

THEOREM. Let S be a compact space and let $f: S \rightarrow T$ be a continuous map of S onto a topological space T Then, T is compact.

Proof:

Let $\{ O _{a}$ be a covering of T by open sets. Since f is continuous, each $f^{-1}(O_{a})$ is an open set in S. There is a finite covering of S by sets of the collection $\{ f^{-1}(O_{a})\}$ and this gives a corresponding finite covering of T by sets of $O_{a}$. This completes the proof.

COROLLARY. If f is a continuous map of S into T then $f(S)$ is a compact subset of T.

1.7.1 THEOREM

Let S be a compact space and $\{ D_{a}\}$ a collection of closed subsets such that $\bigcap_{a}D_{a}$ is empty. Then there is some finite set $D_{a_{1}}, \ldots, D_{a_{n}}$ such that $\bigcap_{i}D_{a_{i}}$ is empty.

Proof:

The complement of $\bigcap_{a}D_{a}$ is $\bigcup_{a}(S-D_{a})$; if the intersection set is empty. the union covers S. There is a finite set of indices $a_{i}$ such that $S \subset \bigcup_{i}(S - D_{a_{i}})$ and conequently $\bigcup_{i}D_{a_{i}}$ is empty for the same finite set of indices.

COROLLARY 1.

Let $D_{n}$, $n \in \mathcal{N}$ be a sequence of non empty closed subsets of the compact space S with $D_{n+1} \subset D_{n}$. Then, $\bigcap_{n}D_{n}$ is not empty.

APPLICATION:

The Cantor Middle Third Set D

From $R_{1}$, “delete” the middle third: $\frac{1}{3} < x < \frac{2}{3}$. Let $D_{1}$ denote the residue: it is a union of two closed intervals. Let $D_{2}$ denote the closed set in $D_{1}$ complementary to the union of the middle third intervals: $\frac{1}{9} < x < \frac{2}{9}$ and $\frac{7}{9} < x < \frac{8}{9}$. Continuing inductively, define $D_{n} \subset D_{n-1}$ consisting of $2^{n}$ closed mutually exclusive intervals Let $D = \bigcap_{n}$. This is homeomorphic to the space of Example 3 of 1.6.

COROLLARY 2.

A lower semi-continuous (upper semi-continuous) real-valued function on a compact space has finite glb (greatest lower bound), lub (least upper bound) and always attains these bounds at some points of space.

This follows from the preceding corollary and the fact that the set where $f(x) \leq r$ is closed, for every r (similarly, $f(x) \geq r$.

1.7.2 THEOREM

A topological space with the property: “every collection of closed subsets with empty set-intersection has a finite subcollection whose set-intersection is empty” is compact.

Proof:

The proof, like that of the Theorem of 1.7.1 is based on the duality between open and closed sets.

DEFINITION.

If a point x of a topological space S belongs to an open subset of S whose closure is compact, then S is called locally compact set at x; S i locally compact if it has this property at every point.

COROLLARY.

A closed subset of a locally compact space is locally compact in the induced topology. Similarly, a closed subset of a compact space is compact. The union of a finite number of compact subsets is compact.

Proof: HW.

A set U in a topological space is called a neighbourhood of a point z if there is an open set O such that $z \in O \subset U$; z is called an inner point of U. A set F is covered by a collection $\{ U_{i}\}$ if each point of F is an inner point of some set $U_{i}$.

1.7.3.

A space S is called a Hausdorff space if for every x, y $\in S$, $x \neq y$, there exist open sets U and V including x and y respectively such that $U \bigcap V = \phi$ where $\phi$ is the empty set; an equivalent property is the existence of a closed neighbourhood of x not meeting y.

HW: Show that a compact subset of a Hausdorff space is closed.

Lemma: Let S be a compact Hausdorff space, let F be a closed set in S, and x a point not in F. Then there is a closed neighbourhood of x, such that $W \bigcap F = \Phi$

For each $y \in F$ let $U_{y}$ be a neighbourhood of y and $W_{y}$ a neighbourhood of x, such that $U_{y} \bigcap W_{y}= \Phi$. There is a covering of F by sets $U_{y_{1}},\ldots, U_{y_{n}}$. Let $W_{x}$ be the intersection of the associated $W_{y_{i}}$ where $i = 1, 2, \ldots, n$ and let W be the closure of $W_{x}$. The union of the $U_{y_{i}}$ does not meet W. Then $W \bigcap F = \Phi$ which completes the proof.

THEOREM. Let U be a compact Hausdorff space and let $F_{n}$ be a sequence of closed subsets of U. If U is contained in the union of sets $F_{n}$, then at least one of the sets $F_{n}$ has inner points.

Proof. Take a sequence $C_{1} \supset C_{2} \supset \ldots$ of non empty compact neighbourhoods such that for each n, $(\bigcup_{1}^{n}F_{i}) \bigcap C_{n} = \Phi$. This leads to $\bigcap C_{n}=\Phi$ a contradiction. QED.

A set is nowhere dense if its closure has no inner points. A space is said to be of the second category if it cannot be expressed as the union of a countable number of nowhere dense subsets. Hence, a compact Hausdorff space is of the second category. Complete metric spaces, to be defined later, are also of the second category.

1.7.4 THEOREM:

Let S be a locally compact space. There exists a compact space $S^{*}$ and a point z in $S^{*}$ such that $S^{*}-z$ is homeomorphic to S.

Proof:

Let z denote a “new” point, not in S, and let $S^{*}$ denote the set union of S and z. If Y is a subset of S, let $Y^{*} \subset S^{*}$ denote the union of Y and z. We topologize $S^{*}$. Any open set in S is also open in $S^{*}$. In addition, if X is a compact subset of S and Y is the complement S-X, then $Y^{*}$ is open in $S^{*}$. These open sets are taken as a sub-basis for open sets in $S^{*}$.

Suppose now that we have some covering of $S^{*}$ by a family of open sets. Then z belongs to one of these open sets, say $z \in U^{*}$. The complement of $U^{*}$ is a compact subset of S . Hence, the complement is covered by a finite subset of the given covering sets, because of the compactness in S. Together with $U^{*}$, this gives a finite covering of $S^{*}$.

QED.

1.8 Tychonoff Theorem

THEOREM:

Let $S_{a}$ where $a \in \{ a \}$, be compact spaces and let P be the topological product of the $S_{a}$. Then, P is compact.

Proof:

Let $P = S_{1} \times S_{2}$ and let F denote a family of open sets of P covering P. For each point 1$x_{1}$ of $S_{1}$, the closed subset $x_{1} \times S_{2}$ of P is homeomorphic to $S_{2}$ amd is therefore compact. Each point of $x_{1} \times S_{2}$ belongs to a set in F because F is a covering. Because of the way in which a product is topologized, it follows that each point of $x_{1} \times S_{2}$ belongs to some open set $U \times V$ of P such that $U \times V$ is a subset of some set of F. It follows from its compactness that $x_{1} \times S_{2}$ is contained in the union of a finite number of sets

$U_{1} \times V_{1}, \ldots, U_{n} \times V_{n}$

each of which is a subset of some set of F. Let $U^{'} = \bigcap U_{i}$. Then, $U^{'} \times S_{2}$ is covered by a finite number of sets of F.

Since $x_{1}$ is an arbitrary point of $S_{1}$ amd $S_{1}$ is compact, there exist a finite number of open sets of $S_{1}$

$U_{1}^{'}, \ldots, U_{m}^{'}$

which cover $S_{1}$ and which are such that there is a finite number of sets of F covering $U_{1}^{'} \times S_{2}$ where $i=1, \ldots, m$. The totality of sets of F thus indicated is a finite number which covers $S_{1} \times S_{2}$. This completes the proof for the case of two factors. The case for a finite number of factors follows by a simple induction.

EXAMPLE.

Let $R_{n} = R_{1} \times R_{1} \times \ldots R_{1}$, n factors. Then, $R_{n}$ is compact and it follows that $E_{n}$ = product of n real lines is locally compact.

1.8.1

To consider the general case, let $\{ a\}$ be an arbitrary collection of at least two indices: let $S_{a}$ be compact topological spaces, let P be the topological product, and let F be a collection of open subsets of F covering P. The proof that P is compact is by contradiction. Accordingly, we shall suppose that no finite subcollection of sets of F covers P.

It was shown by Zermelo that it is possible to well-order the set of all subsets of a given set by the use of an axiom-of-choice of appropriate power, namely the cardinal number of the set of all subsets of the given set. A well-ordering of objects permits them to be inspected systematically.

Using such a well-ordering we can enlarge the given family F to a family $F^{*}$ of open sets, where $F^{*}$ has the following properties:

1. $F^{*}$ is a covering of P by open sets.
2. No finite subcollection of P by open sets.
3. If we adjoin to $F^{*}$ any open subset of P not already in $F^{*}$, then the enlarged collection does contain a finite subcollection which covers P. Of course, it is in (3) that $F^{*}$ has a property not necessarily true of F

Using this enlarged family the proof for the general case becomes similar to the proof for two factors. Let b denote an arbitrary index in $\{ a \}$ which shall be fixed temporarily, and let $P_{b}$ denote the product of all factors $S_{a}$ excepting $S_{b}$. Then P is homeomorphic to $S_{b} \times P_{b}$.

Suppose for a moment that to each point $x_{b} \times S_{b}$ there exists an open $U_{b} \subset S_{b}$ containing $x_{b}$ such that

*) $U_{b} \times P_{b} \in F^{*}$

There must then exist some finite covering of $S_{b}$ by sets $U_{b}^{1}, U_{b}^{2}, \ldots, U_{b}^{n}$ each satisfying *). The producgt P is covered by the union of $U_{b}^{i} \times P_{b}$, $i=1, \ldots, n$. This is impossible by the contradiction of $F^{*}$. Hence in each $S_{b}$ there is at least one $x_{b}$ which does not satisfy the first sentence of this paragraph.

It follows by the axiom of choice that P contains at least one point $x =$ PROD $x_{b}$ such that if $U_{b}$ is an open set in $S_{b}$ and $x_{b} \in U_{b}$ then *) is false. This holds for each coordinate $x_{b}$ of x. This implies for each coordinate $x_{b}$ of x that if $x_{b}$ is in an open set $U_{b}$ of $S_{b}$ then there is a finite collection of sets in $F^{*}$

$O_{b}^{1}, O_{b}^{2}, \ldots, O_{b}^{n_{b}}$

which together with $U_{b} \times P_{b}$ forms a covering of P.

The point x belongs to an open set $O_{x} \in F^{*}$. There is some open set contained in $O_{x}$ which contains x and is of the form

$U_{a_{1}} \times U_{a_{2}} \times \ldots U_{a_{n}} \times P_{a_{1}a_{2}a_{3}\ldots a_{n}}$

for some finite set of indices $a_{i}$ and where the last set is the product of all $S_{n}$ with the exception of $S_{a_{i}}$, where $i=1,2, \ldots, n$. For each $a_{i}$ there exists a finite collection of sets of $F^{*}$ which together with $U_{a_{i}} \times P_{a_{i}}$ covers P, say these sets are

**) O_{a_{i}}^{1}, O_{a_{i}}^{2}, \ldots, O_{a_{i}}^{n_{i}}, where $i=1, 2, \ldots, n$

Then P is covered by the union of $O_{x}$ and the sets of **). This contradiction completes the proof.

QED.

1.8.2 EXAMPLE

The infinite-dimensional torus described in 1.6 whose “dimension” equals the cardinal power of the set of indices A is compact. It is a commutative group where the addition of two points is carried out by adding the respective coordinates in each factor $S_{n} = C_{1}$ each of these factors being itself a commutative group. The group addition is continuous in the topology and this defines a topological group (1.11) In fact, this is a universal compact commutative topological group (depending on the cardinal power of the group). See for example the following paper: Discrete Abelian groups and their character groups, Ann. of Math., (2) 36 (1935) pp. 71-85.

The principal theorem of this section is due to Tychonoff (see: Uber einen Funktionenraum Math. Ann. III (1935), pp. 762-766). The present proof is dual to a proof given by Bourbaki (see: Topologie generale, Paris, 1942).

1.9 Metric Spaces

DEFINITION.

A set S of points is called a metric space if to each pair $x, y \in S$ there is associated a non-negative real number $d(x,y)$ the distance from x to y, satisfying

1. $d(x,y)=0$ if and only if $x=y$.
2. $d(x,y)=d(y,x)$
3. $d(x,y) + d(y,z) \geq d(x,z)$ where $x,y,z \in S$.

The distance function $d(x,y)$ also called the metric induces a topology in S as follows. For each $r >0$ let $S_{r}(x)$ denote the sphere of radius r, that is, the set of $y \in S$ such that $d(x,y) . Now let $S_{r}(x)$, for all positive r and all $x \in S$ constitute a basis for open sets. This choice of basis makes S a topological space. A space is called metrizable if a metric can be defined for it which induces in it the desired topology. It is clear that a metric space has a countable basis at each point x, namely $S_{r}(x)$ where r is rational.

EXAMPLE 1.

If $S_{1}$ and $S_{2}$ are metric spaces then $S_{1} \times S_{2}$ is a metric space in the metric

$d((x_{1}, x_{2}),(y_{1}, y_{2})) = max (d(x_{1},y_{1}), d(x_{2},y_{2}))$

where $x_{1},y_{1} \in S_{1}$, and $x_{2}, y_{2} \in S_{2}$. The topology determined by this metric is the same as the product topology.

EXAMPLE 2.

The set F of continuous functions defined on a compact space S with values in a metric space M becomes a metric space by defining for $f, g \in F$

$d(f,g) =$ lub $(x \in S)$ [$d_{M}(f(x), g(x))$] where $d_{M}$ is the metric in M. See corollary 2, section 1.7.1

THEOREM:

The collection of open sets of a compact metric space S has a countable basis.

Proof:

For each $n \in \mathcal{N}$ there is a covering of the space by a finite number of open sets each of diameter at most $\frac{1}{n}$. The countable collection of these sets for all n is a basis.

EXAMPLE 3:

If S is a compact metric space and $E_{1}$ denotes the real line, then $S \times E_{1}$ is a metrizable locally compact space with a countable basis for open sets.

By Example 1 above, the space is metrizable. If $\{ U_{m}\}$ and $\{ V_{m}\}$ are countable bases in S and $E_{1}$ respectively then $\{ U_{m} \times V_{m}\}$ forms a countable base in $S \times E_{1}$. If $E_{1n} = \{ x \in E_{1}, |x| \leq n\}$ then $S \times E_{1n}$ is a compact subset of $S \times E_{1}$ and any point of the product is interior to $S \times E_{1n}$ for n large enough. This proves the local compactness.

1.9.1

The following is of interest:If $(x,y)$ is a metric for a space M then the following equivalent metric:

$(x,y)^{'} = \frac{(x,y)}{1+(x,y)} \leq 1$

is a bounded metric. Properties (1) and (2) above are obviously satisfied. For (3) , one uses the fact that the function $\frac{t}{1+t}$ increases with t. Thus

$(x,y)^{'}+(y,z)^{'} \geq \frac{(x,y)}{1+(x,y)+(y,z)} + \frac{(y,z)}{1+(y,z)+(x,y)} = \frac{(x,y)+(y,z)}{1+(x,y)+(y,z)} \geq \frac{(x,z)}{1+(x,z)} = (x,z)^{'}$

This has the following consequence:

Lemma:

Let M be a space which is the union of a system $M_{a}$ where $a \in \{ a\}$ of open mutually exclusive sets. Suppose each $M_{a}$ of open mutually exclusive sets. Suppose each $M_{a}$ is a metric space and carries a metric $d_{a}$ bounded by 1. Define a function $d(x,y)$ which is equal to 2 if x and y are not in the same $M_{a}$; otherwise let d agree with the appropriate $d_{a}$. Then d is a metric for M.

Proof: HW.

1.9.2

A sequence of points $x_{n}$ in a metric space is said to converge to a point x, symbolically $x_{n} \rightarrow x$ if $\lim d(x,x_{n}) =0$. A sequence of points $x_{n}$ satisfies the Cauchy convergence criterion if when $\epsilon >0$ is given there is an N such that for $m,n >N$ $d(x_{n},x_{m})<\epsilon$. A metric space is called complete if every sequence of points satisfying the Cauchy criterion converges to a point of the space. A subset of a space is called dense (everywhere dense) in the space if every point of space is a limit of some sequence of points of the subsets.

To be continued in next blog,

Cheers,

Nalin Pithwa

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