Topological Spaces and Groups: Part 2: Fast Review

Reference: Chapter 1, Topological Transformation Groups by Deane Montgomerry and Leo Zippin

1.10 Sequential Convergence:

The proof of the principal theorem of this section illustrates a standard technique. It will involve choosing an infinite sequence, then an infinite subsequence, then again an infinite subsequence and so on repeating this construction a countably infinite number of times. A special form of this method is called the Cantor diagonalization procedure. To facilitate the working of this technique we shall sometimes use the following notation:


The letter I will denote the sequence of natural numbers 1, 2, 3, …When subsequences of I need to be chosen, they will be labelled in some systematic way: I_{1}, I_{2}, \ldots or I^{'}, I^{''}, \ldots or I^{*}, I^{**}, \ldots and so on. Then given a sequence of elements x_{n}, n \in I, we can refer to a subsequence as : x_{n}, n \in I_{1}, or : x_{n}, n \in I^{*} and so on.


Let S denote a metric space and let K_{n}, n \in I be a sequence of subsets of S. The sequence K_{n} is said to converge to a set K if for every \epsilon >0

1.10.3 K_{n} \subset S_{\epsilon}(K) and K \subset S_{\epsilon}(K_{n})

for n sufficiently large (depending only on \epsilon)

If K_{n} is given and if a K exists satisfying relation 1.10.3 then \overline{K} also satisfies 1.10.3. If two closed sets K^{'} and K^{''} satisfy 1.10.3 for the same sequence K_{n} then K^{'}=K^{''}. In the special case that the sets K_{n} are single points, the set K if it exists is a point and is unique.

1.10.4 THEOREM

Every sequence of non-empty subsets of a compact metric space S has a convergent subsequence.


Let K_{n}, n \in I be an arbitrary sequence of subsets of S and let W_{n}, n\in I be a basis for open sets in S. (Theorem 1.9)

Let I be called I_{0} and suppose a sequence I_{m-1} has been defined. Consider W_{m} \bigcap K_{n}, n \in I_{m-1} m fixed. Then either W_{m} \bigcap K_{n} is not empty for an infinite subsequence of integers n \in I_{m-1} or on the contrary W_{m} \bigcap K_{n} is empty for almost all n \in I_{n-1}. In the first of these cases define I_{m} as the set of indices n in I_{m-1}. In the first of these cases define I_{m} as the set of indices n in I_{m-1} such that W_{m} \bigcap K_{n} is not empty for n \in I_{m}. Then in all possible cases I_{m} \subset I_{m-1} is uniquely defined. We now consider I_{m} to be defined by induction for all m \in I.

We can now specify what subsequence of K_{n} we may take as convergent. Let I^{*} \subset I denote the diagonal sequence of the sequences I_{m}, that is I^{*} contains the m-th element of I_{m} for each m. We shall show that K_{n}, n \in I^{*} is convergent. It follows from the definition of I^{*} that for each m \in I, if we except at most the first m integers in I^{*},

W_{m} \bigcap K_{n}, n \in I^{*},

is always empty or is never empty depending on m.

We next define the set K to which the sets K_{n}, n \in I^{*} will be shown to converge. Let W denote the union of those sets W_{m}, m \in I for which W_{m} \bigcap K_{n}, n \in I^{*} is almost empty. Let K = S - W Since each W_{m} forming W meets at most a finite number of the sets K_{n} no finite number of these W_{m} can cover S. Therefore W cannot cover S, since S is compact, and hence K is not empty. The set K is closed and therefore compact.

Let \epsilon >0 be given. There is a covering of K by sets W_{k_{1}}, W_{k_{2}},  \ldots, W_{k_{s}}, k_{i} \in I

each meeting K and each of diameter less than \epsilon. None of the sets W_{k_{i}} can belong to W and each must intersect almost all the K_{n}, n \in I^{*}. Therefore for sufficiently large n, n \in I^{*}

W_{k_{i}} \bigcap K_{n} \neq \Phi

It follows that K \subset S_{\epsilon}(K_{n}), n \in I^{*} for all n sufficiently large. Finally it can be seen that the closed set S - S_{\epsilon}(K) \subset W. It follows that the complement of S_{\epsilon}(K) is covered by sets W_{j_{1}}, W_{j_{2}}, \ldots, W_{j_{t}} each of which is an element in the union defining W. Therefore W_{j_{k}} \bigcap K_{n}, n \in I^{*} k=1,2, \ldots t

is almost always empty. Therefore for sufficiently large n \in I^{*}

K_{n} \subset S_{\epsilon}(K).

This completes the proof of the theorem. QED.


Let X and Y be closed subsets of a compact metric space S. Define Hausdorff metric: d(X,Y) as the greatest lower bound of all \epsilon such that symmetrically X \subset S_{\epsilon}(Y), Y \subset S_{\epsilon(Y)}

This is a metric for the collection of closed subsets of S.


The set F of all closed subsets of a compact metric space S is a compact metric space in the metric defined above.


The set F is a metric space in the metric defined above. It A_{n} is in F then A_{n} has a subsequence converging to a set A in the sense of convergence defined above and the set A may be assumed closed. It follows that the subsequence also converges to A in the sense of the metric of F. Hence, every sequence in F has a convergent subsequence.

Let W_{n} where n \in \mathcal{N} be a basis for open sets in S as in the preceding Theorem. For each n and each choice of integers k_{1}, k_{2}, \ldots, k_{n} let W(k_{1}, k_{2}, \ldots, k_{n}) denote the union of the sets W_{k_{1}}, W_{k_{2}}, \ldots, W_{k_{n}}. Now in F let W^{*}(k_{1}, \ldots, W_{k_{n}}) consist of all the compact sets in S which belong to W(k_{1}, \ldots, k_{n}) and meet each W_{k_{i}}. This gives a countable collection of subsets of F. The proof of the preceding theorem shows that this collection is a basis for open sets in F.

Finally, let \{ O_{n}\} where n=1, 2, …be a countable collection of open sets of F which cover F. We have to find an integer m such that \bigcup_{1}^{m}O_{i} covers F. If no such integer existed we could find a sequence of points x_{n} where x_{n} \in F - \bigcup_{1}^{n}U_{i}. This sequence would have to have a susbsequence converging to some point x. Since x \in O_{m} for some m, it follows that infinitely many of the x_{n} belong to O_{m}; this contradiction proves the theorem.



Note that separability implies that every collection of covering sets has a countable covering subcollection. It also implies that there exists a countable set of points which is everywhere dense in the space.


A metric space S is compact if and only if every infinite sequence of points has a convergent subsequence.


Suppose that every infinite subsequence of points of S has a convergent subsequence. We shall prove that S is separable. The last paragraph of the preceding section then shows that S is compact. The converse is shown in 1.10.4

For each positive integer n construct a set P_{n} such that (1) every point of P_{n} is at a distance at least \frac{1}{n} from every other point of P_{n} (2) every point of S not in P_{n} is at a distance less than 1/n from some point of P_{n}. It is easy to see that no sequence of points in any one P_{n} can be convergent and it follows that P_{n} is a finite point set. Let P= \bigcup P_{n}. Then P is countable and every point of S is a limit point of P. For each rational r >0 and each point of P construct the “sphere” with that point as centre and radius r. The set of these spheres is countable and is a basis for open sets. This concludes the proof.



Let S be a compact metric space let H be the space of real continuous functions defined on S with values in R_{1}. Each continuous function f(x) determines a closed subset of S \times R_{1} namely the graph consisting of the pairs (x, f(x)), x \in S. Hence H is a subset of a compact metric space (see examples 2 and 3 in Sec 1.9).

Example 2 of Section 1.9 reproduced below:

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. Recall also the following here in this example: 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. From this theorem it follows that: A lower semi-continuous (upper semi-continuous) real valued function on a compact space has finite g.l.b. (and L.u.b) and always attains these bounds at some points of space.

Example 3 of Section 1.9 reproduced below:

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.

Remarks: Notice that the metric defined for H in example 2 of 1.9 and the metric which it gets from S \times H_{1} are topologically equivalent. This proves that H itself is a separable metric space.



Topological groups were first considered by Lie, who was concerned with groups defined by analytic relations (to be discussed later). Around 1900-1910 Hilbert and others were interested in more general topological groups. Brouwer showed that the Cantor middle third set can be made into an abelian topological group. Later Schreier and Leja gave a definition in terms of topological spaces whose theory had been developed in the intervening time.

A topological group is a topological space whose points are elements of an abstract group, the operations of the group being continuous in the topology of the space. A detailed definition containing some redundancies is as follows:


A topological group G is a space in which for x, y \in G there is a unique product xy \in G, and

i) there is a unique identity element e in G such that xe = ex =x for all x \in G

ii) to each x \in G there is an inverse x^{-1} \in G such that xx^{-1} = x^{-1}x=e

iii) x(yz) = (xy)z for x, y, z \in G

iv) the function x^{-1} is continuous on G and xy is continuous on G \times G

Familiar examples are the real or complex numbers under addition with the usual topologies for E_{1} and E_{2} respectively, and the complex numbers of absolute value one under multiplication with their usual topology as a subset of E_{2}. The space of this last group is homeomorphic to the circumference of a circle.


Of course, properties 1, 2 and 3 define a group in the customary sense and a topological group may be thought of as a set of elements which is both an abstract group and a space, the two concepts being united through 4. Wnen a subset H of G is itself a group we shall call H a subgroup of G, but we shall understand that H is to be given the relative topology. It is easy to see that then H becomes a topological group.

If H is a subgroup of G and if x and y are points of G belonging to the closure of H, then every neighbourhood of the product xy contains points of H. For, let U be a neighbourhood of xy. Then by property 4, there exists neighbourhoods V of x and W of y such that every product of an element of V and an element of W is contained in U. We see from this that xy belongs to the closure of H. Similarly, x^{-1} belongs to the closure. Thus, \overline{H} is a group, and we shall call it a closed subgroup.


If G_{a} where a \in \{ a\} is a collection of topological groups and if G denotes the product space defined in 1.6, (previous blog), then G can be regarded as a group (the product of two elements of G being defined by the product of their components in each factor G_{a}). Because each neighbourhood of the PROD G_{a} depends on only a finite number of the factors it is easy to see that G becomes a topological group. We shall call it the topological group of the factors G_{a}.

By way of examples, note that the product of an arbitrary number of groups each isomorphic to C_{1}: the group of reals modulo one (isomorphic to the complex numbers of modulus one under multiplication) is a compact topological group. The product of an arbitrary number of factors each isomorphic to the group of reals is a topological group which is locally compact if the number of factors is finite.

A finite group with the discrete topology is compact and the topological product of any collection of finite groups is therefore compact.


If x and y are in topological group, x \neq y, then as will be seen (section 1.16) we may choose a neighbourhood W of e such that

y \notin WW^{-1}

Hence, yW and xW are disjoined, and thus two distinct points of a topological group are in disjoined open sets. This is called the Hausdorff property (see section 1.10); it implies that a point is a closed set.

It is easy to see that if H is an abelian subgroup of a topological group G then \overline{H} is also abelian. Thus, if x, y \in \overline{H} and xy \neq yx there are neighbourhoods U_{1} of xy and U_{2} of yx with U_{1} \bigcap U_{2} = \Phi. There exist neighbourhoods V of x and W of y such that for every v \in V and w \in W vw \in U_{1} and wv \in U_{2}. However, if v, w \in H then wv = vw and we are led to a contradiction proving that the closure of H is abelian.



Important examples of topological groups are given below:


The sets M_{n}(R) and M_{n}(C) of all n \times n matrices of real and complex elements under addition with the distance of A = (a_{ij}), B = (b_{ij}) defined by

d(A, B) = max_{i,j}|a_{ij} - b_{ij}|

The spaces of these two groups are homeomorphic to E_{n^{2}} and E_{2n^{2}}. They are in fact the sets of real or complex vectors with n^{2} co-ordinates, and hence are vector spaces as well as groups. Another example is the set H of continuous real valued functions on a metric space under addition.


The sets of non-singular real or complex n \times n matrices GL(n,R), GL(n,C) under multiplication; these are subsets of M_{n}(R) and M_{n}(C) respectively and are given the induced topology. They are open subsets and are therefore locally compact and locally euclidean. (see 1.27 later)


Let S be a compact metric space and let G be the group of all homeomorphisms of S onto itself topologized as a subspace of the space of continuous maps of S into itself. (Section 1.9 previous blog, example 2)


The spaces associated with two topological groups may be homeomorphic but the groups essentially different; for example, one abelian and the other not.


The matrices \left |\begin{array}{cc}a & 0 \\ 0 & b \end{array}\right | where a and b are real numbers under addition. This is an abelian group with E_{2} as space.


The matrices \left| \begin{array}{cc} e^{a} & b \\ 0 & e^{-a} \end{array} \right | where a and b are real under multiplication. This is a non-abelian group with E_{2} as space.

If we give the space in this example (or example 1) the discrete topology we obtain a new topological group with the same algebraic structure.


In the additive group of integers, for each pair of integers h and k, k \neq 0 let the set \{ h \pm nk\} where n=0, 1,2, …, be called an open set and let the collection of all these sets be taken as a basis for open sets.


Introduce a metric into the additive gorup of integers, depending on the prime number p, defined thus:

d(a,b) = \frac{1}{p^{n}}

if a \neq b and p^{n} is the highest power of p which is a factor of a-b.


Let G be the integers under addition with any set called open if it is the complement of a finite set (or is the whole space or the null set). Algebraically G is a group and it is also a space. However, it is not a topological group because addition is now not simultaneously continuous. It is true however that addition is continuous in each variable separately.

For some types of group spaces separate continuity implies simultaneous continuity. It is not known whether this is true for a compact Hausdorff group space.


Two topological groups will be called isomorphic if there is a one-one correspondence between their elements which is a group isomorphism (preserves products and inverses) and a space homeomorphism (preserves open sets).

An isomorphic map of G onto G is called an automorphism.

In examples (3) and (4) the abstract group structure of the additve group of integers is embodied in infinitely many non-isomorphic topological groups.


If G is a group A \subset G let A^{-1} denote the inverse set \{ a^{-1}\} where a \in A. Clearly, (A^{-1})^{-1}=A. If B \subset G let AB denote the set \{ ab\} where a \in A, b \in B. It is understood that the product set is empty if either factor is empty. We shall write AA = A^{2} and so on. It can be seen that (AB)C = A (BC) and (AB)^{-1} = B^{-1}A^{-1}. The set AA^{-1} satisfies


that is, it is symmetric. Similarly, the set intersection A \bigcap A^{-1} is symmetric. The intersection of symmetric sets is symmetric.

A set H in G is called invariant if gH = Hg for every g \in G equivalently if gHg^{-1}=H.


Let G be a topological group and let A \subset G be an open set. Then A^{-1} is open.


Let a^{-1} be in A. By the continuity of the inverse there exists an open set B containing a such that b \in B implies b^{-1} \in A. This means that B^{-1} \subset A and therefore B \subset A^{-1}. Thus A^{-1} is a union of open sets and is open.


The map x \rightarrow x^{-1} is a homeomorphism.


Let G be a topological group, A an open subset, b an element. Then Ab and bA are open.


Let a \in A and let c=ab. Then a = cb^{-1}. Because “a” regarded as a product is continuous in c there must exist an open set C containing c such that if c^{'} \in C then c^{'}b^{-1} \in A. But then c^{'} \in Ab and it follows that C \subset Ab. Therefore Ab is a union of open sets and is open. The proof that bA is open is similar.



For each a \in G, the left and right translations : a \rightarrow ax and x \rightarrow xa are homeomorphisms.


Let G be a topological group and let A and B be subsets. If A or B is open then AB is open.


Since AB is a union of sets of the form Ab, b \in B, it is open if A is open. SimilarlyAB is a union of sets aB, a \in A and is open if B is open.


Let A be a closed subset of a topological group. Then Ab and bA are closed.


This is true because left and right translations by the constant b are homeomorphisms of G onto G.

Now, a function f(x) taking a group G_{1} into another group G_{2} will be called a homomorphism if

*) f(x)f(y) = f(xy) where x, y \in G_{1}

When G_{1} and G_{2} are topological we shall ordinarily require that f be continuous. The most useful case is where f is open as well as continuous. In many situations (see 1.26.4 and 2.13 in later blogs) continuity implies openness but this is not true in general. The set of elements going into e is a subgroup and if f is continuous it is a closed subgroup (a point in a topological group is a closed set). This is the kernel of the homomorphism.


Let V_{1} be the additive group of real numbers and G be a topological group. A continuous homomorphism h(t) of V_{1} into G is called a one-parameter group in G. If h is defined only on an open interval around zero satisyfying the definition of homomorphism so far as it has meaning, then h(t) is called a local one-parameter group in G. If h(t) is a one-parameter group, the image of V_{1} may consist of e alone and then h(t) is a trivial one-parameter group. If this is not the case and if for some t_{1} \neq 0 and h(t_{1})=e, then the image of V_{1} is homeomorphic to a circumference. In case h(t)=e only for t =0 the image of V_{1} is a one-one image of the line which may be homeomorphism of the line or a very complicated imbedding of the line. To illustrate this let G be a torus which we obtain from the plane vector group V_{2} by reducing mod one in both the x and y directions. In V_{2} any line through the origin is a subgroup isomorphic to V_{1} and after reduction the line y=ax is mapped onto the torus G thus giving a one-parameter group in G. If a is rational the image is a simple closed curve but if a is irrational the image is everywhere dense on the torus.

1.14 PRODUCTS of Closed Sets

If A and B are subgroups of a group G, AB is not necessarily a subgroup. However if A is an invariant subgroup (that is, g^{-1}Ag = A where g \in G) and B is a subgroup then AB is a subgroup.

The product of closed subsets, even if they are subgroups, need not be closed. As an example let G be the additive group of real numbers, H_{1} the subgroup of integers \{ \pm n\} and H_{2} the subgroup \{ \pm n\sqrt{2}\}. The product H_{1}H_{2} is countable and a subgroup but it is not a closed set.

It will be shown later that if A is a compact invariant subgroup and B is a closed subgroup then AB is a closed subgroup (corollary of 2.1 in later blogs). (Remark: I think in I N Herstein’s language of Topics in Algebra, an “invariant subgroup” is a normal subgroup. Kindly correct me if I am mistaken).

1.15 Neighbourhoods of the identity

Let G be a topological group and U an open subset containing the identity e. We showed in 1.13 that xU is open and clearly x \in xU. Conversely, if x \in O, O is open, then U = x^{-1}O is an open set containing e.

If a collection of open sets \{ U_{a}\} is a basis for open sets at e then every open set of G is a union of open sets of the form x_{a}U_{a}, where x_{a} \in G, U_{a} \in \{ U_{a}\} and the topology of G is completely determined by the basis at e. In particular the collection \{ xU_{a}\} is a basis for open sets at x so also is \{ U_{a}x\}

If U is a neighbourhood of e, U^{-1} is a neighbourhood of e and U \bigcap U^{-1} is a symmetric neighbourhood of e.


Let G be a topological group and U a neighbourhood of e. There exists a symmetric neighbourhood W of e such that W^{2} \subset U.


Since e.e = e and the product is simultaneously continuous in x and y there must exist neighbourhoods V_{1} and V_{2} of e such that V_{1}V_{2} \subset U. Define W = V_{1} \bigcap V_{1}^{-1} \bigcap V_{2} \bigcap V_{2}^{-1}. Then W^{2} \subset U and this completes the proof.



Let G be a topological group. If x \neq e there exists a neighbourhood W of e such that W \bigcap xW is empty.


Since G is a topological space there is a neighbourhood V_{1} of e not containing x or there is a neighbourhood V_{2} of e not containing x. In either case there is a neighbourhood of e not containing x. Let W be a symmetric neighbourhood of e, W^{2} \subset U. If W \bigcap xW were not vacuous there would exist w, w^{'} \in W with w^{'}=w. But this gives x = w^{'}w^{-1} \in W^{2} \subset U which is false.



If G is a topological group then a set S of open neighbourhoods \{ V\} which forms a basis at the identity e has the following properties:

(a) the intersection of all V in S is \{ e\}

(b) the intersection of two sets of S contains a third set of S

(c) given U in S there is a V in S such that VV^{-1} \subset U

(d) if U is in S and a \in U then there is a V in S such that Va \subset U

(e) If U is in S and a is in G there is a V in S such that aVa^{-1} \subset U.

Conversely, a system of subsets of an abstract group having these properties may be used to determine a topology in G as will be formulated in the following theorem, the proof of which is contained in main part in remarks already made.


Let G be an abstract group in which there is given a system S of subsets satisfying (a) to (e) above. If open sets in G are defined as unions of sets of the form Va, a \in G then G becomes topological with S a basis for open sets at e. This is the only topology making G a topological group with S a basis at e.

1.16 COSET Spaces

Let G be a group and H a subgroup. The sets xH and yH where x, y \in G either coincide or are mutually exclusive; and xH = yH if and only if x^{-1}y \in H. Each set xH is called a coset of H, more specifically a left coset. Right cosets Hx, Hy will be used infrequently. We use the notation G/H for the set of all left cosets. When G is a (topological space) G/H will be made into a space (see below 1.16.2) but we speak of it as a space in the present case also, although it carries no topology at present.

If H is an invariant subgroup, that is if x^{-1}Hx=H equivalently if xH=Hx for any x \in G, then

xHyH = xyH where x, y \in G

is a true equation in sets of elements of G. In this case, the coset space becomes a group, the factor group G/H.



By the natural map T of a group G onto the coset space G/H, H being a subgroup of G, we mean the map

T : x \rightarrow xH, where x \in G, xH \in G/H

For any subset U \subset G we have

T^{-1}(T(U)) = UH \subset G

Let G be a topological group, H a subgroup. It is useful to topologize the coset space and to do this so that the natural map T is continuous. It will become clear as we proceed that unless the group H is a closed subgroup of G, it will not be possible in general to have T continuous and G/H a topological space; for this reason only the case where H is closed will be considered.



Let G be a topological group and let H be a closed subgroup of G, that is a subgroup which is a closed set. By an open set in G/H we mean a set whose inverse under the natural map I is an open set in G/H.


With open sets defined as above, G/H is a topological space and the map T is continuous and open. If xH \neq yH, there exist neighbourhoods W_{1} and W_{2} of xH and yH respectively such that W_{1} \bigcap W_{2} is empty.


Let U denote an arbitrary open set in G. Then U/H is open (1.13) and is the inverse of T(U). Since T(U) is open in G/H, T is an open map.

Let x, y \in G and xH \neq yH. Then x \in yH and yH is closed because H is closed. There exists a neighbourhood U of e such that Ux \bigcap yH is empty. Let W be open, e \in W and W^{*} \subset U (Theorem in 1.15 above). Then if $latex WxH \bigcap WyH is not empty we can find w, w_{1} \in W and h, h_{1} \in H so that w_{1}xh = wy h_{1}. But this leads to w^{-1}w_{1}x = yh_{1}h^{-1} and implies that Ux meets yH. Therefore WxH \bigcap WyH is empty. The sets Wx and Wy are open. Therefore W_{1}=T(Wx) and W_{2} = T(Wy) are open in G/H; xH \in W_{1} yH \in W_{2} and W_{1} \bigcap W_{2} is empty. This is the main part of the proof and depends on the fact that H is closed. We have proved more than condition (4) of 1.1 (The T_{0} separation axiom) The remaining three conditions of 1.1 are easy to verify. The fact that T is a continuous map is stated in the definition of open set in G/H. The fact that T is open was proved in the last paragraph.



If G is a topological group, x, y \in G where x \neq y, then there exist neighbourhoods W_{1} of x and W_{2} of y such that W_{1} \bigcap W_{2} is empty.


To see this it is only necessary to take H=e. A space in which every pair of distinct points belong to mutually exclusive open sets is called a Hausdorff space. Therefore it has been shown that a topological group G and a coset space G/H, H closed in G, are Hausdorff spaces.


Suppose that G is a topological group and H a closed invariant subgroup. Then with the customary definition of product : (xH)(yH) = xyH,, G/H becomes a topological group. The natural map of G onto G/H is a continuous and open homomorphism.


Suppose that we are given a topological group G and a sequence of neighbourhoods of e: Q_{0}, Q_{1}, \ldots By repeated of the following Theorem 1.15: ( Let G be a topological group and U a neighbourhood of e. There exists a symmetric neighbourhood W of e such that W^{2} \subset U) .We can choose a sequence of symmetric open neighbourhoods of e : U_{0}, U_{1}, \ldots with U_{0}=Q_{0} such that

(1) U_{n+1}^{2} \subset U_{n}\bigcap Q_{n} where n=0, 1, ….

In this section we shall show how to imbed the sets U_{n} in a larger family of neighbourhoods possessing a multiplicative property which generalizes (1). We shall use this family in the next section to construct a real and non-constant function which is continuous on G. In 1.22 we shall use a similar family in order to construct a metric in a metrizable group. We remark in passing that in groups which do not satisfy the first countability axiom the set: \bigcap U_{n} may be of considerable interest (to be discussed in a future blog section 2.6); it is a closed group (if x, y \in \bigcap U_{n} then for every n, xy \in U_{n+1}^{2} \subset U_{n}).

Now, for each dyadic rational r = \frac{k}{2^{n}}, for n=0, 1, …and k=1, \ldots, 2^{n} we define an open neighbourhood V_{r} of e as follows:

(2) V_{1/2^{n}}=U_{n} for all n

and then using (3) and (4) alternately by induction on k.

(3) V_{2k/2^{n+1}}=V_{k/2^{n}}

(4) V_{(2k+1)/2^{n+1}} = V_{1/2^{n+1}}V_{k/2^{n}}

Each V_{n} depends on the dyadic rational r, and not on the particular representation by k/2^{n}. The entire family has the property:

(5) V_{1/2^{n}}V_{m/2^{n}} \subset V_{m+1/2^{n}} where m+1 \leq 2^{n}

For m=2k, (5) is an immediate consequence of (3) and (4). For m=2k+1 the left side of (5) becomes: V_{1/2^{n}}(V_{1/2^{n}}V_{k/2^{n-1}}) \subset V_{1/2^{n-1}}V_{k/2^{n-1}}

The right side of (5) becomes V_{(k+1)/2^{n-1}}. This sets up an induction on n and since (5) holds for n=1, we have proved that (5) is true for all n. It follows from (5) and also more directly that:

(6) V_{r} \subset V_{r^{'}} if r<r^{'} \leq 1



Suppose that G is a topological group and that F is a closed subset of G not containing e. Then one can define on G a continuous real function f, 0 \leq f(x) \leq 1, f(e) =0, f(x)=1 if x \in F.


In virtue of the property described in the theorem, G is called a completely regular space at the point e. The theorem is due to Pontrjagin.

Set Q_{a}=G-F and setting Q_{n}=Q_{0} for every n, construct a family of sets V_{r} as in the preceding section.

Define f(x), x \in G as follows:

(1) f(x)=0 if x \in V_{r} for every r

(2) f(x)=1 if x \in V_{1}

and in all other cases

(3) f(x)=lub_{r} where \{ r \leq 1, x \in V_{r}\}

It is clear that e belongs to every V_{r} and that F = G - V_{1} and does not meet V_{1} so that there remains only to prove that f is continuous; let \epsilon>0 be given and let n be a positive integer such that 1/2^{n}<\epsilon.

Now suppose that f(x) <1 at some point x \in G. Then there is a pair of integers m and k such that k >n (same n as above), m < 2^{n} and (interpreting V_{0}, if it occurs, as the null set),

x \in V_{m/2^{k}}-V_{m-1/2^{k}}

Let y be an arbitrary element in the neighbourhood V_{1/2^{k}}x. Then,

y \in V_{1/2^{k}}V_{m/2^{k}} \subset V_{(m+1)/2^{k}}

By the choice of y, yx^{-1} \in V_{1/2^{k}}; therefore xy^{-1} \in V_{1/2^{k}} and x \in V_{1/2^{k}}y. It follows from this that y cannot belong to V_{(m-2)/2^{k}} and this shows that

(m-2)/2^{k} \leq f(y) \leq (m+1)2^{k}

Concerning x we know that

(m-1)/2^{2^{k}} \leq f(x) \leq m/2^{k} and we may conclude from both inequalities that

|f(x)-f(y)| \leq 2/2^{k} \leq 1/2^{n}<\epsilon

Suppose next that f(y)=1 and choose k>n as before. Let y be an arbitrary element in V_{1/2^{k}}x. Now y cannot belong to V_{m/2^{k}} with m < 2^{k}-2 without implying that f(x)<1. It follows that

1-2/2^{k} \leq f(y) \leq 1

and again we get |f(x) - f(y)| \leq e

This concludes the proof and f(x) is continuous on G.



A topological group is completely regular at every point.



Let G be a topological group, H a closed subgroup. Each element of G determines a homeomorphism of the coset space G/H onto itself, and G becomes a group of homeomorphisms of this space (topological space); furthermore G is transitive on the space: that is, each point may be carried to any other by an element of G.


Let a \in G. Associate to a the mapping T_{a}: xH \rightarrow axH.

This is a one-one transformation of G/H onto itself with T_{a^{-1}} as inverse. These transformations are open and each transformation is a homeomorphism.

Since T_{b}T_{a} is given by

T_{b}T_{a}(xH) = T_{b}(axH)=baxH = T_{ba}(xH).

the association of a \in G and T_{a} makes a group of transformations of G/H. It is clear that xH is carried to yH by T_{a} with a = yx^{-1}. This completes the proof.


A space is called HOMOGENEOUS when a group of homeomorphisms is transitive on it. We have shown that G/H is homogeneous with G being the transitive group of homeomorphisms. This implies that the unit segment R_{1}, for example, cannot be the underlying space of a group or even a coset space G/H since an end point of R_{1} cannot go to an interior point to a homeomorphism of R_{1}.

It follows from the simultaneous continuity of ax \in G in the factors a and x that the image point axH of xH under T_{a} is continuous simultaneously in the counter point xH and the element T_{a} of G. This makes G an instance of what is called a topological transformation group of a space M which will be defined below.

However, we shall be principally concerned with transformation groups which are locally compact and separable, acting on spaces which are topologically locally euclidean.


An open neighbourhood of the identity of a topological group when it is regarded as a space in the relative topology has some of the properties of a group. There will usually be pairs of elements for which no product element exists in the neighbourhood. A structure of this kind is called a local group and will be defined below. Local groups often arise in a natural way, especially in the case of analytic group (Lie groups of transformations) and they have been intensively studied in that form


A space G is called a local group if a product xy is defined as an element in G for some pairs x, y in G and the following conditions are satisfied:

i) there is a unique element e in G such that ex and xe are defined for each x in G and ex = xe=x.

ii) If x, y are in G and xy exists then there is a neighbourhood U of x and a neighbourhood V of y such that if x^{'} \in U and y^{'} \in V then x^{'}y^{'} exists. The product xy is continuous wherever defined.

iii) The associative law holds whenever it has meaning.

iv) If ab=e then ba=e. An element b satisfying this relation is called an inverse and is denoted by a^{-1}. We assume that a^{-1} is unique and continuous where defined and that if it exists for an element a it exists for all elements in some neighbourhood of a. Note that a^{-1} always exists in some neighbourhood of e. In fact, there exists a symmetric open neighbourhood U of e such that U^{2} is defined.

The above definition is somewhat redundant.


Any neighbourhood O of the identity oa topological group is a local group if the neighbourhood is open.

We shall call two local groups isomorphic if there is a homeomorphism between their elements which carries inverse to inverse and product to product in so far as they are defined. However, in some applications, it is natural to regard two local groups as equivalent if they belong to the same local equivalence class, that is, a neighbourhood of e in one is isomorphic to a neighbourhood of e in the other. In this book an isomorphism and preserves group operations so far as they are defined.


Let G be a local group with U the symmetric open neighbourhood of e described in the definition. Given any neighbourhood V of e, V \subset U, there is a symmetric neighbourhood W of e, W^{3} \subset V. The product sets AB, BC, (AB)C, A(BC) exist for A, B, C \subset W and A(BC) = (AB)C. The set AB is open if either A or B is open. The sets A, bA, and Ab are homeomorphic for b \in W. Any two points of W have homeomorphic neighbourhoods.

The proof is omitted, the details being as in 1.13, 1.14 and 1.15.


Nalin Pithwa

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