##### Document Text Contents

Page 70

§18] COMPACTA 59

of closed sets satisfying the condition

n(R '" 0 .. ) = o.

The condition that we can select a finite subcovering from {O .. } is equiva-

lent to the fact that the system of closed sets {R '" O .. } cannot have the

finite intersection property if it has a void intersection.

We shall now prove that Condition 1 is necessary and sufficient that R

be a compactum.

Necessity. Let R be a compactum and let {O .. } be an open covering of R.

We choose in R for each n = 1, 2, ... a finite (l/n)-net consisting of the

points ak (n) and we enclose each of these points with the sphere

S(ak(n), lin)

of radius lin. It is clear that for arbitrary n

R = UkS(A k (n), lin).

We shall assume that it is impossible to choose a finite system of sets

covering K from {O .. }. Then for each n we can find at least one sphere

S(ak(n)(n), lin) which cannot be covered by a finite number of the sets 0 ...

We choose such a sphere for each n and consider the sequence of their

centers {ak(n) (n)}. Since R is a compactum, there exists a point ~ E R which

is the limit of a subsequence of this sequence. Let Ofj be a set of {O .. } which

contains~. Since Ofj is open, we can find an E > 0 such that S{~, E} C Ofj .

We now choose an index n and a point ak(n) (n) so that p(~, ak(n) (n») < E/2,

lin < E/2. Then, obviously, S(ak(n)(n), lin) c Ofj, i.e. thesphereS(ak(n)(n),

lin) is covered by a single set Ofj . The contradiction thus obtained proves

our assertion.

Sufficiency. We assume that the space R is such that from each of its

open coverings it is possible to select a finite subcovering. We shall prove

that R is a compactum. To do this it is sufficient to prove that R is complete

and totally bounded. Let E > O. Take a neighborhood O(x, E) about each

of the points x E R; we then obtain an open covering of R. We choose from

this covering a finite subcovering O(Xl , E), ... ,O(xn , E). It is clear that the

centers Xl, ••• , Xn of these neighborhoods form a finite E-net in R. Since

E > 0 is arbitrary, it follows that R is totally bounded. Now let {Sn} be a

sequence of nested closed spheres whose radii tend to zero. If their inter-

section is void, then the sets R '" S" form an open covering of R from which

it is impossible to select a finite subcovering. Thus, from Condition 1 it

follows that R is complete and totally bounded, i.e. that R is compact.

THEOREM 5. The continuous imO{Je of a compactum is a compactum.

Proof. Let Y be a compactum and let Y = cp(X) be its continuous image.

Further, let {O .. } be an open covering of the space Y. Set U .. = cp-l(O .. ).

Page 71

60 METRIC SPACES [CH. II

Since the inverse image of an open set under a continuous mapping is open,

it follows that {U a} is an open covering of the space X. Since X is a com-

pactum, we can select a finite subcovering U1 , U2, ... , Un from the

covering {Ua }. Then the corresponding sets 0 1 , O2 , ••• , 0" form a finite

subcovering of the covering {O a}.

THBOREM 6. A one-to-one mapping of a compactum which is continuous in

one direction is a homeomorphism.

Proof. Let cp be a one-to-one continuous mapping of the compactum X

onto the compactum Y. Since, according to the preceding theorem, the

continuous image of a compactum is a compactum, the set cp(M) is a com-

pactum for an arbitrary closed M C X and consequently cp(M) is closed in

Y. It follows that the inverse image under the mapping cp -1 of an arbitrary

closed set M C X is closed; i.e. the mapping cp -1 is continuous.

RI<;MARK. The following result follows from Theorem 6: if the equation

(3) dy/dx = f(x, y),

where the function f(x, y) is continuous in a closed bounded region G

which contains the point (xo, Yo) for every Yo belonging to some closed

interval [a, b], has a unique solution satisfying the initial condition y(xo) =

Yo , then this solution is a continuous function of the initial value Yo .

In fact, since the function f(x, y) is continuous in a closed bounded

region, it is bounded and consequently the set P of solutions of equation

(3) corresponding to initial values belonging to the closed interval [a, b]

is uniformly bounded and equicontinuous. Moreover, the set P is closed.

In fact, if {cp,,(x)} is a sequence of solutions of equation (3) which con-

verges uniformly to a function cp(x), then cp(x) is also a solution of

equation (3) since, if

cp,,' = f(x, CPn(x)),

then passing to the limit as n ~ 00, we obtain

cp' = f(x, cp(x)).

We have

cp(xo) = lim cp,,(xo) E [a, b].

In virtue of Arzela's theorem and Theorem 1 of this section it follows

from this that P is a compactum.

We set the point cp(xo) of the closed interval [a, b] into correspondence

with each solution cp(x) of equation (3). By assumption this correspondence

is one-to-one. Moreover, since

Page 140

Product of operator::! 98

Proper subset 1

Quadratic metric 19

Radius of a sphere 23

Range of variation of a function 13

Real functions in metric spaces 62 ff.

Rectifiable curves 69

Reflexive space 89

Reflexivity 12

Regular point 110

Residual spectrum 111

Resolvent 110

Schwartz 105

Schwarz inequality 17, 19

Second

- conjugate space 88

- countability axiom 28

Segment 74

Semi continuous function 64

Set 1

-, Cantor 32

-, closed 26

-, compact 51

-, compact in itself 58

-, convex 74

-, dense 25

-, denumerable 4, 7

-, derived 26

-, everywhere dense 25

-, finite 4

-, infinite 4, 8

-, open 27

-, nondenumerable 5

-, totally bounded 51

-, void 1

Simplex 76

Space

-, complete 37

-, conjugate 82

-, connected 29

-, Hausdorff 31

-, linear 71

-, metric 16

--, n-dimensional R" (Euclidean) 17

-, n-dimensional Ro n 17

-, n-dimensional Rp" 20

-, normed 71

D1DEX 129

- of continuous function::; C 18

- of continuous functions C[a, b) 17,72

- of continuous functions C2[a, b) 19, 72

-- of continuous functions with quad-

ratic metric 19

-- of real numbers Rl 16, 72

- of sequences c 72

- of sequences l2 18, 72

- of sequences lp 22

- of sequences 1/1. 72

-, reflexive 89

-, separable 25

-, topological 30

- with countable basis 28

Spectrum

-, continuous 111

- of an operator 110

-, point 111

-, residual 111

Subset 1

-, proper 1

Subspace 73

Sum

- of operators 97

- of sets 1,2

Symmetric difference of "eta :2

Symmetry 12

System of sets with finite intersection

property 58

Theorem on nested spheres :39

Topological space 30

Total

- boundedness 51

- variation 64

Totally bounded set 51

Transitivity 12

Uniform bounded ness of n family of

functions 54

Union 1

Upper limit 64

Variation, see Total variation

Vertices of a simplex 76

Volterra

-- integral equation 50

--- operator 114

Weak convergence 90,93

Weak* convergence 93

§18] COMPACTA 59

of closed sets satisfying the condition

n(R '" 0 .. ) = o.

The condition that we can select a finite subcovering from {O .. } is equiva-

lent to the fact that the system of closed sets {R '" O .. } cannot have the

finite intersection property if it has a void intersection.

We shall now prove that Condition 1 is necessary and sufficient that R

be a compactum.

Necessity. Let R be a compactum and let {O .. } be an open covering of R.

We choose in R for each n = 1, 2, ... a finite (l/n)-net consisting of the

points ak (n) and we enclose each of these points with the sphere

S(ak(n), lin)

of radius lin. It is clear that for arbitrary n

R = UkS(A k (n), lin).

We shall assume that it is impossible to choose a finite system of sets

covering K from {O .. }. Then for each n we can find at least one sphere

S(ak(n)(n), lin) which cannot be covered by a finite number of the sets 0 ...

We choose such a sphere for each n and consider the sequence of their

centers {ak(n) (n)}. Since R is a compactum, there exists a point ~ E R which

is the limit of a subsequence of this sequence. Let Ofj be a set of {O .. } which

contains~. Since Ofj is open, we can find an E > 0 such that S{~, E} C Ofj .

We now choose an index n and a point ak(n) (n) so that p(~, ak(n) (n») < E/2,

lin < E/2. Then, obviously, S(ak(n)(n), lin) c Ofj, i.e. thesphereS(ak(n)(n),

lin) is covered by a single set Ofj . The contradiction thus obtained proves

our assertion.

Sufficiency. We assume that the space R is such that from each of its

open coverings it is possible to select a finite subcovering. We shall prove

that R is a compactum. To do this it is sufficient to prove that R is complete

and totally bounded. Let E > O. Take a neighborhood O(x, E) about each

of the points x E R; we then obtain an open covering of R. We choose from

this covering a finite subcovering O(Xl , E), ... ,O(xn , E). It is clear that the

centers Xl, ••• , Xn of these neighborhoods form a finite E-net in R. Since

E > 0 is arbitrary, it follows that R is totally bounded. Now let {Sn} be a

sequence of nested closed spheres whose radii tend to zero. If their inter-

section is void, then the sets R '" S" form an open covering of R from which

it is impossible to select a finite subcovering. Thus, from Condition 1 it

follows that R is complete and totally bounded, i.e. that R is compact.

THEOREM 5. The continuous imO{Je of a compactum is a compactum.

Proof. Let Y be a compactum and let Y = cp(X) be its continuous image.

Further, let {O .. } be an open covering of the space Y. Set U .. = cp-l(O .. ).

Page 71

60 METRIC SPACES [CH. II

Since the inverse image of an open set under a continuous mapping is open,

it follows that {U a} is an open covering of the space X. Since X is a com-

pactum, we can select a finite subcovering U1 , U2, ... , Un from the

covering {Ua }. Then the corresponding sets 0 1 , O2 , ••• , 0" form a finite

subcovering of the covering {O a}.

THBOREM 6. A one-to-one mapping of a compactum which is continuous in

one direction is a homeomorphism.

Proof. Let cp be a one-to-one continuous mapping of the compactum X

onto the compactum Y. Since, according to the preceding theorem, the

continuous image of a compactum is a compactum, the set cp(M) is a com-

pactum for an arbitrary closed M C X and consequently cp(M) is closed in

Y. It follows that the inverse image under the mapping cp -1 of an arbitrary

closed set M C X is closed; i.e. the mapping cp -1 is continuous.

RI<;MARK. The following result follows from Theorem 6: if the equation

(3) dy/dx = f(x, y),

where the function f(x, y) is continuous in a closed bounded region G

which contains the point (xo, Yo) for every Yo belonging to some closed

interval [a, b], has a unique solution satisfying the initial condition y(xo) =

Yo , then this solution is a continuous function of the initial value Yo .

In fact, since the function f(x, y) is continuous in a closed bounded

region, it is bounded and consequently the set P of solutions of equation

(3) corresponding to initial values belonging to the closed interval [a, b]

is uniformly bounded and equicontinuous. Moreover, the set P is closed.

In fact, if {cp,,(x)} is a sequence of solutions of equation (3) which con-

verges uniformly to a function cp(x), then cp(x) is also a solution of

equation (3) since, if

cp,,' = f(x, CPn(x)),

then passing to the limit as n ~ 00, we obtain

cp' = f(x, cp(x)).

We have

cp(xo) = lim cp,,(xo) E [a, b].

In virtue of Arzela's theorem and Theorem 1 of this section it follows

from this that P is a compactum.

We set the point cp(xo) of the closed interval [a, b] into correspondence

with each solution cp(x) of equation (3). By assumption this correspondence

is one-to-one. Moreover, since

Page 140

Product of operator::! 98

Proper subset 1

Quadratic metric 19

Radius of a sphere 23

Range of variation of a function 13

Real functions in metric spaces 62 ff.

Rectifiable curves 69

Reflexive space 89

Reflexivity 12

Regular point 110

Residual spectrum 111

Resolvent 110

Schwartz 105

Schwarz inequality 17, 19

Second

- conjugate space 88

- countability axiom 28

Segment 74

Semi continuous function 64

Set 1

-, Cantor 32

-, closed 26

-, compact 51

-, compact in itself 58

-, convex 74

-, dense 25

-, denumerable 4, 7

-, derived 26

-, everywhere dense 25

-, finite 4

-, infinite 4, 8

-, open 27

-, nondenumerable 5

-, totally bounded 51

-, void 1

Simplex 76

Space

-, complete 37

-, conjugate 82

-, connected 29

-, Hausdorff 31

-, linear 71

-, metric 16

--, n-dimensional R" (Euclidean) 17

-, n-dimensional Ro n 17

-, n-dimensional Rp" 20

-, normed 71

D1DEX 129

- of continuous function::; C 18

- of continuous functions C[a, b) 17,72

- of continuous functions C2[a, b) 19, 72

-- of continuous functions with quad-

ratic metric 19

-- of real numbers Rl 16, 72

- of sequences c 72

- of sequences l2 18, 72

- of sequences lp 22

- of sequences 1/1. 72

-, reflexive 89

-, separable 25

-, topological 30

- with countable basis 28

Spectrum

-, continuous 111

- of an operator 110

-, point 111

-, residual 111

Subset 1

-, proper 1

Subspace 73

Sum

- of operators 97

- of sets 1,2

Symmetric difference of "eta :2

Symmetry 12

System of sets with finite intersection

property 58

Theorem on nested spheres :39

Topological space 30

Total

- boundedness 51

- variation 64

Totally bounded set 51

Transitivity 12

Uniform bounded ness of n family of

functions 54

Union 1

Upper limit 64

Variation, see Total variation

Vertices of a simplex 76

Volterra

-- integral equation 50

--- operator 114

Weak convergence 90,93

Weak* convergence 93