Page 323 - Advanced Linear Algebra

Metric Spaces 307

Theorem 12.3 Let be a nonempty subset of a metric space 4 .
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1 %M²:³ if and only if there is a sequence ²% ³ in for which % £ % for

all and % ² ³ ¦ . %

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2 : ) is closed if and only if ² M : ³ ‹ : . In words, is closed if and only if it
contains all of its limit points.
3)cl²:³ ~ : r M²:³ .
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4 %  cl²:³ if and only if there is a sequence ²%³ in for which ²%³ ¦ % .
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Proof. For part 1 , assume first that %M²:³ . For each , there exists a point
% £ % such that %  )²%Á ° ³ q :. Thus, we have

²% Á %³  °

and so ²% ³ ¦ % . For the converse, suppose that ²% ³ ¦ % , where % £ %  : .

%
If )²%Á ³ is any ball centered at , then there is some 5 such that € 5
%
implies %  )²%Á ³ . Hence, for any ball )²%Á ³ centered at , there is a point

%£ % such that % : q )²%Á ³. Thus, % is a limit point of :.

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As for part 2 , if is closed, then by part 1 , any % )  M ² : ³ is the limit of a
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sequence ²% ³ in and so must be in . Hence, M²:³ ‹ : . Conversely, if
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M²:³ ‹ :, then : is closed. For if ²%³ is any sequence in : and ²%³ ¦ %, then

there are two possibilities. First, we might have %~ % for some , in which

case %~%  : . Second, we might have % £% for all , in which case

²% ³ ¦ % implies that %  M²:³ ‹ :. In either case, %  : and so : is closed

under the taking of limits, which implies that is closed.
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For part 3 , let ;~ : r M²:³ . Clearly, : ‹ ; . To show that is closed, we
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show that it contains all of its limit points. So let %M²;³ . Hence, there is a
sequence ²% ³  ; for which % £ % and ²% ³ ¦ % . Of course, each % is

either in , or is a limit point of . We must show that  % ; , that is, that is
%
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either in or is a limit point of .
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Suppose for the purposes of contradiction that %¤: and % ¤M²:³ . Then there
is a ball )²%Á ³ for which )²%Á ³ q : £ J . However, since ²% ³ ¦ % , there

must be an % )²%Á ³ . Since % cannot be in , it must be a limit point of .
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Referring to Figure 12.1, if ²% Á %³ ~  , then consider the ball

)²% Á ² c ³° ³. This ball is completely contained in )²%Á ³ and must contain

&
an element of : , since its center % is a limit point of : . But then
& :q )²%Á ³, a contradiction. Hence, %  : or %  M²:³. In either case,
%; ~ : r M²:³ and so is closed.
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Thus, ; is closed and contains and so cl ² : ³ ‹ ; . On the other hand,
:
cl
; ~ : r M²:³ ‹ ²:³ and so cl ²:³ ~ ; .