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4
hand,
E
=
A
∪
B
, and so
A
∪
B
⊆
B
; in particular,
A
⊆
B
. But
A
∩
B
=
∅
, and this is only
possible if
A
=
∅
, contra our assumptions. Whence,
A
1
±
=
∅
, and indeed, neither
A
1
nor
B
1
are empty. Therefore,
E
=
A
1
∪
B
1
is a separation of
E
, which is therefore disconnected
as claimed.
±
Proof of Proposition 14.
As
C
is a connected component, it is nonempty. Fix any point
x
∈
C
– so
C
=
C
x
. By Lemma 15,
C
x
is also connected, and hence is a connected set
containing
x
. But by the de±nition of
C
x
, this means the connected set
C
x
is included in
the union that is
C
x
, and hence
C
x
⊆
C
x
. But any set is contained in its closure, and hence
C
x
=
C
x
, meaning that
C
=
C
x
is closed, as advertised.
±
In many cases, components provide examples of socalled
clopen
sets – sets that are both
open and closed.
Corollary 16.
Let
X
be a metric space with Fnitely many components. Then its components are
clopen.
Proof.
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This note was uploaded on 05/06/2010 for the course MATH Math2009 taught by Professor Koskesh during the Spring '09 term at SUNY Empire State.
 Spring '09
 Koskesh

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