Page 81 - DMTH401_REAL ANALYSIS

Unit 5: Connectedness

Proposition: If C, C j (j Î J) connected subspaces of topological T and if c j  C  " j ÎJ then Notes
K = C   C j
j J
Î
is connected.
Proof: Suppose K disconnected. Hence  U, V  T open that separate K.
C connected so cannot be separated by U,V, so does not meet one of them. Suppose w.l.o.g C 
V = . Then C  U. Since V open C  V = , so K  C  U. Then C  U  " j.
j
C connected so C  U or C  V. C  U  so C  U.
j j j j j
Then K  U contradicting V  K .

Corollary: C  T connected and C  K  C . Then K connected.
Proof: K = C  {x} and {x}  C  " x.
Î
x K
Proposition: Product of connected spaces is connected.

Proof: Let T, S connected, so Î S. Define C = T  {s } and C = {t}  S (for some t Î T). Then C, C
0 t t
homeomorphic to T and S are connected. C  C  and T S = C  t T C connected.
t Î t
æ 1ö 2
Î
-
Example: Sin ç ÷  {(0 , t) Î : ( 1, 1)} is connected.

t è ø
I
Proof:
æ æ 1 ö ö ü
>
C = ç t, sin ç ÷ ÷ : t 0 ý
 è t è ø ø þ
æ æ 1 ö ö ü
<
D = ç t, sin ç ÷ ÷ : t 0 ý
 è t è ø ø þ
C, D, I cts images of intervals so connected.

æ 1 ö 1
(0, 0) Î I is in C as (t ) sin  (0, 0) when t = . Then I C connected. Similarly I  D.
k ç ÷ k
t è k ø kp

5.4 Connected Components

Definition: x  y if x, y belong to a common connected subspace of T. Equivalence classes are
connected components of T.

Are maximal connected subsets of T. Number of connected components is topological invariant.
Property T\{x} connected " x ÎT topological invariant.

5.5 Path Connectedness

Definition: a, b Î T. : [0, 1]  T cts with (0) = a, (1) = b called a path from a to b.
Definition: T path connected if any two points can be joined by a path.

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