Page 82 - DMTH401_REAL ANALYSIS

Real Analysis

Notes Proposition: Path connected  connected.
Proof: a Î T. " x Î T image C of path a to x is connected, and all C contain a. Then T =  x T C
x x Î x
connected by 4.7.

5.6 Open Sets in  n

n
Theorem: Any U C  open, connected is path connected.
Proof: Let a ÎU, V = {x ÎU :  path from a to x}.
Let z Î U  V . Find  > 0 s.t. B(z,)  U. z V so y Î V  B(x, ).

Then B(z, )  V since join path from a to y to path from y to z.
n
Theorem: All components of open U   open.
Proof: C component of U, x ÎC. Find  > 0 with B(x, )  U. B(x, ) connected and C union of
all connected subsets of U containing x so B(x, )  C, so C open.
Theorem: U   open iff disjoint union of countably many open intervals.
Proof: () Any union of open sets open.
() U   open, C (j Î J) its components. C connected and open so are open intervals. For each
j j
j  rational r ÎC. C mutually disjoint so j  r injection into , so can order J into a sequence.
S
j j j j
Self Assessment

Fill in the blanks:
1. Topological T connected if for every decomposition ................. into disjoint open A, B
either A or B is empty.
2. T  M (M metric) disconnected iff  ............... U,V  M s.t. TU  T  V and T  U  V.
3. .................... is an interval iff " x, yÎI, " z Î, x < z < y z Î I

4. Suppose f : T  S cts, T connected. If f(T) disconnected  U, V  S open separating f(T). Then
–1
f (U), f (V) open, disjoint, cover T. Contradiction as ...........................
–1
5. C  T connected and C  K  C . Then ........................
6. x  y if x, y belong to a common connected subspace of T. ........................ are connected
components of T.

5.7 Summary

 Topological T connected if for every decomposition T = AB into disjoint open A, B either
A or B is empty.

 T  S separated by sets U, V  S if T  UV, U  V  T = , U  T , V  T .
 T  C S disconnected iff T is separated by some U, V  S.
 Proof. () If disconnected  A, B  T, A, B  s.t. T = A  B and A  B = . T  S so U, V open
in S s.t. A = U T, B = V T. Then U, V separate T.
 Suppose K disconnected. Hence  U, V  T open that separate K.

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