Page 80 - DMTH401_REAL ANALYSIS

Real Analysis

Notes 2. T has subset which is open, closed, different from , T
3. T admits non-constant cts function to two point discrete space.
Proof: (1.  2.)  decomposition T = A B with A, B open, non-empty. Hence A = T\B is open and
closed, different from , = T.
0 x Î A
(2.  3.) , T  A  T open, closed. Define f : T  {0,1} by f(x) = 
 1 x Ï A
This cts as pre-images open
–1
(3.1.) f: T  {0, 1} non-constant and cts. Define A = F (0), B = f –1
5.2 Connectedness in Metric Spaces

Theorem: T  M (M metric) disconnected iff  disjoint open U,V  M s.t. TU  T  V and
T  U  V.
Proof: () Clear
() T = A U B. Let

U = {x Î M : d(x, A) < d(x, B)}
V = {x Î M : d(x, A) > d(x, B)}
U, V disjoint, open.
Going to prove A  U: Let x Î A. A open in T so  > 0 s.t. B(x, )T  A. B  T disjoint from A
so B(x, )B = , so d(x, B)   > 0. Since d(x, A) = 0 we have x ÎU. Similarly B  V.
Lemma: I  is an interval iff " x, yÎI, " z Î,

x < z < y  z Î I
Proof: Intervals clearly have this property. Conversely suppose I has above property, non-empty,
not single point. Let a = inf I, b = sup I.

Show (a, b)  I: If z Î (a, b)  x, y Î with x < z < y so z ÎI. Hence (a, b)  I  (a, b)  {a, b}.
Theorem: T  connected iff it is an interval.
Proof: () Suppose I not interval. Then by lemma 4.4  x, y Î I, z Î  s.t. x < z < y and z Ï I. Let
A = (–, z)  I, B = (z, )  I. A, B disjoint, non-empty, open and I = A B.
() Suppose I not connected. Then 3 cts non-constant f : I  {0, 1} where {0, 1} has discrete
contradicting IVT.

() I partitioned into non-empty A, B open. Choose a Î A, b Î B, a < b. A, B open cover of [a, b].
é  ù é  2 ù
Let  be its Lebesgue number. Then a, a +  A, a + , a + A, .... until we get to an
ê ú ê ú
ë 2 û ë 2 2 û
interval containing b. So b Î A and A, B not disjoint.
5.3 Connected Spaces from Others

Proposition: Cts image of connected space connected.
Proof: Suppose f : T  S cts, T connected. If f(T) disconnected  U, V  S open separating f(T). Then
–1
–1
f (U), f (V) open, disjoint, cover T. Contradiction as T connected.

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