Download Topology (Part 2) - Department of Mathematics, University of Toronto

Topology (Part 2)
Original Notes adopted from April 9, 2002 (Week 26)
© P. Rosenthal , MAT246Y1, University of Toronto, Department of Mathematics typed by A. Ku Ong
Let X be a topological space.
Let S ⊂ X
The relative topology on S (as a subset of X) is the topology where a subset U of S is open if there exists
V, an open subset of X such that U = V @6
Eg. X = R usual topology
S = [0,1].
0
¾
1
7
U = (3/4, 1]. (is an open subset on relative topology but not open in R)
Recall: If f: X→Y is continuous, if f-1(U) is open (in X) whenever U is open (in Y).
Definition: The topology X is disconnected if there exists U,V open subsets of X such that X = U U V
& U @9 ∅ (and neither U nor V is ∅), If there is no such disconnection, then X is connected.
Theorem: If X is connected topological space, and f:X→
→Y is continuous, then f(X) is a connected
topological space in the relative topology as a subset of Y0
Proof: Suppose f(x) was disconnected.
Then f(x) = U, Union U2 , each Ui is open subset of f(x)
U1 @U2 = ∅, U1 ≠ 0, U2 ≠ 0.
There exists V1 ,V2 open subsets of Y such that
Ui = V1 @f(X) & U2 = V2 @f(X)
Let W1 = f-1(V1) & W2 = f-1(V2)
f continous ⇒ W1,W2 open (definition – inverses are open)
If x ∈ W1@W2 , then f(x) ∈ V1 @V2 . But also f(x) ∈ f(X)
∴ f(x) ∈ V1 @9 2 @I ( x ) = (V1 @ f ( x ) ) @9 2 @ f ( x ) ) = U1 @8 2
But U1 @8 2 ≠ 0, so nu such x.
∴ W1@W2 = ∅, clearly W1U W2 = X.
U1 ≠ 0 ⇒ W1 ≠∅, U2 ≠ 0 W2 ≠∅
∴ W1U W2 is a disconnection of X . Contradiction.
Corollary: If f:[a,b]→
→ R, then f assumes all values between f(a) & f(b) (Intermediate Value Theorem)
Proof: [a,b] is connected (in its relative topology) – this really requires proof.
By above theorem, f([a,b]) (image) is connected.
Suppose f(a) < f(b) (Etc. if other way as usual).
Suppose f(a) <k f(b) & no x ∈ [a,b]
satisfies f(x) = k.
Must show: not possible.
Let U1 = {y: y <k} U2 = {y: y >k}
U1 & U2 open , f continuous ⇒ f-1(U1) & f-1(U2) are open.
Claim: f-1(U1) & f-1(U2) is a disconnection of [a,b].
Since f(x) ≠ k ∀ x, [a,b] = f-1(U1) U f-1(U2)
f-1(U1) ≠ ∅ (a ∈ f-1(U1)), f-1(U2) ≠ ∅ (b ∈ f-1(U2))
f-1(U1) @I -1(U2) =∅.
Contradicits [a,b] being connected.
Homeomorphic Examples
D O – yes
X T – no
\ _ _ - no because connectedness is not preserved.
__ O No it is not, if you take a point out of ___, and get _ _, this is disconnected but the circle will still be
connected.
Definition: A component of topological space is a connected subset of the space that is not properly
contained in any other connected subset (ie., a maximal connected subset of the space).
Theroem: If each of Sα is a connected subset of a topological space x, and if there exists x ∈ @6 2 ,
then
U Sα is connected.
Proof: If U Sα is disconnected,
U Sα = U1 U U2 , each Ui open, non empty, U1 @8 2 = ∅ (disjoint).
There exists V1 ,V2 open in X with U1 = V1@8 Sα)
U2 = V2@8 Sα)
V1@6 2 V2@6 2
x ∈ U1 U U2. Suppose x ∈ U1 (etc if x ∈ U2 )
there exists α0 such that Sα0 @8 2 ≠ ∅.
Then (V1@ ( U Sα0) U (V2@ ( U Sα0) is a disconnection of Sα0
Definition: A collection {Uα } of sets covers a subset S if S ⊂ U Uα . A subcovering of a covering
{Uα } is a subcollection of the Uα 's which still covers S.