Download TOPOLOGY, DR. BLOCK, SPRING 2016, NOTES, PART 5 501

TOPOLOGY, DR. BLOCK, SPRING 2016, NOTES, PART 5
501. Lemma. Let X be a normal space, and let A and B be disjoint closed
subsets of X. There is a collection of open sets Ut for each t of the form t = 2mn ,
where n = 1, 2, 3, . . . and m = 1, 2, . . . , 2n − 1 such that if r and s are of this form
and r < s, then
A ⊂ Ur ⊂ Ur ⊂ Us ⊂ Us ⊂ X − B.
502. Theorem. (Urysohn Lemma) Let X be a normal space, and let A and B
be disjoint closed subsets of X. There exists a continuous function f : X → [0, 1]
such that for all x ∈ A, f (x) = 0 and for all x ∈ B, f (x) = 1.
Hint: Using the previous lemma, let f (x) = 0 if x is in every Ut and f (x) =
lub{t : x ∈
/ Ut } otherwise. Use Lemma 128.
503. Corollary. (Urysohn Lemma) Let X be a normal space, and let A and B
be disjoint closed subsets of X. If [a, b] is any closed interval on the real line, there
exists a continuous function f : X → [a, b] such that for all x ∈ A, f (x) = a and
for all x ∈ B, f (x) = b.
504. Theorem. (Tietze Extension Theorem) Let X be a normal space, F a closed
subset of X, and f : F → [a, b] continuous. Then f has a continuous extension
f ∗ : X → [a, b]. (This means that there exists a continuous function f ∗ : X → [a, b]
such that for all x ∈ F, f ∗ (x) = f (x).)
Hints:
1. We may assume that a = −1 and b = 1.
2. Let f0 = f, let A0 = {x ∈ F : f0 (x) ≤ − 31 }, and let B0 = {x ∈ F : f0 (x) ≥ 31 }.
There is a continuous g0 : X → [− 31 , 13 ] such that for all x ∈ A0 , g0 (x) = − 13 and
for all x ∈ B0 , g0 (x) = 13 .
3. Define f1 on F by f1 = f0 − g0 . Note that for all x ∈ F, |f1 (x)| ≤ 23 .
4. Let A1 = {x ∈ F : f1 (x) ≤ −( 13 )( 23 )} and B1 = {x ∈ F : f1 (x) ≥ ( 31 )( 23 )}.
There is a continuous g1 : X → [−( 13 )( 23 ), ( 13 )( 23 )] such that for all x ∈ A1 , g1 (x) =
−( 13 )( 23 ) and for all x ∈ B1 , g1 (x) = ( 13 )( 23 ).
5. Define f2 on F by f2 = f1 − g1 = f0 − (g0 + g1 ). Note that for all x ∈
F, |f2 (x)| ≤ ( 32 )2 . Continue, and let f ∗ = g0 + g1 + g2 + . . . .
TOPOLOGY, DR. BLOCK, SPRING 2016, NOTES, PART 5
505. Lemma. Suppose X is a regular space. Let A be a closed subset of X, and
let x ∈ X − A. Then there exists a neighborhood V of x such that V ∩ A = ∅.
506. Theorem. Every regular Lindelof space is normal.
507. Corollary. Every regular second countable space is normal.
508. Definition. Let [0, 1]ω denote the countable product of the space [0, 1]. This
space is sometimes called the Hilbert cube. (Observe that the Hilbert cube is a
compact metrizable space.)
509. Theorem. (Urysohn Metrization Theorem) If X is a second countable,
regular space, then there is a homeomorphism f of X onto a subspace of [0, 1]ω ,
and X is therefore metrizable.
Hints:
1. We can assume that X is infinite.
2. X has a countably infinite basis {G1 , G2 , . . . } where each Gn is neither X nor
∅.
3. The set of all ordered pairs (Gi , Gk ) such that Gi ⊂ Gk is countably infinite,
and so can be arranged in a sequence P1 , P2 , P3 , . . . .
4. For each Pn = (Gi , Gk ) there is a continuous fn : X → [0, 1] such that
fn (Gi ) = 0 and fn (X − Gk ) = 1.
5. Define f : X → [0, 1]ω by f (x) = (f1 (x), f2 (x), . . . ). Show that f is one to
one, and that both f and f −1 are continuous.
510. Theorem. Let X be a compact metric space, Y a Hausdorff space, and
f : X → Y a continuous, onto function. Then Y is metrizable.
Hint. Prove that Y is second countable. To do this, first show that X has a
countable basis B. Let A be the collection of all finite unions of members of B. Then
A is countable. For each element A of A let A∗ = Y − f (X − A). Prove that the
collection of sets A∗ is a countable basis for Y.
Q
511. Problem. Let I be an uncoutable index set, and let Y = i∈I {1,2}. (where
{1, 2} has the discrete topology, and Y has the product topology). Is Y normal? Is
Y second coutable? Is Y metrizable?
512. We recall this Definition. Let I denote [0, 1]. A path in a topological space
X is a continuous function p : I → X. The points p(0) and p(1) are called the
endpoints of the path. p(0) is called the initial point of the path, and p(1) is called
the terminal point of the path. If p is a path in X, the reverse path is the path q
defined by q(t) = p(1 − t).
TOPOLOGY, DR. BLOCK, SPRING 2016, NOTES, PART 5
A space X is path connected, if and only if for every pair of points x, y ∈ X
there is a path in X with initial point x and terminal point y.
513. We recall this Proposition. Any interval on the real line is path connected.
514. We recall this Proposition. If a space X is path connected, then X is
connected.
515. We recall the Pasting Lemma. Let A and B be closed subsets of a space
X with X = A ∪ B. Let f : A → Y and g : B → Y be continuous functions with
f (x) = g(x) for all x ∈ (A ∩ B). Then there is a unique function h : X → Y with
h(x) = f (x) if x ∈ A and h(x) = g(x) if x ∈ B, and h is continuous.
516. Definition. Let p and q be paths in a space X with p(1) = q(0). The path
product p ∗ q is the path f defined by f (t) = p(2t) if 0 ≤ t ≤ 12 and f (t) = q(2t − 1)
if 12 ≤ t ≤ 1.
It is easy to verify that p ∗ q is a path in X.
517. Definition. A path component of a space X is a path connected subspace
P of X such that if B is a path connected subspace of X with P ⊂ B, then P = B.
518. Lemma.
Let {Aj : j ∈ J} be a collection of path connected subspaces of X
T
such that j∈J Aj 6= ∅.
S
Then j∈J Aj is path connected.
519. Proposition.
1. Each point x ∈ X belongs to exactly one path component. The path component containing x is precisely the union of all path connected subspaces of X which
contain x.
2. Any two path components of X are either identical or disjoint.
3. Every path connected subspace of X is contained in a path component.
4. X is path connected if and only if X has exactly one path component.
520. Problem. Let S denote the subset of the plane given by
1
S = {(x, y) : 0 < x ≤ 1, y = sin( )}.
x
Let X be the closure of S. Consider X as a topological space (with the relative
topology on X as a subset of the plane). Prove that X is connected, but not path
connected. Prove also that X has a path component which is not closed. (The
space X is called the topologist’s sine curve.)
521. Problem. Prove that an open, connected subspace of the Euclidean space
R is path connected.
n
TOPOLOGY, DR. BLOCK, SPRING 2016, NOTES, PART 5
522. Definition. Let (a1 , . . . , an ), (b1 , . . . , bn ) ∈ Rn . The line segment joining
these two points is defined to be the image of the function f : I → Rn given by
f (t) = (a1 + (b1 − a1 )t, . . . , an + (bn − an )t).
523. Definition. A subset S of Rn is convex if and only if given any two points
in S, the line segment joining these two points lies entirely in S.
524. Proposition. Any convex subset of the Euclidean space Rn is path connected. In particular, Rn is path connected.