Download 4. Connectedness 4.1 Connectedness Let d be the usual metric on

4. Connectedness
4.1 Connectedness
Let d be the usual metric on R2 , i.e. d(x, y) =
(x1 , x2 ), y = (y1 , y2 ). Let
p
(x1 − y1 )2 + (x2 − y2 )2 , for x =
X = {x ∈ R2 | d(x, 0) ≤ 1 or d(x, (4, 1)) ≤ 2}
and
Y = {x = (x1 , x2 ) ∈ R2 | − 1 ≤ x1 ≤ 1, −1 ≤ x2 ≤ 1}.
So X is
X =A
S
B and Y is
Are X and Y homeomorphic? Both are Hausdorff (subspaces of R2 ) and both are compact
since closed and bounded. However, it seems unlikely that they are homeomorphic since
X is in two bits and Y is not.
Definition
A topological space X is disconnected if there exist two disjoint nonempty
S
T
open sets U, V such that X = U V (and U V = ∅). A topological space is connected if
it is not disconnected.
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S
(i) X as in the above example. X = A B and A is a closed subset of R2
T
(its a closed ball). Hence X = X A is a closed subset of X (in the subspace topology).
S
Hence B = CX (A) is open in X. Similarly A = CX (B) is open. Hence X = A B, for
Examples
nonempty disjoint open sets A, B. Hence X is disconnected.
(ii) Y (above) is connected (see later).
(ii) Let X be a discrete topological space with at least 2 elements. Pick x ∈ X. Then
S
X = {x} CX ({x}) and so X is disconnected.
(4.1a) Lemma
A topological space X is disconnected if and only if there exists a proper
subset (i.e. neither ∅ nor X) M, say, such that M is both open and closed.
Proof
(⇒) Assume X is disconnected.
Then X = A
S
B with A, B disjoint and
nonempty open sets. So A is a proper subset. Now A is closed and A = CX (B) is
closed. Hence A is open and closed.
(⇐) Suppose X has a subset M which is both open and closed. Then X = M
expresses X as the union of disjoint open sets. Hence X is disconnected.
S
CX (M)
(4.1b) A topological space X is disconnected if and only if there exist non-null, disjoint
S
closed sets A, B such that X = A B.
Proof
Exercise.
(4.1c) Lemma A topological space X is connected if and only if every continuous map
from X into a discrete topological space is constant.
Proof
(⇒) Suppose X connected and let f : X → D be a continuous map into a discrete
S
space D. Let x0 ∈ X and let α = f(x0 ). Put A = {α} and B = D\{α}. Then D = A B
and hence
X = f −1 D = f −1 A
[
f −1 B.
Now A, B are open in D (every subset of D is open) and hence f −1 A, f −1 B are open in X.
T
S
Note also that f −1 A f −1 B = f −1 ∅ = ∅. Hence X = f −1 A f −1 B expresses X as the
disjoint union of two open sets. Since X is connected, one of these must be empty. But
x0 ∈ f −1 A so f −1 A 6= ∅ and hence f −1 B = ∅ and we have X = f −1 A. Thus f(x) ∈ A for
every x ∈ X, i.e. f (x) = α for every x ∈ X. Hence f is the constant function.
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(⇐) We suppose that every continuous map from f into a discrete space is constant and
prove X is connected. Assume for a contradiction that X is not connected. Thus we may
S
write X = A B, with A, B nonempty and open. Let D{α, β} be the discrete space with
two elements α, β. Define f : X → D by
f (x) =
½
α, if x ∈ A;
β, if x ∈ B.
Check that f is continuous and hence nonconstant. This is a contradiction, so X is
connected.
(4.1d) Proposition
The image of a connected space is connected, i.e. if f : X → Y is
continuous and X is connected then f(X) is connected.
Proof
(i) Let Z = f (X) = Im(f ) and let g : X → Z be the restriction of f , i.e.
g(x) = f (x) for all x ∈ X (the only difference between f and g is that the codomain of f
is Y and the codomain of g is Z).
“Direct Proof”
If Z is disconnected then we can write Z = U
S
V for disjoint nonempty
S
S
open sets U, V but then we would have X = g −1 Z = g −1 (U V ) = g−1 U g −1 V , a
disjoint union of nonempty open sets. But X is connected so this can’t happen. So Z is
connected.
“Another Proof”
Let h : Z → D be a continuous map, with D discrete. Then h ◦ g :
X → D is constant (since X is connected, see (4.1c)). But then if z1 , z2 ∈ Z we can write
z1 = g(x1 ), z2 = g(x2 ) for some x1 , x2 ∈ X. So h(z1 ) = h(g(x1 )) = h(g(x2 )) = h(z2 )
(since h ◦ g is constant). But then h(z1 ) = h(z2 ) so h is constant and Z is connected, by
(4.1c).
(4.1e) Corollary
Proof
Connectedness is a topological property.
We must show that if X is connected and X is homeomorphic to Y then Y
is connected. Assume X is connected and X is homeomorphic to Y . Thus there is a
homeomorphism f : X → Y . The map f is in particular a surjective (onto) continuous
map. By (4.1e), Y = f(X) is connected.
Example
Let X be an infinite set regarded as a topological space with the cofinite
topology. Then X is connected.
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S
T
If not we can write X = A B with A, B nonempty and open and with A B = ∅. But
S
then X = CX (A) CX (B), a union of two finite sets and hence finite. But X is infinite
so this is impossible. Hence X is connected.
We shall need the following general result.
(4.1f ) Lemma Let X be a topological space and let Y be a subspace of X. Let A be a
subset of Y (so A ⊂ Y ⊂ X).
(i) A is closed in Y if and only if it has the form K
T
Y for some closed set K of X.
(ii) If A is open in Y (in the subspace topology on Y ) and Y is an open subset of X then
S is an open subset of X.
(iii) If A is closed in Y (in the subspace topology on Y ) and Y is a closed subset of X then
A is a closed subset of X.
T
(i) Suppose A is closed in Y (in the subspace topology). Then CY (A) = Y U ,
S
for some open set U in X. Let K be the closed set CX (U ). Then X = U K and so
T
S T
T
T
Y = (Y U ) (Y K). Thus Y K is the complement in Y of CY (A) = Y U . But
T
the complement of CY (A) in Y is A, i.e. we have A = Y K.
T
(ii) We have A = U Y , with U open in X. So A is a union of two open sets in X hence
Proof
open in X.
(iii) We have A = Y
T
K for some closed set K in X, by (i). Thus A is the intersection of
two closed sets in X and hence A is closed in X.
(4.1g) Theorem
Proof
The closed interval [a, b] is connected (for a < b).
[a, b] is homeomorphic to [0, 1] so by (4.1e) it suffices to prove [0, 1] is connected.
Let A be a nonempty subset of [0, 1] which is open and closed (in the subspace topology
on [0, 1]). Let α be the greatest lower bound of A. Now A is closed in R (by (4.1f)(iii) since
A is closed in [0, 1] and [0, 1] is closed in R). So α ∈ A by (3.2l). We claim that in fact
T
α = 0. Suppose not. Since A is open in [0, 1] we have A = [0, 1] U for some open set U
in R. So we have (α − r, α + r) ⊂ U for some r > 0. Choose s > 0 such that s < min{r, α}.
T
Then (α − s, α] ⊂ U [0, 1] = A. But then α − s/2 ∈ A, contradicting the fact that α is
a lower bound of A. Hence α = 0. We have shown that if A is any nonempty subset of
[0, 1] which is open and closed then 0 ∈ A. But now if [0, 1] is disconnected then we can
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write [0, 1] = A
S
B for nonempty disjoint subsets A, B which are open and closed. But
T
then 0 ∈ A and 0 ∈ B contradicting the fact that A B = ∅. Hence [0, 1] is connected.
(4.1h) Corollary (The Intermediate Value Theorem)
Let f : [a, b] → R be con-
tinuous and suppose that f (a) < k < f (b). Then there exists some c ∈ (a, b) such that
f (c) = k.
Proof
Suppose for a contradiction that there is no such c. Let U = (∞, k), V = (k, ∞).
Then we have
[a, b] = f −1 U
[
f −1 V
and this is a disjoint union of nonempty open subsets of [a, b]. But [a, b] is connected so
this is impossible.
(4.1i) Corollary
(a, b) is connected and R is connected.
(a, b) is homeomorphic to R so it is enough to show that R is connected. Suppose
S
not. Then we have R = U V , for U, V open, nonempty disjoint subsets of R. Pick x ∈ U
Proof
and y ∈ V . We may suppose that x < y (otherwise relabel U and V ). Hence we have
\
\
[
\
[x, y] = [x, y] R = ([x, y] U) ([x, y] V )
and this disconnects [x, y] contrary to (4.1g).
(4.1j) Corollary
Proof
[a, b) is connected (for real numbers a < b).
[a, b) is homeomorphic to [0, 1) so it is enough to prove [0, 1) connected. Define
f : (−1, 1) → [0, 1) by
f(x) =
½
x, if x ≥ 0;
0, if x ≤ 0.
Then f is continuous and onto so [0, 1) = f(−1, 1) is connected by (4.1i) and (4.1e).
We still haven’t resolved the question : Is [0, 1) homeomorphic to (0, 1)? Both spaces
are Hausdorff, neither is compact, both are connected. However, removing {0} from [0, 1)
leaves a connected space whereas removing any point from (0, 1) disconnects the space.
(4.1k)
Suppose that X and Y are homeomorphic spaces and that x1 , . . . , xn are distinct
points of X. Then there exist distinct points y1 , . . . , yn of Y such that X\{x1 , . . . , xn } is
homeomorphic to Y \{y1 , . . . , yn }.
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Proof
Let f : X → Y be a homeomorphism with inverse g : Y → X. Let yi = f(xi ),
1 ≤ i ≤ n. Let X0 = X\{x1 , . . . , xn } and let Y0 = Y \{y1 , . . . , yn }. Then f0 : X0 → Y0 , the
restriction of f , is continuous, (1.4b), and is a bijection with continuous inverse g0 : Y0 →
X0 , the restriction of g. Hence f0 is a homeomorphism (and X0 and Y0 are homeomorphic
spaces).
(4.1l)
[0, 1) is not homeomorphic to (0, 1).
Proof
Well suppose these spaces are homeomorphic. In the Lemma take X = [0, 1),
Y = (0, 1), take n = 1 and x1 = 0. Then X0 = X\{0} = (0, 1) is connected. But
S
Y0 = Y \{y1 } = (0, y1 ) (y1 , 1) is disconnected. These spaces are not homeomorphic and
so X is not homeomorphic to Y .
We are now going to show that Rn is connected. We do this by proving generally that
if X and Y are connected then X × Y is connected.
(4.1m)
Let X, Y be topological spaces and let y0 ∈ Y . The map α : X → X × Y , given
by α(x) = (x, y0 ) is continuous.
Proof
By (3.3e), α is continuous provided that p ◦ α : X → X and q ◦ α : X → Y are
continuous (where p : X × Y → X and q : X × Y → Y are the projections). But we
p ◦ α : X → X is the identity, which is continuous, and q ◦ α : X → Y is the constant map
with value y0 , and this map is continuous. Hence α is continuous.
(4.1n) Let X, Y, Z be topological spaces and let φ : X × Y → Z be a continuous map.
For each y ∈ Y the map φy : X → Z, given by φy (x) = φ(x, y) is continuous.
Proof
We have φy (x) = φ(x, y) = φ(α(x)), where α : X → X × Y is the map α(x) =
(x, y). Thus φy = φ ◦ α, a composite of continuous maps and hence continuous.
(4.1o) Proposition
Proof
If X, Y are connected then X × Y is connected.
By (4.1c) it suffices to show that every continuous map φ : X ×Y → D is constant,
where D is a discrete space.
For each y ∈ Y the map φy : X → D, given by φy (x) = φ(x, y) is constant, by (4.1c),
since X is connected. In other words, we have
φ(x, y) = φ(x0 , y)
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(1)
for all x, x0 ∈ X and y ∈ Y .
Similarly the function φx : Y → D, given by φx (y) = φ(x, y), is constant, in other
words
φ(x, y) = (x, y 0 )
(2)
for all x ∈ X, y, y 0 ∈ Y . Now by (1) and (2), for (x, y), (x0 , y 0 ) ∈ X × Y we have
φ(x, y) = φ(x0 , y) = φ(x0 , y 0 )
so that φ is constant. Hence X × Y is connected.
(4.1p) Rn is connected, for n > 0.
Proof
We have already shown that R = R1 is connected. For n > 1 we have that Rn is
homeomorphic to R × Rn−1 . So the result follows by induction on n, and (4.1o).
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