Download Connectedness - GMU Math 631 Spring 2011

Connectedness and disconnectedness
These notes discuss the same topic as Sections 23 to 25 of Munkres’ book.
1. Definition and basic properties
Definition 1. A space X is called disconnected if X can be represented in the form
X = X1 ∪ X2 so that both X1 and X2 are non empty, clopen, and X1 ∩ X2 = ∅.
A space which is not disconnected is called connected.
In other words, X is connected iff the only clopen sets in X are ∅ and X. In the
definition above, instead of requesting that both X1 and X2 are clopen, one can
request only that they both are closed (then automatically they will be clopen) or
both open.
Examples 2. The following spaces are disconnected:
• A discrete space consisting of more than one point.
• Q
• In general, every zero-dimensional space consisting of more than one point.
• ...
The following spaces are connected:
• R (we will see this later)
• Rn
• Every set with trivial topology;
• Every infinite set with cofinite topology;
• ...
Proposition 3. A continuous image of a connected space is connected.
Proof. Let f : X → Y be a continuous surjection. Suppose Y is disconnected.
Then there are disjoint non empty clopen sets Y1 , Y2 ⊂ Y such that Y = Y1 ∪ Y2 .
Put X1 = f −1 (Y1 ) and X2 = f −1 (Y2 ). Then X1 and X2 are disjoint non empty
clopen sets in X, and X = X1 ∪ X2 . So X is disconnected. 2
Proposition 4. Every Tychonoff space of cardinality more than 1 but less than c
is disconnected.
Proof. Suppose X is Tychonoff space such that 1 < |X| < c. Fix distinct points
p, q ∈ X. There is a continuous function f : X → R such that f (p) = 0 and
f (q) = 1. Since |X| < c, |f (X)| < c. Since |[0, 1]| = c there is a ∈ (0, 1) such that
a 6∈ f (X). Then X1 = f −1 ((−∞, a)) and X2 = f −1 ((a, ∞)) are disjoint non empty
clopen subsets of X, and X = X1 ∪ X2 . So X is disconnected. 2
Proposition 5. Let X be a topological space, and Y and Z subsets of X such that
Y ⊂ Z ⊂ Y . If Y is connected then Z is connected.
Proof. Suppose Z is disconnected. Then there are non empty disjoint subsets
Z1 , Z2 ⊂ Z such that Z = Z1 ∪ Z2 and Z1 and Z2 are clopen in Z. Put Y1 = Z1 ∩ Y
and Y2 = Z2 ∩Y . Then Y1 and Y2 are disjoint clopen subsets of Y , and Y = Y1 ∩Y2 .
Since Y is dense in Z, the sets Y1 and Y2 are non empty. So Y is disconnected. A
contradiction. 2
Corollary 6. The closure of a connected set is connected.
1
2
Definition 7. Say that a family of sets A is linked if for every A, B ∈ A, A∩B 6= ∅.
Proposition 8. Suppose A is a linked S
family of subsets of a topological space X.
If each element of A is connected, then A is connected.
S
Proof. Put Y = A. Suppose Y is disconnected. Then there are non empty
disjoint sets Y1 , Y2 such that Y = Y1 ∪ Y2 and Y1 and Y2 are clopen in Y . Pick
y1 ∈ Y1 , y2 ∈ Y2 . There are A1 , A2 ∈ A such that y1 ∈ A1 and y2 ∈ A2 . Since A is
linked we can pick z ∈ A1 ∩ A2 . Then either z ∈ Y1 or z ∈ Y2 .
In the case z ∈ Y2 the sets Y10 = Y1 ∩ A1 and Y20 = Y2 ∩ A1 are non empty disjoint
clopen subsets of A1 such that A1 = Y10 ∪ Y20 . This contradicts the connectedness
of A1 .
In the case z ∈ Y1 we arrive to a contradiction with the connectedness of A2
similarly. So Y is connected. 2
Corollary
9. Suppose A is a family of subsets of S
a topological space X such that
T
A 6= ∅. If each element of A is connected, then A is connected.
Exercise 10. Suppose A is a family of subsets of a topological space
X such that
T
T
A 6= ∅. If each element of A is connected, does it follow that A is connected?
2. Connectedness of products
Theorem 11. A (Tychonoff ) product of any family of connected spaces is connected.
Proof. First we prove that the product of two connected spaces, say X and Y is
connected. For every x ∈ X and y ∈ Y denote Tx,y = ({x} × Y ) ∪ (X × {y}). Then
({x} × Y ) and (X × {y}) are homeomorphic to Y and X, respectively, and thus are
connected. Next, the intersection ({x} × Y ) ∩ (X × {y}) contains point hx, yi and
thus is non empty. Thus by Corollary 9, Tx,y is connected. Next, note that:
• The family {Tx,y : x ∈ X, y ∈ Y } is linked. (Indeed, for every x, x0 , y, y 0 ,
Tx,y ∩ Tx0 ,y0 3 hx, y 0 i. )
S
• {Tx,y : x ∈ X, y ∈ Y } = X × Y .
Therefore by Proposition 8, X × Y is connected.
Second, by induction the statement extends to any finite product of connected
spaces.
Q
Last, let X = a∈A Xa be an arbitrary (possibly infinite) product of connected
spacesQXa . Fix a point p = hpa : a ∈ Ai ∈ X. For a finite subset F ⊂ A, put
PF = a∈A Xa,F where
Xa if a ∈ F
Xa,F =
{pa } otherwise.
Note that:
• Each Xa,F is homeomorphic ti a finite product of connected spaces, and
thus, by the previous step, connected.
• The family {PF : F is a finite subset of A} has nonS empty intersection
(point p is in each element). Hence by Corollary 9, {PF : F is a finite
subset of A} is connected.
S
Q
• {PF : F is a finite subset of A} is a σ-product in X = a∈A Xa . As we
know from the previous lectures, it follows that it is dense in X.
3
Therefore by Corollary 6, X is connected since it has a dense connected subspace.
2
Remark 12. A box product of infinitely many connected spaces may be disconnected (see Munkres, Section 23 Example 6.)
3. Connectedness of R
Example 13. R (with the usual order topology) is connected.
Proof. Suppose R is represented as the union of two non empty disjoint sets:
R = H1 ∪ H2 . Pick a ∈ H1 and b ∈ H2 . Without loss of generality we can assume
that a < b. Put z = inf([a, b] ∩ H2 ). Note that z ∈ H1 ∩ H2 . So (since z must
belong to either H1 or H2 ) it is not possible that both sets H1 and H2 are closed.
Thus R is connected. 2
The same argument shows that any interval of the real line is connected.
4. Components and quasicomponents
Definition 14. Let X be a topological space and x ∈ X. The component of x in
X is the union of all connected subsets of X that contain x.
By Corollary 9, the component of x is a connected set. Moreover, the component
of x is the greatest connected set containing x. Furthermore, y belongs to the
component of x iff x belongs to the component of y. Thus X is partitioned into
pairwise disjoint connected components.
(the rest of this section is optional)
Definition 15. Let X be a topological space and x ∈ X. The quasicomponent of
x in X is the intersection of all clopen subsets of X that contain x.
Similarly to the discussion after Definition 14, one can see that X is partitioned
into pairwise disjoint quasicomponents.
Proposition 16. In general, components are subsets of quasicomponents.
Proof. Let X be a topological space and x ∈ X. Let Cx be the component and
Qx the quasicomponent of x. Suppose Y ∈ Cx . Then there is a connected subset
A of X that contains both x and y. It is clear that A is contained in every clopen
set of X that contains x. Thus A ⊂ Qx . So y ∈ Qx . This proves that Cx ⊂ Qx . 2
Example 17. A space in which components do not equal quasicomponents.
Consider S = {0} ∪ {1/n : n ∈ N} with the topology inherited from the real line
and X = ({0, 1} × {0}) ∪ (I × {1/n : n ∈ N}) ⊂ I × S. Let p = h0, 0i and q = h1, 0i.
Then the component of p is {p} while the quasicomponent of p is {p, q}. 2
Proposition 18. In a compact space, components are equal to quasicomponents.
We will use a lemma which may be of independent interest.
Lemma 19. Let F beTa family of closed subsets of a compact space X, and let U
be
T a neighborhood of F. Then there exists a finite subfamily F0 ⊂ F such that
F0 ⊂ U .
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Proof. Put H = X \ U . Then H is a closed subset ofTa compact space X and
thus is H compact. Put O = {X \ F : F ∈ F}. Since F ⊂ U , O is a cover of
H by open sets of X. Since H is compact, there is a finite subcover O0 ⊂ O of H.
Put F0 = {F : X \ F ∈ O0 }. Then F0 is as desired. 2
Lemma 20. Let X be a topological space, F a clopen subset of X, and U and V
disjoint open subsets of X such that F ⊂ U ∪ V . Then the set F ∩ U is clopen in
X.
Proof. This is an exercise! 2.
Proof of Proosition 18. Let X be compact and x ∈ X. It suffices to show that
the quasicomponent Qx is connected. Suppose Qx = Q1 ∪ Q2 where Q1 and Q2
are closed and disjoint. Being compact X is normal, so there are open (in X)
sets U and V such that Q1 ⊂ U , Q2 ⊂ V and U ∩ V = ∅. Then Qx ⊂ U ∪ V
and thus
T by Lemma 19 there is a finite family F0 of clopen sets of X such Tthat
Qx ⊂ F0 ⊂ U ∪ V . Being a finite intersection of clopen sets the set F = F0
is clopen. Furthermore, by Lemma 20 the set F ∩ U is clopen. But then it follows
that x ∈ Qx ⊂ F ∩ U ⊂ Q1 So Q2 is empty. 2
5. Pathwise connectedness
Definition 21. Let X be a topological space, and x, y ∈ X. A continuous mapping
p : I → X such that p(0) = x and p(1) = y is called a path from x to y. X is called
pathwise connected if between every two points there is a path.
Proposition 22. Every pathwise connected space is connected.
Proof. Fix x ∈ X. For every y ∈ X fix a path py from x to y and denote
Ay = py (I). Then by Proposition 3, Ay is a connected subspace of X. The family
A = {Ay : y ∈ X} S
consists of connected subsets of
S X and has a common point x.
So by Corollary 9, A is connected. But clearly A = X. 2
Example 23. (The topological sin curve). Consider the following subspace of R2 :
X = X1 ∪ X2 where X1 = {hx, sin x1 i : x ∈ (0, 1]} and X2 = {h0, yi : y ∈ [−1, 1].
Then X is connected but not pathwise connected.
Proof. X1 is a continuous image of a connected space (0, 1] and thus by Proposition 3 X1 is connected. Further, X1 = X, so by Corollary 6, X is connected.
Next, there is no path between a = h1, sin 1i and b = h0, 0i. Indeed, if such a
path p : I → X would exist, then the y-coordinate of p(t) must tend to 0 as t → 1.
On the other hand, it must repeatedly take values 1 and -1, a contradiction. 2
Example 24. (The Long Line). Denote X0 = ω1 × [0, 1) where ω1 and [0, 1) bear
the standard order topology, and × denotes the lexicographic product. Next, put
X = X0 ∪{ω1 } where X0 is an open subspace and a basic neighborhood of the point
ω1 is this point together with arbitrary final interval of X0 . Then X0 is pathwise
connected, and X is connected but not pathwise connected.
Proof. We skip the proof that X0 is pathwise connected.
X is connected because it contains a dense connected subspace X0 . X is not
pathwise connected because there is no path from the leftmost point to the point
ω1 . Indeed, if there would be such a path, then the whole space X would be a
continuous image of I. But this is impossible because:
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• I is separable;
• X is not separable;
• a continuous image of a separable space is separable.
2
Proposition 25. A continuous image of pathwise connected space is pathwise connected.
it Proof. Let X be a pathwise connected space, f : X → Y a continuous
surjection, and x, y ∈ Y . Pick a ∈ f −1 (x) and b ∈ f −1 (y). Since X is pathwise
connected there is a path p : I → X such that p(0) = a and p(1) = b. Then f ◦ p
is a path between x and y in Y . 2
Proposition 26. Any product of pathwise connected spaces is pathwise connected.
Q
Proof. Let X = a∈A Xa be a product such that each Xa is pathwise connected.
Let x, y ∈ X. For each a ∈ A, there is a path pa : I → Xa such that pa (0) = x(a)
and pa (1) = y(a). Then the diagonal product p = ∆{pa : a ∈ A} is a pass between
x and y. 2
6. Local connectedness
Definition 27. A topological space X is called locally connected if for every x ∈ X
and every neighborhood U 3 x there is a connected neighborhood V such that
x ∈ V ⊂ U.
Example 28. The Topological Sin Curve is connected but not locally connected.
(And of course any more than one point discrete space is locally connected but
not connected.)
Proposition 29. The following conditions are equivalent:
(1) X is locally connected;
(2) Components of open subspaces of X are open.
Proof. (1) ⇒ (2) Let U be an open set in X. Then X is locally connected.
Let C be a component of U and let x ∈ C. Since U is locally connected there is
a connected open set V such that x ∈ V ⊂ U . By the definition of component,
V ⊂ C. So C contains every point together with some neighborhood, i.e. C is
open.
(2) ⇒ (1) Let x ∈ U ⊂ X where U is open, and let C be a component of x in
U . Being a component, C is connected; by (2), C is open in U and thus in X. 2
Proposition 30. In a locally connected space, components are equal to quasicomponents.
Proof. Let X be locally connected x ∈ X; let C be the component of x, and
Q the quasicomponent of x. By Proposition 29, C is open. But also for every
y ∈ X \ C, the component of y in X is open and is a subset of X \ C. Thus X \ C
is open. So C is closed and thus clopen. Thus Q \ C, and so by Proposition 16,
Q = C. 2
Example 31. A continuous image of a locally connected space does not have to
be locally connected.
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Proof. Take X0 from Example 23 and put X10 = {h−1, yi : y ∈ [−1, 1]} and
X = X0 ∪ X10 ⊂ R2 . Then X 0 is locally connected and X from Example 23 is
a continuous image of X 0 (under a condensation); we know that X is not locally
connected. 2
0
Proposition 32. (1) Any finite product of locally connected spaces is locally connected.
(2) An infinite product of locally connected spaces is locally connected if the
following two conditions are satisfied:
• All factors are locally connected;
• All except may be finitely many factors are connected.
We will discuss the proof in class.