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Math 541 Fall 2008
Connectivity
Transition from Math 453/503 to Math 541
Ross E. Staffeldt-August 2008
Closed sets
We have been operating at a fundamental level at which a topological space is a set together
with a distinguished family of subsets. This family of subsets is called a topology for the
given set.
Definition. Let X be a set. A collection of subsets U = {U} of X is called a topology for
X, if it satisfies the following properties:
1. X ∈ U and ∅ ∈ U.
2. The intersection of finitely many members of U is again in U.
3. The union of arbitrarily many members of U is again in U.
A topological space is a set together with a topology. The family U is generally called the
family of open sets of the topological space.
We have also observed that a topology is also determined by specifying the family of
closed sets, namely, the family of sets whose complements are open.
Proposition 1. Let X be a set with a topology U. Define a family of subsets C = {U c },
where U c denotes X − U, the complement of U in X. Then the family C has the following
properties.
1. X ∈ C and ∅ ∈ C.
2. The union of finitely many members of C is again in C.
3. The intersection of arbitrarily many members of C is again in C.
The family C is generally called the family of closed sets of the topological space X.
In some respects it is a matter of convenience whether or not one chooses to work with
closed sets or open sets.
However, for connectivity, we need to develop a few more properties of closed sets. First
we define an operation called closure.
Definition. Let X be a topological space and let A ⊂ X. Define
\
C
A=
C closed, C ⊃ A
We call A the closure of A.
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Thus, A is the smallest closed set in X containing A, and, if C is closed, C = C. Often,
formal properties of the closure operation follow easily from this conceptual defintion.
For the sake of calculation of set closures in examples, we characterize the elements of a
closed set in terms of open sets.
Proposition 2. Let X be a topological space and let A ⊂ X. A point x ∈ X is in A if, and
only if, for every open set U with x ∈ U, U ∩ A 6= ∅.
Proof. One way to see this result is to convert it to the contrapositive equivalent. A point
x ∈ X is not in A if, and only if, there is an open set U, x ∈ U, U ∩ A = ∅. Now, if x ∈
/ A,
then there is a closed set C, C ⊃ A, and x ∈
/ C. Then x ∈ U = X − C, which is an open
set, and, since A ⊂ C, U ∩ A = ∅.
On the other hand, if, given x, there is an open set U such that U ∩A = ∅, then C = X −U
is a closed set, A ⊂ C, and x ∈
/ C. So x ∈
/ A.
Here is a characterization of continuous functions in terms of the closure operation that is
occasionally useful. In fact, Hatcher uses this on page 35 in his calculation of the fundamental
group of high-dimensional spheres.
Proposition 3. Let X and Y be topological spaces and let f : X → Y be a function. The
following conditions are equivalent.
1. The function f is continuous.
2. For any B ⊂ X, f (B) ⊂ f (B).
Proof. If f is continuous, we know that the preimage of a closed subset of Y is closed. In
particular, for any B ⊂ X, f −1 (f (B)) is closed. Obviously, B ⊂ f −1 (f (B)). Since B is
the smallest closed set containing B, we have B ⊂ f −1 (f (B)), which is a restatement of
condition 2.
On the other hand, assume condition 2 holds, and let V ⊂ Y be open. Let B =
X − f −1 (V ) = f −1 (Y − V ). By condition 2, we have
f (B) ⊂ f (f −1 (Y − V )) ⊂ Y − V = Y − V,
since V is open. Equivalently, B ⊂ f −1 (Y − V ) = B. Since we always have B ⊂ B, we have,
in fact, B = B. That is, B = f −1 (Y − V ) = X − f −1 (V ) is closed. In turn, f −1 (V ) is open,
which means that f is continuous.
This characterization of continuity reflects the intuitive idea that a continuous function
is one that preserves limits. Here we have that B and its limit points are carried into f (B)
and its limit points, if f is continuous.
There is a dual notion, called the (set-theoretic) interior of a set, defined as
[
◦
IntA =A=
U
U open, U ⊂ A
and described as the largest open subset of A. We won’t use this very often.
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Topological connectedness
Essentially, we start this section by defining what we mean for a subset of a topological space
to be disconnected. Roughly speaking, we know how points are separated by open sets in
Hausdorff space, so we try to extend the separation idea to subsets.
Definition. Let X be a topological space, A ⊂ X a subset, possibly X itself. A separation
of A is a pair U, V of open subsets of X, such that
1. A ⊂ U ∪ V .
2. U ∩ V = ∅.
3. neither A ∩ U nor A ∩ V is empty.
Once we know when parts of a subset are separated, then, by definition, we also “know”
when a subset is connected. This may be true in a strict logical sense, but it may take some
effort to accept some of the examples.
Definition. Let X be a topological space and let A be a subset. The set A is called connected
if it has no separation. That is, A is connected, if, for every pair of open subsets U, V such
that
1. A ⊂ U ∪ V .
2. U ∩ V = ∅.
3. either A ∩ U or A ∩ V is empty.
The empty subset is always connected. Take A = X. If X has a separation X =
U ∪ V , then U and V are both open and closed in X. On the other hand, if X has
no separation, then ∅ and X are the only subsets of X that are both open and closed.
Sometimes we exploit this fact of connected spaces to prove that a property holds everywhere
in a connected topological space by proving that set of points having the property is both
open and closed. We will see examples later. The first part of the following proposition says
connectedness is a topological property, preserved by continuous functions. The second part
is a characterization of connected sets in terms of maps to discrete spaces.
Proposition 4. Let f : X → Y be a map and let A be a connected subset of X. Then f (A)
is a connected subset of Y .
Let f : X → {0, 1} be a map, where {0, 1} has the discrete topology. If A ⊂ X is
connected, then the map f |A is constant.
Proposition 5. If A is a connected subset of X and B satisfies A ⊂ B ⊂ A, then B is also
connected.
Proof. Let U, V be open, suppose B ⊂ U ∪ V and U ∩ V = ∅. If x ∈ B, then x ∈ A, and
for every open subset W containing x, W ∩ A 6= ∅. Now, if x ∈ B, either x ∈ U or x ∈ V .
If x ∈ U, then U ∩ A 6= ∅. But A is connected, and has no separation, so we must have
A ∩ V = ∅. That is, A ⊂ X − V , which implies B ⊂ A ⊂ X − V . Thus, B ∩ V = ∅. We
conclude that B has no separation, and, so, B is connected.
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Proposition 6. Let X be a topological space and let {Ai : i ∈ I} be a family of connected
subsets. If ∩i∈I Ai 6= ∅, then ∪i∈I Ai is connected as well.
Proof. Write ∪i∈I Ai = U ∪ V , where U and V are open in X. Choose an index i0 . We have
Ai0 ⊂ U ∪ V , and Ai0 is connected. Then all of Ai0 is in one or the other of the open sets,
so we may assume Ai0 ⊂ U.
Now let x ∈ ∩i∈I Ai . We have x ∈ Ai0 ⊂ U, so x ∈ U. Then, for all i ∈ I, Ai ∩ U 6= ∅.
Since each Ai is connected, we must have Ai ⊂ U and Ai ∩ V = ∅. Then ∪i∈I Ai ⊂ U and
∪i∈I Ai ∩ V = ∅. In other words, ∪i∈I Ai has no separation, so it’s connected.
So far we don’t know any examples of topologically connected spaces, except for trivial
ones. The following proposition cures this. The essential ingredients of the proof are that the
real numbers have the least upper bound property and that between any two real numbers
there is another.
Proposition 7. Let I = [a, b] be an interval in R. Then I is connected.
Proof. Let U, V be open subsets in R that provide a separation of I. That is, suppose also
I ⊂ U ∪ V and U ∩ V = ∅. Without loss of generality suppose a ∈ U. Let
S = {x ∈ I : The interval [a, x] is in U.}
The set S is non-empty, because a ∈ S, and S is clearly bounded above by b. Therefore, S
has a least upper bound c.
Now we want to show two things: First, that c ∈ S and, second, that c = b. Once we
have these facts, I ⊂ U and and I ∩ V = ∅, so our “separation” was not real.
Toward showing these facts, we first show c ∈ U. If not, then c ∈ V , and we obtain a
contradiction, as follows. Since V is open, there is an interval (c−ǫ, c+ǫ) ⊂ V . Then the
interval [a, c− 2ǫ ] 6⊂ U, and no interval [a, y] with c− 2ǫ ≤ y < c can be in U. This contradicts
that c is the least upper bound for S. We conclude that c ∈ U.
If c ∈ U, which is open, there is an interval (c−ǫ′ , c+ǫ′ ) ⊂ U. Since c is the least upper
bound of S, there is an x ∈ (c−ǫ′ , c) such that x ∈ S. That is, the closed interval [a, x] ⊂ U.
Then [a, c] = [a, x] ∪ [x, c] ⊂ U, so that c ∈ S.
Now it is easy to see that c = b. We have c ∈ S, so, if c < b, then there are numbers
c < y < b such that y ∈ (c−ǫ′ , c+ǫ′ ) ⊂ U. Then [a, y] = [a, c] ∪ [c, y] ⊂ U, which contradicts
that c is an upper bound for S.
Thus, we have our first non-trivial example of a connected space.
Examples
Proposition 8. The set R is connected. So are open intervals (c, d), half-lines (c, ∞) and
(−∞, c) and rays [c, ∞) and (−∞, c].
Proof. For each n ≥ 1, the interval [−n, n] is connected, by proposition 7. Now ∩∞
n=1 [−n, n]
is non-empty, containing the point 0, for example. Therefore, by proposition 6,
∞
[
[−n, n] = R
n=1
is connected. The other examples are handled similarly.
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Proposition 9. Let X be a connected space, and let f : X → R be a map. If f (x) < 0 and
f (y) > 0, then there is z ∈ X such that f (z) = 0.
Proof. Consider U = f −1 ((−∞, 0)) and V = f −1 ((0, ∞)). We have x ∈ U and y ∈ V , so
neither is empty. Both are open in X, because f is continuous. If X = U ∪ V , then X has a
separation, contradicting the hypothesis that X is connected. Consequently, there must be
z ∈ X such that f (z) = 0.
Definition. Define a relation ∼ on X by x ∼ y if and only if there is a connected A ⊂ X
such that x, y ∈ A.
Proposition 10. This relation is an equivalence relation. Therefore X is partitioned into
disjoint subsets, each of which is called a connected component of X.
Proof. Proof of transitivity of ∼ requires proposition 6, but is otherwise straightforward.
A somewhat more intuitive definition of connectedness, based on the idea of mapping
connected test spaces into the space under investigation is called path-connectedness
Definition. Let X be a topological space. We say X is path-connected, if, for any pair of
points x, y ∈ X, there is a map f : I → X such that f (0) = x and f (1) = y. We call f a
path in X from x to y.
Alternatively, given a topological space X, define an relation ∼p on X by x ∼p y if there
is a path f in X from x to y.
The proof the following proposition is an exercise.
Proposition 11. For any space X, the relation ∼p is an equivalence relation. The space X
is now partitioned into equivalence classes, called the path-components.
Here are two examples that show connectedness and path-connectedness are different
notions.
Example 1. Let I = [0, 1], as usual, and let A = {1, 1/2, 1/3, 1/4, . . .} ⊂ I. Define X = I ×
{0}∪A×I∪{(0, 1)}. X is called the deleted comb space. The space Y = I×{0}∪A×I∪{0}×I
is called the comb space. Make a sketch showing both Y and X.
Clearly Y is connected, by multiple applications of proposition 6. We see that X is
connected, because X ⊂ I × {0} ∪ A × I, and I × {0} ∪ A × I is connected. Now we see
that X is not path-connected.
To show that X is not path-connected, let f : I → X be a path with f (0) = P = (0, 1).
Obviously, f −1 (P ) is closed in I, since f is continuous.
We will also show that f −1 (P ) is open in I. Once we have this additional fact we argue
as follows. Since I is connected, its only subsets that are both open and closed are the empty
set and I itself. Since f −1 (P ) is not empty, f −1 (P ) = I and the path f starting at P is
constant.
To see that f −1 (P ) is open, let V be an open ball around P that does not touch the xaxis. f −1 (V ) is open in I. Then, for any t ∈ f −1 (V ) there is a small interval J = (t − ǫ, t + ǫ)
such that f (J) ⊂ V ∩ X. Now let (1/n, y) ∈ V ∩ X. Choose r, 1/(n + 1) < r < 1/n, and
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define HL = (−∞, r) × R and HR = (r, ∞) × R. We note that P ∈ HL and (1/n, y) ∈ HR .
Since the ball V does not touch the x-axis, we must have that f (J) ⊂ HL . This is true
because J is connected, its image under f is connected, and the only way to reach HR in X
is by taking a path through (r, 0) on the x-axis, which is specifically excluded from V . Thus,
for any point (1/n, y) ∈ V , ( (1/n, y) 6= P !), (1/n, y) is not in the image f (J). We conclude
that f −1 (V ) = f −1 (P ) for an open ball V as above. Thus, f −1 (P ) is also open.
Example 2. Let Γ = {(x, sin(1/x) : 0 < x ≤ 1} be a part of the graph of the function
f (x) = sin(1/x), defined for x 6= 0. Let W = Γ ∪ {0} × [−1, 1] ⊂ R2 . It is easy to verify that
W = Γ, so W is connected by proposition 5. An argument similar to the one proving the
deleted comb space is not path-connected shows that W is not path-connected. This example
is sometimes called the “Warsaw sine curve” or the “Polish sine curve”.
Product spaces and connectedness
We will see that it is straightforward to demonstrate that a finite product of connected spaces
is connected. For infinite products, the result is not so obvious.
Proposition 12. Let X and Y be connected non-empty topological spaces. Then X × Y is
also connected.
Proof. We make two applications of proposition 6. Choose x0 ∈ X and let y ∈ Y . The
subsets Ay = X × {y} and B = {x0 } × Y are connected, since they are homeomorphic
to X and Y , respectively. And Ay and B have a point in common. So, by proposition 6,
Cy = Ay ∪ B is connected. Now consider the family of subsets {Cy : y ∈ Y }. Each of these is
connected, and ∩Cy = B is not empty. Therefore, the union ∪Cy = X × Y is connected.
Inductively, any finite product of connected spaces is connected.
Proposition 13. LetQ{Xα : α ∈ A} be a family of non
Q empty connected spaces indexed by the
infinite set A. Then α∈A Xα is connected, where α∈A Xα is given the product topology.
Proof. This result requires a few minutes of consideration of the definition of open sets in
the product topology. Recall that the basic open sets, whose unions are the general open
sets of the product topology, are sets
Y
U=
Uα where Uα is open in Xα and where Uα = Xα for all but finitely many α.
α∈A
Q
Back to the proof of the proposition, we know
that α∈F Xα , is connected for any finite
Q
subset F of A.QNow, choose a point (x0α ) ∈ α∈A Xα . For
Q each finite subset F of A we
have a copy of α∈F Xα sitting in side of the full product α∈A Xα as the set of points (xα )
defined by the equations xα = x0α , for α ∈
/ F . That is, we restrict the coordinates, if the
coordinate index is not in F . All subsets constructed this way are connected, by the finite
case, and these subsets have the point (x0α ) in common. Therefore the union W is connected.
However, we are not yet finished. The union W is not all of the full product. Indeed, it
consists of all points of the product for which at most finitely many of the coordinates differ
from those of the “basepoint” (x0α ). At this point we really need the definition of open sets
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in the product topology. Given the description of the basic open sets containing any point
(xα ), it is easy to see that any of these basic open sets has non-empty
intersection with W .
Q
But this says that,
Q in the product topology the closure W = α∈A Xα , the full product. By
proposition 5, α∈A Xα is connected.
Local path-connectedness:
We have seen by means of the deleted comb space example and the Warsaw sine curve that
connectedness and path-connectedness are distinct notions. However, because the interval
is connected, every path-connected space is connected.
From the properties of equivalence classes in general, it follows that every connected
component of a topological space is a union of path components. However, path components
are not necessarily open or closed subsets of a connected component. Just identify the path
components in the two examples to see this.
There is a useful condition that we can impose to make the situation a little better. That
is the condition of being “locally path connected”. Read the definition very carefully; it may
not coincide with your first idea.
Definition. We say a space X is locally path-connected, if for every point x, and for every
open set V containing x, there is a path-connected open set U such that x ∈ U ⊂ V .
This says that every point has path-connected neighborhoods that are arbitrarily small,
which is more than saying every point has a path-connected neighborhood. This is not the
only way to define “locally path-connected”; see Hatcher’s remarks at the top of page 62.
Proposition 14. Let X be a connected space that is locally path-connected. Then X is
path-connected.
Proof. Let x0 ∈ X be a point, and define
U0 = {x ∈ X : There is a path in X from x0 to x.}
We will show that U0 , which is not empty, is both open and closed in X. Since X is
connected, U0 must be all of X. Let x ∈ U0 . The set X is an open set containing x, so there
is a path-connected open set U, such that x ∈ U ⊂ X. Following a path from x0 to x by a
path from x to any other point y ∈ U shows that, if x ∈ U0 , then there is an open subset U
such that x ∈ U ⊂ U0 . Thus, U0 is open.
Now let y ∈ U0 . We show that y ∈ U0 . Recall the characterization of points of a closure
given in proposition 2. If y ∈ U0 , then for any open set V containing y, V ∩ U0 6= ∅. By
local path-connectedness, any open set V containing y contains a path-connected open set
U, which also contains y. And U ∩ U0 6= ∅. So, if z ∈ U ∩ U0 , we can connect x0 to z, and
we can connect z to y by a path in U. Joining the paths together shows that y ∈ U0 . Thus,
U0 is also closed.
As indicated, U0 is a non-empty subset of X that is both open and closed. Since X is
connected, U0 = X.
This argument can be extended to show that if a space is locally path-connected, then
the path components coincide with the components. Moreover, each path component is an
open subset of the whole space.
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