Download Connectedness and path-connectedness

Connectedness and path-connectedness
Saul Glasman
October 3, 2016
Recall the following lemma from the end of last time:
Lemma 1. Let (Ui )i∈I be a collection of connected subspaces of a space X,
and suppose that
\
Ui 6= ∅.
i∈I
Then
U=
[
Ui
i∈I
is connected.
This will allow us to prove the following:
Proposition 2. If X and Y are connected spaces, then X × Y is connected.
Proof. Choose some point x ∈ X. For any y ∈ Y , the set
Ax,y = {x} × Y ∪ X × {y}
is connected; indeed, it’s the union of a copy of Y and a copy of X, both of
which are connected, and which intersect in (x, y) (draw picture). Now
[
X ×Y =
Ax,y
y∈Y
(draw picture) but
\
Ax,y = {x} × Y,
y∈Y
which is nonempty). So X × Y is connected.
Another source of connected spaces is as follows:
Lemma 3. Any quotient of a connected space is connected.
Proof. Remember we proved last time that if f : X → Y is a continuous function
and X is connected, then the image f (X) is connected. In particular, if f is
surjective, then f (X) = Y and Y is connected.
But if S is an equivalence relation on X, then the quotient map q : X → X/S
is continuous and surjective. It follows that X/S is connected.
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Example 4. Since [0, 1] is connected and S 1 is a quotient of [0, 1], S 1 is also
connected.
So what connected spaces do we know now? We know that R is connected,
and since an open interval (a, b) is homeomorphic to R, (a, b) is also connected remember that homeomorphic spaces all share the same topological properties.
In the course of proving this, we proved that a closed interval was connected.
We’d also like to assert that a half-open interval is connected - to do this, we
could go through the whole proof again, but it’s simpler to prove the following
more general result from the book:
Lemma 5. Let A be a connected subspace of X. If
A ⊆ B ⊆ A,
then B is also connected.
This would imply that [a, b) is connected, since
(a, b) ⊆ [a, b) ⊆ [a, b].
Proof. Let B = C ∪ D be a separation of B. Since A is connected, either
A ⊆ C or A ⊆ D; suppose that A ⊆ C. Then D ⊆ A but D ∩ A = ∅. This
is a contradiction, since D is open, and any open neighborhood of a point in A
intersects A.
So now we know that open, half-open and closed intervals, their products
and their quotients are connected. That’s a fair number of spaces.
Now I’d like to introduce another perspective on connectedness. In the
examples of disconnected spaces we’ve seen, with separated components A and
B, it seems like A and B are separated by an unbridgeable gulf - in other words,
there’s no path from a point in A to a point in B. We can turn this into a
definition which is almost, but not quite, the same thing as connectedness.
Definition 6. Let X be a space, and let x, y be points in X. Then a path from
x to y is a continuous map
p : [0, 1] → X
such that p(0) = x and p(1) = y.
Definition 7. Let X be a space. We say X is path-connected if for any points
x, y ∈ X, there exists a path from x to y.
Lemma 8. If X is path-connected, then it is connected.
Proof. Suppose A, B form a separation of X, and take a ∈ A and b ∈ B. Let
p : [0, 1] → X be a path from a to b. Then p−1 (A) and p−1 (B) are disjoint open
sets whose union is [0, 1], and since 0 ∈ p−1 (A) and 1 ∈ p−1 (B), neither of them
is empty. That would give us a separation of [0, 1], which is a contradiction.
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This would tempt you to think that connectedness is really the same thing
as path-connectedness, but unfortunately that’s not the case: there really are
connected spaces which are not path-connected. These spaces are fairly nasty,
but also fairly interesting. The most famous of these spaces is the famous
“topologist’s sine curve”.
Definition 9. Define f : (0, 1] → R2 by
f (x) = (x, sin(1/x)).
f is continuous. Let S = f ((0, 1]) be the image of f (draw a picture); since
S is the image of the connected space (0, 1] under a continuous function, it is
connected. Define the topologist’s sine curve to be
X = S ⊆ R2
with its subspace topology. Since the closure of a connected subset of a space
is always connected, X is connected.
Theorem 10. X is not path-connected. Therefore there exist connected spaces
which are not path-connected.
Proof. First we claim that
X = S ∪ ({0} × [−1, 1].
Indeed, suppose y ∈ [−1, 1], and let
a = sin−1 (y) ∈ [−π, π].
Then for all n ≥ 1,
1
,y
a + 2πn
∈ S,
and this is a sequence of points converging to (0, y), so (0, y) ∈ X. But it’s
clear pictorially that there are the only points in the closure of S apart from
the points of S itself. However, we claim that there is no path from (0, 0) to
(1/π, 0) ∈ S. Indeed, suppose
q : [0, 1] → X, q(t) = (qx (t), qy (t))
were such a path. Well, such a path has to get out of this bar on the left
eventually. q −1 ({0} × [−1, 1]) is a closed subset of [0, 1], so it has a maximum
element m. We can’t have m = 1, since q(1) = (0, 1/π), so we know m < 1.
So we might as well start at m, and [m, 1] is homeomorphic to [0, 1], so we can
instead assume that
qx (t) > 0 for t > 0
and
q(0) ∈ {0} × [−1, 1].
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Now our aim is to describe a sequence (tn )n∈N with tn → 0 but such that
qy (tn ) oscillates wildly. In fact, we’ll be able to make sure that
(
1
n even
n
qy (tn ) = (−1) =
−1 n odd.
For each n, let xn = qx (1/n). There is some tn with sn < xn and sin(1/sn ) =
(−1)n , by using a similar trick to the one we used to identify the closure. By the
intermediate value theorem, there is some tn ∈ (0, 1/n) such that qx (tn ) = sn .
Now tn → 0, so
q(tn ) = (sn , (−1)n ) → q(0).
But this sequence clearly does not converge, and so we have a contradiction.
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