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Math 525 More notes on connectedness (sec 24, 25)
Recall that a subset A of (X, τ ) is connected if there are no open sets U , V which disconnect A (i.e. U ∩ V ∩ A = φ, U ∩ A 6= φ, V ∩ A 6= φ, and A ⊆ U ∪ V ). Connectedness is a
topological property which is not hereditary, but is preserved under continuous maps, finite
products (see Thm 23.6, Munkres), and passage to a coarser topology (Prop A42). Among
other important properties: closures of connected sets are connected (but not interiors), and
each x ∈ X is in Cx , the largest connected subset containing x. Connectedness is often
useful for showing that certain spaces are not homeomorphic, and that certain functions are
not continuous. Along these lines, we state (without proof) the following.
Prop A47 Let f : (X, τ ) → (Y, µ) be a homeomorphism. Then A ⊆ X is connected iff f (A)
is connected. For any x ∈ X, f (Cx ) = Cf (x) . The cardinal number of connected components
in (X, τ ) equals that in (Y, µ).
Example 1 Does there exist a continuous map from (R, τu ) onto any of the following: (a) a
two-element discrete space; (b) Q with its usual, order topology; (c) (R, τs )? In each case,
the answer is NO, because (R, τu ) is connected, while the other spaces are not. Indeed, the
only continuous map from (R, τu ) into any of these spaces is a constant map.
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Example 2 Let A = {(x, y) ∈ R2 ¯ x2 + y 2 ≤ 1},
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B = {(x, y) ∈ R2 ¯ (x − 1)2 + y 2 ≤ 1} ∪ {(x, y) ∈ R2 ¯ (x + 1)2 + y 2 ≤ 1}.
Are A and B homeoporphic when considered as subspaces of (R, τu )2 ? The answer is NO.
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Note that B − {(0, 0)} is disconnected (by U = {(x, y) ¯ x > 0} and V = {(x, y) ¯ x < 0}),
but removal of a single point will not disconnect A. Also note that B o is not connected,
whereas Ao is. This could also be used to show that A and B are not homeomorphic.
Example 3 Is the letter
homeomorphic to the letter , where both are considered as sub-
spaces of (R, τu )? Again, the answer is NO. If there was a homeomorphism from the set of
points forming the
to the set of points forming the , then the removal of the middle
point of the
would result in a disconnected subspace with four connected components. The
homeomorphic image of this subspace would have to have the same properties, but this is
impossible (removal of a single point from the
can not leave four connected components).
Next we define and discuss three topological properties which attempt to further describe
“connectedness” and “disconnectedness”. Let (X, τ ) be a topological space.
Definitions
1. (X, τ ) is totally disconnected iff for all x ∈ X, Cx = {x}.
2. A set A in (X, τ ) is path-connected iff whenever x, y ∈ A, there exists a continuous
map f : ([0, 1], τu ) → A such that f (0) = x and f (1) = y. f (or its image) is called a path
in A from x to y. If X is path-connected, the (X, τ ) is called a path-connected space.
3. (X, τ ) is locally connected iff for all x ∈ X, there is a nbhd basis at x consisting of
connected sets.
Note that all T2 , zero-dimensional spaces (like (R, τs )) are totally disconnected, but (Q, τu )
is a totally disconnected space which is not zero-dimensional.
Prop A48 Any path-connected set is connected, but the converse is false.
Proof : Take any path-connected set A in (X, τ ). Suppose U and V disconnect A. Then
there must exist x ∈ U and y ∈ V . By the path-connectedness of A, there is a continuous
map f : [0, 1] → A with f (0) = x and f (1) = y. Since f ([0, 1]) is the continuous image of a
connected set, it too must be connected. But then U and V must disconnect f ([0, 1]) in A,
which is a contradiction. So there can be no disconnection of a path-connected set.
To show the convers is false, consider the following example:
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S0 = {(x, sin( x1 )) ¯ 0 < x ≤ 1}, A = {0} × [−1, 1], S = A ∪ S0 .
As subspaces of (R, τu )2 , it should be relatively clear that (1) S0 is path-connected, and
therefore connected; (2) S0 = S, so S is connected; (3) S is not path-connected. The details
of this example are mentioned in Munkres, §24, Example 7.
The “topologist’s sine curve”, as the previous example is known, shows that path-connectedness
is not preserved under closures. Path-connected components may be defined in much the
same way as connected components, but path-connected components are not necessarily
closed.
Prop A49 The continuous image of a path-connected set is path-connected.
Proof : Let f : (X, τ ) → (Y, µ) be continuous, and let A ⊆ X be path-connected. To show
that f (A) must be path-connected, take any two points a, b ∈ f (A). Then there are x, y ∈ A
with f (x) = a and f (y) = b. By the path-connectedness of A, there is a continuous function
g : [0, 1] → A such that g(0) = x and g(1) = y. Then f ◦ g : [0, 1] → f (A) is a path in f (A)
from a to b.
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The fact that no space other than a discrete space can be both locally connected and totally
disconnected follows from the next proposition.
Prop A50 If (X, τ ) is locally connected, the every connected component is clopen.
Proof : Let (X, τ ) be locally connected, and let C be any connected component in X. We
already know that C is closed by Prop A44. Now, for every x ∈ C, there must be a connected
nbhd Ux , and it must be the case that Ux ⊆ C, since C = Cx is the largest connected set
[
containing x. Now we can write C =
Ux , which shows that C is a union of nbhds, and
x∈C
is therefore open.
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Now note that if (X, τ ) is totally disconnected and locally connected, then the connected
componenets are singletons, and so the singletons must all be open, which means (X, τ )
must be discrete, as noted earlier.
Recall that in (R2 , τdict ), the connected components are the vertical lines. This space is
locally connected, but not connected. There are also spaces which are connected, but not
locally connected.
Example 4 Define
A0 = {0} × [0, 1], An = { n1 } × [0, 1], B = [0, 1] × {0}.
S
Then C = A0 ∪ B ∪ ( An ), as a subspace of (R, τu )2 , is called the topologist’s comb.
It is path-connected, and therefore connected, but it is not locally connected at the point
z = (0, 1). Take, for example, the nbhd U = (− 21 , 21 ) × ( 12 , 23 ) ∩ C. There is no connected
nbhd of z which fits inside of U , since any nbhd of z which fits inside of U would contain
infinitely many “tine tips”, and must therefore be disconnected.