Download Homework Set #2 Math 440 – Topology Topology by J. Munkres

Homework Set #2
Math 440 – Topology
Topology by J. Munkres
Clayton J. Lungstrum
October 26, 2012
Exercise 1.
Prove that a topological space X is Hausdorff if and only if the diagonal ∆ = {(x, x) :
x ∈ X} is closed in X × X with the product topology.
Solution.
Let X be a Hausdorff topological space. Consider ∆; we know ∆ is closed if and only if
its complement is open. To that end, let (x, y) ∈ (X × X) \ ∆. Since (x, y) is not on the
diagonal, the points are not the same, i.e., x 6= y. Thus, in X, there exist open sets U and
V such that x ∈ U and y ∈ V such that U ∩ V = ∅ (since X is Hausdorff). Since U and V
don’t share any elements in common, (U × V ) ∩ ∆ = ∅, but U × V is open in X × X as it is
the product of open sets. Since the point (x, y) is arbitrary, this implies ∆C is open, hence
∆ is closed.
Conversely, let ∆ be closed. Then, for any (x, y) ∈ ∆C , there exists a basic open set
U × V containing (x, y) such that (U × V ) ∩ ∆ = ∅. Then U and V don’t share any elements
in common, hence U ∩ V = ∅ where x ∈ U and y ∈ V . Thus, X is Hausdorff.
Q.E.D.
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Exercise 2.
Write the 1-point compactification of the natural numbers with the discrete topology as a
subspace of [0, 1].
Solution.
The 1-point compactification of the natural numbers with the discrete topology as a
subspace of [0, 1] is
∞ [
1
.
{0} ∪
n
n=1
It is clear this is the space we want, for consider the function h : N∞ → [0, 1] (where N∞ is
the 1-point compactification of N) defined by
(
1
: n∈N
h(n) = n
.
0 : n=∞
h is clearly injective, it is surjective onto its image, and it and its inverse are clearly continuous
as their domains are discrete topological spaces (where every function is continuous). Thus,
h is a homeomorphism.
Q.E.D.
2
Exercise 3.
Let X be a connected, locally connected, locally compact Hausdorff space. Prove that for
every x, y ∈ X, there is a compact connected subset containing them.
Solution.
Let U be a cover of X, then a chain in U is a finite indexed set U1 , U2 , . . . , UN ∈ U such
that, for all i = 1, . . . , N − 1, we have that Ui ∩ Ui+1 6= ∅. It is called a chain from x to y in
U when we have x ∈ U1 and y ∈ UN .
Now, a space X is connected if and only if for every open cover U of X, we have a chain
between any pair of points of X.
Fix x0 ∈ X and define O to be the set of all y ∈ X such that there is a chain from x0 to
y. O is nonempty, as any x ∈ X is covered by some U ∈ U, then U1 is a chain from x to x,
so x ∈ O. Now I will show O is open and closed, thus it must be the whole space (otherwise
we could separate X by it and its complement).
Open: Let y ∈ O and let x ∈ U1 , . . . , UN be a chain (from U) for it. Then for every z in UN ,
that same chain will be a chain from x to z, so that z ∈ O as well. Thus, UN ⊆ O,
and every point of O is an interior point. Note that we do not even need the cover to
be open, just that the interiors of the sets cover X.
Closed: Suppose that y ∈
/ O, and let U be an element from U that covers y. Suppose that
some z ∈ U is in O, and again, let x ∈ U1 , . . . , UN be a chain from x to z so we have
z ∈ UN . But then the chain x ∈ U1 , . . . , UN , U is a chain from U as well, because
all intersections are non-empty in the beginning by assumption, and UN ∩ UN +1 is
non-empty, as both contain z, and this would be a chain from U from x to y, so that
y ∈ O, a contradiction to what we assumed. Thus, U misses O entirely, so O is closed.
But now the connectedness of X forces O = X (there is only one nonempty closed and
open set), and we have what we wanted in the chain condition, as x was arbitrary.
Having this, we can quickly see that between any two points, we can find such a chain,
and using the properties of locally compact and locally connected, we can find compact
and connected neighborhoods inside such a chain, and then we take the union of these
neighborhoods. Being a finite union, the set is still compact, and since each subsequent set
has nonempty intersection with the previous one, the union is connected, thus we’re done.
Q.E.D.
3
Exercise 4.
Prove that a countable closed subset of a locally compact Hausdorff space has an isolated
point.
Solution.
Let A be a closed subset of a locally compact Hausdorff space with no isolated points.
Let x ∈ A. Then, since X is locally compact, x has a compact neighborhood, call this set
V . As a subset of a Hausdorff space, clearly V is closed, so V ∩ A is closed. We claim V ∩ A
is compact. This is immediate as we take any open cover of V ∩ A, say {Bi }i∈I . Then
[
(V ∩ A)C ∪ Bi
i∈I
is an open covering of V , a compact set, hence there is a finite subcover. Therefore, we have
a finite subcover of the open cover of {Bi }i∈I , hence V ∩ A is compact. As a subspace of X,
V ∩ A is also Hausdorff. Thus, V ∩ A is a compact Hausdorff set with no isolated points,
hence V ∩ A must be uncountable. Since V ∩ A ⊆ A, A is uncountable.
Taking the contrapositive of this, we see that if we have a countable closed subset of a
locally compact Hausdorff space, it must have an isolated point.
Q.E.D.
4
Exercise 5.
Let X be a compact Hausdorff space, C ⊆ U , where C is a component of X and U is
open. Prove that there is a set V which is both open and closed such that C ⊆ V ⊆ U .
Solution.
First, suppose that a component C can be characterized by the intersection of all closed
and open sets containing C. Thus, suppose
\
C=
Fi
i∈I
where each Fi is a closed and open set containing C. Then U C is covered by {FiC }i∈I . Since
U is open, its complement is closed, and a closed subset of a compact space is compact,
hence U C is compact. Thus, there must be a finite collection, relabeling if necessary, such
that
N
[
C
U ⊆
FiC .
i=1
Thus,
C⊆
N
\
Fi ,
i=1
which is a finite intersection of closed and open sets, hence is closed and open.
To show that a component can be characterized by the intersection of all closed and open
sets containing it in a compact, Hausdorff topological space is clear, though, as the following
argument shows. Let x ∈ C, a component, and let E be the intersection of all closed and
open sets containing C, that is, the Fi above. Clearly E is closed as it is the intersection
of closed sets. Suppose E = A ∪ B with A and B closed and disjoint in E (thus they are
closed in X). Without loss of generality, assume x ∈ A. Recall that compact and Hausdorff
implies normal, so there exists disjoint open subsets U and V such that A ⊆ U and B ⊆ V ;
hence, we have E ⊆ U ∪ V . Now, the union of the FiC ’s form an open cover (as each FiC is
C
C
also closed and open)
T subcover.
TNof (U ∪ V ) . Since (U ∪ V ) is compact, there is a finite
Thus, we have E ⊆ i=1 Fi ⊆ U ∪ V by construction. For simplicity, let F = N
i=1 . Now we
claim that U ∩ F is closed and open. It is clearly open as a finite intersection of open sets.
Also, any point in its closure would be in F , thus in U ∪ V , but it cannot be in V as U and
V are disjoint. Hence, U ∩ F is closed. Therefore, x ∈ U ∩ F since x ∈ A ⊆ U . We also
have x ∈ E ⊆ F , thus U ∩ F is just one of the closed and open sets containing x, therefore
E ⊆ U ∩ F . But the B must be empty as U and V are disjoint, and in particular, U ∩ F
and V are disjoint. Thus, there is no separation of E, hence E is connected. Conversely,
a connected set must be contained in some component, as a component is a maximally
connected subset.
Q.E.D.
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