Download Topology I Final Solutions

Topology I Final Solutions
1. Which of the following properties are topological? (Circle those that are.):
Path-connectedness, local connectedness, compactness, convexity, separability.
Solution
Everything but convexity is.
2. (a) Give an example of a subspace A of a space X, in which there is a
relatively open set S of A that is not an open set of X.
(b) Give an example of a closed mapping that is not continuous.
(c) Give an example of subsets A and B of R2 such that A and B are
disconnected, but A ∪ B is connected.
(d) Give an example of a connected space that is not path connected.
Solution
(a) Let X be the real line, A = [2, 4] and S = (3, 4]. S is not open in X,
but S = (3, 5) ∩ A so is relatively open in A.
(b) Let Y be the finite set {a, b} with the discrete topology, so any map
from any space to Y is closed. Define f : R → Y be defined by f (x) = a
if x ∈ Q and f (x) = b otherwise (R with the usual topology.) This is
not continuous.
(c) A = {(x, y) | x = ±1}, B = {(x, y) | y = ±1}.
(d) The topologist’s sine curve.
3. Let X be a set and T be a family of subsets whose union equals X. Show
that T is a subbasis for a topology on X.
Solution
Recall that T is a subbasis of some topology if it generates, by finite intersection, a basis of the topology. Let B be the family of sets generated from finite
intersections of sets of T. We show that this is a basis of some topology.
As X is the union of sets of T and so the union of sets of B, it is enough to
show that for B1 and B2 in B, B1 ∩ B2 is also in B. But this is clear: for B1
T
T
and B2 in B, we have that B1 = n
Ti for Ti ∈ T and B1 = m
i=1 Si for
Tn
Tm i=1
Si ∈ T. So B1 ∩ B2 = i=1 Ti ∩ i=1 Si is an intersection of a finite number
of sets from T, so by definition is in B. This is enough.
4. (a) Define what it means for a space to be connected. (Give any of the
equivalent defintions.)
(b) Prove that being connected is a topological property.
Solution
A space X is connected if there is no continouous surjection from X onto D2 ,
the discrete topology on two points.
If X is connected and h : X → Y is a homoeomorphism, then Y must also be
connected. Indeed, if it were not then there would be a continuous surjection
f : Y → D2 , but then f ◦ h is a continous surjection from X to D2 , which
cannot exist, as X is connected. So by contradiction Y is connected.
5. Prove that a homeomorphism h : X → Y between spaces X and Y induces
a one-to-one correspondence between components of X and components
of Y .
Solution
Let C be a component of X containing x. As connectedness is a continuous
invarient, and h is continuous, h(C) is connected in Y . We claim it is, infact
a component of Y . Indeed, if it isn’t then it is properly contained in some
component B of Y . But then by the continuity of h−1 , h−1 (B) is a connected
set of X which properly (by injectiveness) contains h−1 (h(C)) = C. This
contradicts the fact that C is a component, so h maps components of X to
components of Y . Similarily h−1 maps components of Y to components of
X.
Now since h : X → Y is a homeomorpishm, h ◦ h−1 and h−1 ◦ H are identity
maps, this holds when h and h−1 are viewed as maps of components, and so
h is a bijection from the components of X to the components of Y .
6. Let {Ai }∞
i=1 be a sequence of path connected subsets of a space X such
that for each integer i ≥ 1, Ai has at least one S
point in common with one
∞
of the preceeding sets A1 , . . . , Ai−1 . Show that i=1 Ai is path connected.
Solution
S
We show by induction on n that the set n
i=1 Ai is path connected. The
statement follows by
transfinite
induction.
The base case n =S1 is clear.
S
Assume that S = n−1
connected. We show that n
i=1 Ai is.
i=1 Ai is path S
Indeed, let x and y be two points in n
i=1 Ai . We must show there is a path
from x to y. If x and y are both in S or both in An then there is a path
between them by the connectedness of S or An . So we may assume that x ∈ S
and y ∈ An . By the assumption there is a point z in both An and Ai ⊂ S
for some i < n. By the path connectedness of S we have a path P1 from x to
z and by the path connectedness
of An we have a path P2 from z to y. So
S
P1 ∗ P2 is a path in n
i=1 Ai from x to y. This is enough.
An alternate way to do the induction is to prove (as in the induction step
above) the statement (*)
If A and B are path connected spaces sharing a common point,
then A ∪ B is path connected.
Then use this in the induction step by taking A = An and B = A1 ∪· · ·∪An−1 .
Many people did this, but without proving (*). You need to prove (*)!
7. (a) Define what it means for a space to be compact.
(b) Prove that a space X is compact if and only if it has a basis B for
which every cover of X by members of B has a finite subcover.
Solution
A space X is compact if every open cover of the space has a finite subcover.
(⇒) Assume that X is compact and let B be any basis. Then every cover of
X by members of B is an open cover so has a finite subcover.
(⇐) Assume that X has a basis B = {Bi }i∈I such that any cover of X by
members of B has a finite subcover. To show that X is compact we let S
be an open cover of X, and show that it has a finite subcover. Indeed, for
S ∈ S is open, so can be written S = ∪i∈IS Bi for some subset IS ⊂ I. Now
∪S∈S ∪i∈IS Bi = ∪S∈S S is a cover of X by members of B so has a finite
subcover B1 , . . . , Bn . Now for each i = 1, . . . n let Si be a set in S such that
i ∈ ISi , (so Bi is one of the basis element whose union is Si ). We claim that
n
S1 , . . . , Sn is a cover of X. Indeed X = ∪n
i=1 Bi ⊂ ∪i=1 Si ⊂ X.