MAT 3500/4500
MANDATORY ASSIGNMENT SUGGESTED SOLUTION
Problem 1
Let X = {1, 2, 3, 4}.
(a) Which of the following collections of subsets are topologies on X ?
(1) ∅, {1}, {1, 2}, {1, 3, 4}, X.
(2) ∅, {2}, {1, 2}, {2, 4}, X.
(3) ∅, {1}, {2}, {1, 2}, {1, 3}, {2, 4}, {1, 2, 3}, {1, 2, 4}, X.
Solution: In all cases X and ∅ are subsets in the collections. Moreover, since the set
X is finite in all cases, it is enough to check wether the union and intersection of any
two subsets are subsets in the given collection. In (2) we see that {1, 2} ∪ {2, 4} =
{1, 2, 4} is not a subset in the collection. So we do not have a topology in this case.
In (1) and (3), we see by inspection that the union and intersection of any two
subsets are in the given collection, so we have a topology in (1) and (3).
(b) In (a), when we have a topology on X, decide whether X is a connected space.
Solution: To decide whether or not the space is connected, we must decide if the
space can be written as a disjoint union of two open non-empty subsets. In (1) we
see that any two non-empty open subsets have 1 in their intersection, so we do not
have a separation of the space in this case. The space is consequently connected in
(1).
In (3) we see that X = {1, 2} ∪ {3, 4} which is a separation of the space and the
space is disconnected.
Problem 2
Let τ be the collection of subsets of R such that U ∈ τ if either U = ∅ or 0 ∈ U
where U is open in the standard topology of R.
(a) Show that τ is a topology on R.
Solution: ∅ ∈ τ and since R is open and 0 ∈ R, R ∈ τ . Let {Ui | i ∈ I} be a a
S
subcollection of sets in τ . Since each Ui is open in the standard topology, Ui is
i
S
open in the standard topology. If Ui = ∅ for all i, Ui = ∅ ∈ τ . If not, Ui 6= ∅ and
i
S
0 ∈ Ui for at least one i, so 0 ∈ Ui and it follows that τ is closed under arbitrary
i
unions. Let U1 , U2 be sets in τ . If Uj = ∅ for either j = 1 or j = 2. U1 ∩ U2 = ∅.
If Uj 6= ∅ for j=1 and j=2, each Uj is open in the standard topology and 0 ∈ Uj .
So U1 ∩ U2 is open in the standard topology and 0 ∈ U1 ∩ U2 . It follows that τ is
closed under finite intersections. All together, this proves that τ is a topology.
(b) Find the closure of the sets {0} and {x}, where x 6= 0, and find the interior of
the set {1, 2} in the space (R, τ ).
1
2
Solution: Let x ∈ R. Let U be a neighbourhood of x in τ . Then 0 ∈ U and we get
that x ∈ {0}. So {0} = R. Let x 6= 0. Then 0 ∈ R − {x} and R − {x} is open in the
standard topology, hence open in τ . So {x} is closed in τ and {x} = {x}. Finally,
since 0 ∈
/ {1, 2}, {1, 2} does not contain any open sets in τ hence int{1, 2} = ∅.
(Note that it was actually my intention to ask you find the interior of the interval
(1, 2) and not of the set {1, 2}, but unfortunately, I made a tex misprint in the
manuscript.)
(c) Let K be a compact set in the standard topology of R. Explain why K also is
compact in the topology τ . Decide whether (R, τ ) is a locally compact space. Are
all compact sets in the space (R, τ ) also closed ?
Solution: Let K be compact in the standard topology, and consider a covering of
K of sets in τ . Then this is also an open covering in the standard topology, and
we can consequently find a finite subcollection the covers K. It follows that K is
compact in τ . Let x ∈ R. Then I = (−|x| − 1, |x| + 1) is open in τ . Moreover
K = [−|x| − 1, |x| + 1] is compact in the standard topology, hence in τ , and we have
x ∈ I ⊂ K. It follows that τ is a locally compact space. The one-point set {0}
is compact in τ , but it is not closed (since we have seen that {0} = R) so not all
compact sets in τ are closed in τ .
(d) Let f : R → R be a continuous map in the standard topology such that f (0) = 0.
Show that f also is continuous in the topology τ . If we drop the condition f (0) = 0,
which functions are then continuous both in standard topology and in τ ?
Solution: Assume f is continuous in the standard topology, with f (0) = 0. Let U
be open in τ . Since U is open and f continuous in the standard topology, f −1 (U )
is open in the standard topology. Since f (0) = 0, 0 ∈ f −1 (U ). So f −1 (U ) ∈ τ ,
hence f is continuous in τ . Assume f is continuous in the standard topology, but
f (0) = y 6= 0. Then U = R − {y} is open in τ , but 0 ∈
/ f −1 (U ). So f −1 (U ) is not in
τ unless f −1 (U ) = ∅, that is f (x) = y for all x. Hence only the constant functions
are continuous in both topologies when f (0) 6= 0.
(e) Find an example of a function f : R → R which is continuous in τ but not
continuous in the standard topology.
Solution: Let f (x) = 0 when x < 1 and f (x) = 1 when x ≥ 1. Obviously, f is not
continuous in the standard topology. Let U 6= ∅ be in τ . Then 0 ∈ U so if 1 ∈
/ U,
then f −1 (U ) = (−∞, 1). If 1 ∈ U , then f −1 (U ) = R. In both cases, f −1 (U ) is a
set in τ , so f is continuous in τ .
Problem 3
(a) Let p : X → Y be a quotient map. Show that Y is Hausdorff if and only if the
following condition is satisfied:
For any x1 , x2 in X with p(x1 ) 6= p(x2 ) , there exist saturated open sets U , V of X
such that x1 ∈ U , x2 ∈ V and U ∩ V = ∅.
3
Solution: Assume Y is Hausdorff. Let y1 = p(x1 ), y2 = p(x2 ) with y1 6= y2 . Since
Y is Hausdorff, there exist disjoint open sets Ũ and Ṽ such that y1 ∈ Ũ and y2 ∈ Ṽ
Then U = f −1 (Ũ ) and V = f −1 (Ṽ ) are open, disjoint and saturated. Moreover
x1 ∈ U and x2 ∈ V .
Let y1 = p(x1 ), y2 = p(x2 ) with y1 6= y2 . Assume U and V are open, disjoint and
saturated, with x1 ∈ U and x2 ∈ V . Let Ũ = f (U ), Ṽ = f (V ). Since U and V are
saturated, U = f −1 (Ũ ), V = f −1 (Ṽ ). Since U and V are open, and f a quotient
map, Ũ and Ṽ are open in Y . Since U and V are disjoint and saturated, Ũ and Ṽ
are disjoint. Since y1 ∈ Ũ and y2 ∈ Ṽ , it follows that Y is Hausdorff.
(b) Let X be a compact Hausdorff space, and A a closed subset. Let x ∈ X − A.
Show that there exist open sets U , V , with x ∈ U and A ⊂ V such that U ∩ V = ∅.
Solution: Since A is a closed subset of a compact space, A is compact. Since X also
is Hausdorff, the conclusion follows from Lemma 26.4 in the text book. (If you do
not want to refer to this lemma, you must reproduce arguments from the proof of
Theorem 26.3 (since the lemma follows from that proof)).
(c) Let X and Y be two topological spaces. Let X
`
Y denote the disjoint union of
X and Y (the union of X and Y where we regard each point of X as different from
`
`
any point of Y ). Let i : X → X Y and j : Y → X Y , denote the inclusion
maps. Let τX and τY be the topologies of X and Y respectively. Let τ be the
`
Y given by U ∈ τ if and only if i−1 (U ) ∈ τX and
`
−1
j (U ) ∈ τY . Explain why τ is a topology on X Y .
collection of subsets of X
`
`
Solution: Since i−1 (X Y ) = X, j −1 (X Y ) = Y and X resp. Y are open in X
`
resp. Y , X Y is a set in τ . Similarly, since i−1 (∅) = ∅ and j −1 (∅) = ∅, ∅ is a
S
S
set in τ . Let {Uk , | k ∈ I}, be a family of sets in τ . Then i−1 ( Uk ) = i−1 (Uk ),
k S
k
is a union of open sets in X, hence open in X. Similarly j −1 ( Uk ) is open in
k
S
Y . It follows that Uk is a set in τ . Let Ul , l = 1, . . . , n be sets in τ .Then
−1
i
(
n
T
l=1
Ul ) =
n
T
k
−1
i
l=1
X. Similarly j −1 (
(Ul ), is a finite intersection of open sets in X, hence open in
n
T
Ul ) is open in Y . This shows that
l=1
n
T
Ul is a set in τ and all
l=1
together we get that τ is a topology.
Let X and Y be topological spaces, let A be a non-empty subset of X and f : A → Y
be a continuous map . With the notation from (c) above, let ∼ be the equivalence
`
relation on X Y generated by i(a) ∼ j(f (a)) for a ∈ A. (So the equivalence
class of i(a) is equal i(f −1 ({f (a)})) ∪ {j(f (a))} and {i(x)}, {j(y)} are equivalence
`
classes when x ∈ X − A and y ∈ Y − f (A) respectively.) Give X Y the topology
described in (c), and denote the quotient space under the relation ∼ by X ∪ Y . Let
f
`
p : X Y → X ∪ Y denote the quotient map, and put ī = p ◦ i and j̄ = p ◦ j.
f
(d) Show that X ∪ Y is a connected space, when X and Y are connected spaces.
f
4
Solution: It is clear that ī and j̄ are continuous maps. So ī(X) and j̄(Y ) are
connected when X and Y are. Note that ī(X) ∩ j̄(Y ) = ī(A) 6= ∅. Therefore
ī(X) ∪ j̄(Y ) = X ∪ Y is connected.
f
Assume now that X and Y above are compact Hausdorff spaces and that A is a
closed subspace.
(e) Show that X ∪ Y is a compact space.
f
Solution: Since ī and j̄ are continuous maps, ī(X) and j̄(Y ) are compact sets when
X and Y are compact. Since a union of two compact sets is compact, ī(X) ∪ j̄(Y ) =
X ∪ Y is compact.
f
(f) Let x ∈ ī(X − A) and y ∈ j̄(Y ) Show that there exist disjoint open sets U and
V in X ∪ Y such that x ∈ U and y ∈ V . (Hint use (b) above.)
f
Solution: Let x = ī(x0 ). Use (b) to find open disjoint sets U 0 and V 0 in X such that
x0 ∈ U 0 and A ⊂ V 0 . Then i(U 0 ) and i(V 0 ) ∪ j(Y ) are open disjoint saturated sets i
`
X Y and U = ī(U 0 ) and V = ī(V 0 ) ∪ j̄(Y ) become open disjoint neighbourhoods
around x and y.
(g) Show that X ∪ Y is a Hausdorff space.
f
Let x and y, x 6= y be points in X ∪ Y . we must show that there exist open disjoint
f
neighbourhoods of x and y. If x ∈ ī(X −A) and y ∈ j̄(Y ) this follows from (f) above.
Note that since A is a closed subset of a compact space, A is compact. So, since f is
continuous, f (A) is compact, hence closed since Y is Hausdorff. If x ∈ j̄(Y − f (A))
and y ∈ ī(X), we can therefore argue as in (f) (interchangeing X and Y and A and
f (A) respectively in our arguments) and find disjoint open neighbourhoods around
x and y. If x and y are two points in ī(X − A), it follows since X is Hausdorff and
A is closed, that we can find disjoint neighbourhoods U 0 and V 0 of x and y such
that U 0 ∩ A = V 0 ∩ A = ∅. So i(U 0 ) and i(V 0 ) becomes open disjoint saturated
`
sets in X Y , and U = ī(U 0 ), V = ī(V 0 ) are disjoint neighbourhoods of x and y
respectively. The case when x and y are two points in j̄(X − f (A)) is similar. The
only (and hard) case which is left is when x and y are points in j̄(f (A)) = ī(A). In
this case let x0 and y 0 be points in f (A) such that x = j̄(x0 ), y = j̄(y 0 ). Since we
must have x0 6= y 0 and Y is Hausdorff, we can find open disjoint neighbourhoods
U 0 and V 0 around x0 and y 0 . Since Y is compact, therefore locally compact and is
Hausdorff, we may (using Theorem 29.2) choose these neighbourhoods such that U 0
and V 0 are compact and disjoint. Now i(f −1 (U 0 ))∪j(U 0 ) and i(f −1 (V 0 ))∪j(V 0 ) are
`
then disjoint and saturated in X Y , but these sets are not necessarily open, since
f −1 (U 0 ) and f −1 (V 0 ) are only open sets in A and not necessarily in X. To fix this we
may find open sets Ũ and Ṽ in X such that Ũ ∩ A = f −1 (U 0 ) and Ṽ ∩ A = f −1 (V 0 ).
However, Ũ and Ṽ may intersect in points outside A, and we need to modify these
sets so they become disjoint. Now, since U 0 and V 0 are compact they are also closed
sets (since Y is Hausdorff) . So f −1 (U 0 ) and f −1 (U 0 ) are closed disjoint subsets of
5
X which is a compact Hausdorff space. So, since X is normal, we can find disjoint
open sets U1 , V1 such that f −1 (U 0 ) ⊂ U1 , f −1 (V 0 ) ⊂ V1 . (Note that you do not
really need to use normality here, because f −1 (U 0 ) and f −1 (V 0 ) also are compact
sets and you can use arguments similar to those in the proof of Lemma 26.4 to find
such open neighbourhoods U1 and V1 .) Now, let U2 = U1 ∩ Ũ and V2 = V1 ∩ Ṽ . Then
U2 and V2 are open and disjoint in X with U2 ∩ A = f −1 (U 0 ) and V2 ∩ A = f −1 (V 0 ).
`
Now i(U2 ) ∪ j(U 0 ) and i(V2 ) ∪ j(V 0 ) are open disjoint and saturated sets in X Y ,
and ī(U2 ) ∪ j̄(U 0 ) and ī(V2 ) ∪ j̄(V 0 ) becomes disjoint neighbourhoods around x and
y respectively.
-End-.