Download Math F651: Take Home Midterm Solutions March 10, 2017 1. A

Math F651: Take Home Midterm Solutions
March 10, 2017
1. A subset A of a topological space X is said to be nowhere dense if Int A = ∅.
a) Let U be an open subset of a topological space. Prove that ∂U is closed and nowhere
dense.
b) Let V be a closed and nowhere dense set. Show that V is the boundary of an open set.
Solution, part a:
Let U be an open set. Its boundary is closed as it is the intersection of the closure of U with the
closure of its complement. So to show that the boundary of U is nowhere dense, it is enough to
show it has empty interior. Let x ∈ ∂U. Let V be an open set containing x. Since x ∈ U, V ∩ U
is nonempty. Note that V ∩ U is open, since U and V are, and hence is contained in the interior
of U. The interior of U and its boundary are disjoint, so V contains a point distinct from ∂U.
That is, ∂U has empty interior.
Solution, part b:
Suppose V is closed and nowhere dense. Let U = V c . I claim that V = ∂U. Certainly V ⊆ U c
since V ⊆ U c . Moreover, since V is nowhere dense, every point in V is a contact point of
V c = U. So V ⊆ ∂U. On the other hand, every point of U is an interior point of U (and
therefore not in ∂U), so ∂U ⊆ U c = V as well.
2. Let f and g be continuous maps from a topological space X to a Hausdorff space Y. Suppose
f = g on a dense subset of X. Prove that f = g.
Solution:
Consider the map F : X → Y × Y defined by
F(x) = ( f (x), g(x)).
The Characteristic Property of the Product Topology ensures that F is continuous. Since Y is
Hausdorff, the diagonal ∆ in Y × Y is closed. Moreover, the set {x : f (x) = g(x)} is exactly
F −1 (∆). Since ∆ is closed and F is continuous, F −1 (∆) is closed. By hypothesis it is dense in X,
and hence it is X itself.
3. Suppose q : X → Y is a quotient map and that each fibre of q is connected. Show that if Y is
connected, then so is X.
Solution:
Suppose to the contrary that {A, B} is a separation of X. Since each fibre f is a connected
set, it must be completely contained in one of A or B; otherwise A ∩ f and B ∩ f would be a
separation of f . So A and B are unions of fibres and are therefore saturated sets. Since they
form a separation, they are in fact saturated open sets. So p(A) and p(B) are open subsets of Y.
They are nonempty and disjoint, so Y is disconnected. This is a contradiction.
4. Lee 4-2 (and probably 4-1 as well)
Lemma A: If U ⊆ R is open and if x ∈ U, then U \ {x} is disconected.
Math F651: Take Home Midterm Solutions
March 10, 2017
Proof. Suppose x ∈ U. Since U is open, there exist points y− , y+ in U such that y− < x < y+ . Now
consider A− = (−∞, x) ∩ U and A+ (x, ∞) ∩ U. These are evidently disjoint open sets in U \ {x} and
their union is U \ {x}. The sets are nonempty since y± ∈ A± . Thus they form a disconnection of
U \ {x}.
Lemma B: If n ≥ 2, the Rn \ {0} is connected.
Proof. For each r > 0 let Ar be the union of the ray R = {(t, 0, . . . , 0) : t > 0} with the sphere of
radius r. A sphere is path connected since by stereographic projection it is a union of two sets that
are homeomorphic to Rn−1 , and these two sets must have a point in common (for otherwise S n−1
would consist of only two points, which is only true if n = 1). The ray R is path connected as it
is homeomorphic to R. Moreover, R intersects the sphere of radius r at the point (r, 0, . . . , 0) and
thus Ar is path connected. Finally, Rn \ {0} is the union of the path connected sets Ar , all of which
contain R, and hence Rn \ {0} is path connected.
Lemma C: Every nonempty open subset of R is not homeomorphic to an open subset of Rn for
some n > 1.
Proof. Suppose to the contrary that U ⊆ R is a nonempty open set homeomorphic to some open set
V in Rn for some n > 1. Since U is nonempty it contains an open set U 0 that is homeomorphic to
a ball in Rn , which homemomorphic to Rn itself. Deleting a point from U 0 we find from Lemma A
that U 0 is disconnected and from Lemma B that U 0 is path connected. This is a contradiction. Solution:
Suppose that M is nonempty and both a 1-manifold and an n-manifold for some n > 1. Let
x ∈ M. We can find neighbourhoods U and V of x that are homeomorphic to R1 and Rn
respectively. But then U ∩ V is open in U and open in V and is hence homeomorphic to open
subsets of R1 and Rn . But Lemma C shows that this is a contradiction.
5. Lee 4-11
Solution, part a:
Given x ∈ X, the set I x = {x} × I in X × I is homeomorphic to to I and is path connected.
Hence its image π(I x ) is path connected. The sets π(I x ) have a point in common, namely the
equivalence class X × {0}, and their union is all of CX. Thus CX is path connected.
Solution, part b:
We provide a proof in the locally conected case, the locally path connected case being essentially similar.
Let P ∈ CX be the equivalence class X × {0} in CX.
Lemma D: CX \ {P} is an open set and is homeomorphic to X × (0, 1].
Proof. Observe that π−1 (CX \ {P}) = X × (0, 1], which is open in X × [0, 1] and hence CX \ {P} is
open. Moreover, since X × (0, 1] is saturated and open, the restriction of π to this set is a quotient
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Math F651: Take Home Midterm Solutions
March 10, 2017
map onto CX \ {P}. But this quotient map makes no identifications, and hence by uniqueness of
quotients is a homeomorphism.
Lemma E: The point P admits a neighbourhood base of path connected sets, namely the set of
path connected open sets containing P.
Proof. Let U be an open set containing P and consider W = π−1 (U). For each x ∈ X we can find an
open set S x ⊆ X and an interval [0, r x ) such that (x, 0) ∈ S x × [0, r x ) ⊆ W. Let R = ∩ x∈X S x × [0, r x ),
which is an open saturated set containing X × {0}. Hence π(R) is an open set containing P and
contained in π(W) = U. Moreover, the sets π(S x ×[0, r x )) are path connected by the same argument
as part a). And the sets π(S x × [0, r x )) all contain the point P. Hence their union, namely π(R), is
path connected.
We have therefore shown that every open set containing P contains a path connected open set that
contains P, which completes the proof.
Now suppose X is locally connected. Since (0, 1] is locally connected (it has a basis of intervals)
so is X × (0, 1]; simply take as a basis the sets that are products of basic open sets in X with
basic open sets in (0, 1]. But then CX \ {P} is a locally connected open set in CX. Let B1 be
a basis of locally connected sets for CX \ {P} and observe that these sets are open in CX since
CX \ {P} is open in CX. Let B2 be the neighbourhood base at P of path connected sets given
by Lemma E. Let B = B1 ∪ B2 , and obseve that B is a collection of open sets in CX. We claim
that B is a basis for CX.
Indeed let U be open in CX. Then U \ {P} is a union of sets from B1 . And by Lemma E, if
P ∈ U then U contains an set from B2 as well, and this additional set contains the point P.
Hence U is a union of set from B.
Conversely, suppose CX is locally connected. Then so is the open subset CX \ {P}; simply take
as a basis the connected open sets that are contained in CX \ {P}. But this implies X × (0, 1] is
locally connected as well, as it is homeomorphic to CX \ {P}.
Let p : X ×(0, 1] → X be the natural projection. Let B be a basis of connected sets for X ×(0, 1],
and let B0 = {p(B) : B ∈ B}. Since p is an open map, B0 is a collection of open subsets of X.
Suppose x ∈ X and U is an open neighbourhood of x. Then p−1 (U) is open in X × (0, 1] and
contains (x, 1/2). Thus there is a connected basic open set B ∈ p−1 (U) that contains (x, 1/2).
But then p(B) ∈ B0 satisfies x ∈ p(B) ⊆ U. So B0 is a basis for X, and its elements are connected
as they are images of connected sets under p.
6. Consider the set Rω . Let `2 be the subset of Rω of square summable sequences.
That is,
p
(x1 , x2 , . . .) ∈ `2 if x12 + x22 + · · · is finite. We can put a metric on `2 by d(x, y) = (x1 − y1 )2 + · · ·
You do NOT need to verify that this is a metric. We take the topology on `2 to be the topology
generated by this metric.
Determine if the embedding of `2 ,→ Rω is continuous if Rω is given
a) the product topology, or
b) the box topology.
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Math F651: Take Home Midterm Solutions
March 10, 2017
The embedding into the product topology is continuous:
Let πk : Rω → R be the projection onto the kth factor. Let U ⊆ R be open and consider the
subbasic open set V = π−1
k (U). To show the embedding is continuous, it suffices to show that
`2 ∩ V is open in `2 .
Let x ∈ `2 ∩ V. Then xk ∈ U, and there is a number r such that (xk − r, xk + r) ⊆ U. Suppose
y ∈ Br (x). Then
X
(xn − yn )2 < r2
n
and in particular,
(xk − yk ) < r2
and
|xk − yk | < r
Hence yk ∈ U and therefore, y ∈ U ∩ `2 . Therefore Br (x) ⊆ V, and therefore V is open.
The embedding into the box topology is not continuous:
Consider the set U = (−1, 1) × (−1/2, 1/2) × (−1/3, 1/3) × · · · , which is open in Rω with the
box topology. To show the embedding is not continuous, it suffices to show that V = U ∩ `2 is
not open in `2 . Since 0 ∈ V, it is enough to show that for no radius r is Br (0) ⊆ V.
Indeed, let r > 0. Pick n ∈ N such that 1/n < r/2. Define x ∈ `2 by



k,n
0
xk = 

r/2 k = n.
Then d(x, 0) = r/2 < r, so x ∈ Br (0). But xn = r/2 > 1/n so xn < V.
7. Lee 3-18
Lemma F: Suppose A ⊆ R and A ∩ [n, n + 1] is closed for every n ∈ Z. Then A is closed.
Proof. Let x be a contact point for A and pick n such that x ∈ [n, n + 1]. Each interval (x − , x + )
for 0 < < 1 contains a point from A and hence a point from A ∩ [n − 1, n + 2]. That is, x is a
contact point of A ∩ [n − 1, n + 2] = ∪ j = −11 A ∩ [n − j, n − j + 1], which is a finite union of closed
sets and is closed. Thus x ∈ A and A is closed.
Solution, part a:
For each n ∈ N let In = [0, 1] and let S n = S 1 . Define fn : In → S n by fn (x) = e2πix , which is
a quotient map. By the characteristic properties of the disjoint union and subspace topologies,
there is a continuous map
f : tIn → tS n
whose restriction to each In is fn . By the closed map lemma, each fn is a closed map. But then
if A is closed in tIn then S n ∩ f (A) = S n ∩ (∪ f (A ∩ In )) = fn (A ∩ In ) which is contained in and
closed in S n . Thus f is also a closed map. Since it is surjective, it is a quotient map, and we can
V
V
follow up with the quotient map from tS n to S n to obtain a quotient map fˆ : tIn → S n
that identifies all of the endpoints of the invervals In and makes no further identifications.
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Math F651: Take Home Midterm Solutions
March 10, 2017
On the other hand, define g : tIn → R by g|In (x) = x + n. Evidently each gn = g|In is continuous,
and by the Glueing Lemma, g is as well. Moreover, by the Closed Map Lemma, each gn is a
closed map. If A ⊆ tIn is closed then each [n, n + 1] ∩ g(A) = gn (A) is closed. By Lemma
F, g(A) is closed as well. We can compose g with the quotient map from R to R/A to obtain a
quotient map ĝ : tIn → R/A. Observe that ĝ identifies all the endpoints of the intervals In and
V
performs no other identifications. That is, ĝ makes the same identifications as fˆ. Hence S n
and R/A are homeomorphic.
Solution, part b:
Let π : R → R/A be the projection and for convenience let p = π(0). Suppose to the contrary
−1
that p admits a countable neighbourhood base V = {Vk }∞
k=1 . For each k, π (Vk ) is an open
set containing A and hence contains an interval (k − k , k + k ) where 0 < k < 1/2. Let
W = ∪(k − k /2, k + k /2). Observe that W is saturated and open and hence R = π(W) is an open
set containing p. Moreover, each xk = k + k /3 satisfies xk ∈ Vk and xk < W (this uses the fact
that k < 1/2). Since W is saturated, it follows that π(xk ) < R. But π(xk ) ∈ Vk and hence Vk * R.
This is true for all k, which contradicts the hypothesis that V is a neighbourhood base at p.
8. Let G be an algebraic group. We say that G is a topological group if in additional G is a T 1
topological space such that that the multiplication map m : G × G → G and the inversion map
i : G → G defined by m(g, h) = g · h and i(g) = g−1 are continuous.
a) Suppose G is an algebraic group and a T 1 topological space. Show that G is a topological
group if and only if the map f : G × G → G defined by f (g, h) = gh−1 is continuous.
b) Let G be a topological group and let H be a subgroup. Show that H is a subgroup.
Solution, part a:
Suppose m and i are continuous. Then f = m ◦ (id × i). Then id × i is continuous as each of id
and i are; this is a consequence of a homework problem. So f is a composition of continuous
functions and is continuous.
Now suppose f is continuous. Define s : G → G × G by s(g) = (e, g) where e is the group
identity. Then s is continuous as its coordinate functions (a constant function and the identity
respectively) are. Notice that i = f ◦ s. So i is a composition of continuous maps and is
continuous. Define k : G × G → G × G by k = id × i. Then k is continuous as id and i are. Then
m = f ◦ k, so m is a composition of continuous maps and is continuous.
Solution, part b:
To show that H is a subgroup, it is enough to show that a · b−1 belongs to H whenever a and
b do. Notice that A = f −1 (H) is closed since f is continuous and H is closed. Moreover, A
contains H × H, for if g, h ∈ H, then g · h−1 ∈ H ⊆ H. So A is a closed set containing H × H
and therefore also contains
H × H = H × H.
That is, if g, h ∈ H, then f (g, h) = g · h−1 ∈ H.
9. Suppose X is a space and Y is compact. Show that the projection πX : X × Y → X is a closed
map.
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Math F651: Take Home Midterm Solutions
March 10, 2017
Solution:
Suppose that A ⊆ X × Y is closed. Suppose x ∈ πX (A)c . Then the fibre π−1
X ({x}) is contained in
c
the open set A . Since Y is compact the Tube Lemma implies that there is an open set V in X
c
c
c
containing x such that π−1
X (V) ⊆ A as well. Hence V is contained in π(A) as well. Thus π(A)
is open and π(A) is closed.
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