Download Math F651: Homework 8 Due: March 29, 2017 Several of the

Math F651: Homework 8
Due: March 29, 2017
Several of the problems on this assignment are repeats. Use nets!
1. (Solution by Alayna Rupert)
Show that a topological space in X is Hausdorff if and only if every convergent net in X
has exactly one limit.
2. (Solution by Aven Bross)
Consider the product space X × Y. Find (and prove) a condition in terms of coordinate
functions that characterizes convergence of nets in the product. Does your condition
also work for an arbitrary product?
Solution:
Let ⟨z α ⟩α∈A be a net in X × Y. I claim z α → z ∈ X × Y if and only if π X (z α ) → π X (z) and
πY (z α ) → πY (z).
Suppose z α → z. Then since π X and πY are continuous π X (z α ) → π X (z) and πY (z α ) →
πY (z).
Conversely suppose π X (z α ) → π X (z) and πY (z α ) → πY (z). Suppose K ⊆ X × Y is an
open set such that z ∈ K. There must exist U ⊆ X, V ⊆ Y such that π X (z) ∈ U, πY (z) ∈ V ,
and U × V ⊆ K.
Since π X (z α ) → π X (z) there exists α X ∈ A such that π X (z α ) ∈ U for all α ≥ α X . Similarly,
since πY (z α ) → πY (z), there must exist αY ∈ A such that πY (z α ) ∈ V for all α ≥ αY .
Since A is a directed set there must exist α0 ∈ A such that α0 ≥ α X and α0 ≥ αY . Thus for
all α ≥ α0 we have π X (z α ) ∈ U and πY (z α ) ∈ V . Thus z α ∈ U × V ⊆ K for all α ≥ α0 . So
z α → z.
Now suppose X = ∏β∈B X β is an arbitrary product of topological spaces and ⟨z α ⟩α∈A is a
net in X. Again I claim z α → z ∈ X if and only if π X β (z α ) → π X α (z) for all β ∈ B.
Suppose z α → z. Just as in the case of the product of two spaces above, π X β is continuous
for each β ∈ B and thus π X β (z α ) → π X β (z).
Now suppose π X β (z α ) → π X β (z) for each β ∈ B. Suppose U ⊆ X is an open set containing
z. By the definition of the product topology there must exist a basic open set ∏β∈B U β ⊆ U
such that z ∈ ∏β∈B U β , each U β ⊆ X β an open set, and U β = X β for all but the finite set
{β1 , . . . , β n } ⊆ B.
For each i ∈ {1, . . . , n} there exists α i ∈ A such that for all all α ≥ α i we have π X β i (z α ) ∈
U β i . By induction may find α0 ∈ A such that α0 ≥ α i for all i ∈ {1, . . . , n}.
Let β ∈ B. For all α ≥ α0 we have π X β (z α ) ∈ U β . Thus z α ∈ ∏β∈B U β ⊆ U. So z α → z.
3. (Solution by Jody Gaines)
Show that a space X is Hausdorff if and only if the diagonal in X × X is closed.
Lemma 1: Let A and B be directed sets. Then the set A × B with the relation ≥, where
Math F651: Homework 8
Due: March 29, 2017
(α2 , β2 ) ≥ (α1 , β1 ) if α2 ≥ α1 and β2 ≥ β1 , is a directed set.
Proof. Note for all (α, β) ∈ A × B that α ≥ α and β ≥ β, so (α, β) ≥ (α, β).
Let (α1 , β1 ), (α2 , β2 ), (α3 , β3 ) ∈ A × B. Suppose (α1 , β1 ) ≥ (α2 , β2 ) and (α2 , β2 ) ≥ (α1 , β1 ).
Then α1 ≥ α2 and α2 ≥ α1 , which implies α1 = α2 . Since β1 ≥ β2 and β2 ≥ β1 as well then
β1 = β2 . Thus (α1 , β1 ) = (α2 , β2 ).
Now suppose (α3 , β3 ) ≥ (α2 , β2 ) and (α2 , β2 ) ≥ (α1 , β1 ). Then α3 ≥ α2 and α2 ≥ α1 , which
implies α3 ≥ α1 . Likewise, β3 ≥ β2 and β2 ≥ β1 , which implies β3 ≥ β1 . Thus (α3 , β3 ) ≥
(α1 , β1 ).
Since A is a directed set then there exists γ ∈ A such that γ ≥ α2 and γ ≥ α1 . However B is a
directed set as well, so there exists δ ∈ B such that δ ≥ β2 and δ ≥ β1 . Thus (γ, δ) ≥ (α2 , β2 )
and (γ, δ) ≥ (α1 , β1 ). Hence A × B is a directed set.
4. (Solution by Jody Gaines)
Let G be a topological group and let H be a subgroup. Show that H is a subgroup.
Solution:
Let a, b ∈ H. By the one-step subgroup test it is sufficient to show ab −1 ∈ H. Note a and b
are contact points of H, so there exist nets ⟨a α ⟩α∈A and ⟨b β ⟩β∈B in H such that a α → a and
b β → b. Since H is a subgroup then for all α ∈ A and β ∈ B, a α b −1
β ∈ H. So consider the set
−1
⟨a α b β ⟩(α,β)∈A×B in H where, for each (α1 , β1 ), (α2 , β2 ) ∈ A×B, (α2 , β2 ) ≥ (α1 , β1 ) if α2 ≥ α1
and β2 ≥ β1 . By Lemma 1 A× B is a directed set, so ⟨a α b −1
β ⟩(α,β)∈A×B a net in H. Since G is a
topological group then the maps m ∶ G × G → G and i ∶ G → G defined by m(g, h) = g ⋅ h
−1
and i(g) = g −1 are continuous. So b β → b implies b −1
β = i(b β ) → i(b) = b . Thus by
−1
−1
−1
−1
−1
Problem 2, a α → a and b −1
β → b imply a α b β = m(a α , b β ) → m(a, b ) = ab . Hence
ab −1 is a contact point of H, which implies ab −1 ∈ H.
5. (Solution by Erika Burr)
Suppose X is a space and Y is compact and Hausdorff.
a) Use nets to show that a function f ∶ X → Y is continuous if and only if the graph
of f is closed. The graph of f is G f = {(x, f (x)) ∶ x ∈ X}.
b) Show that if G f is closed, then f ∶ X → Y is continuous. (Without nets. Delayed
from HW7.)
Solution, part a:
Lemma 3: Let ⟨x α ⟩α∈A be a net in a space X. If every subnet of ⟨x α ⟩α∈A has a (sub-)subnet
that converges to x ∈ X, then x α → x.
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Math F651: Homework 8
Due: March 29, 2017
Proof of Lemma ??. We prove the contrapositive. Suppose x α →
/ x. Then there exists U ∈
V(x) that does not contain a tail of ⟨x α ⟩α∈A . That is, for all α ∈ A there exists α ′ ∈ A with
α ′ ≥ α such that x α′ ∉ U . Let S = {α ∈ A ∶ x α ∉ U}. We claim that S, with the ordering
inherited from A, is a directed set. Note that for any α1 , α2 ∈ S there exists β ∈ A such that
β ≥ α1 and β ≥ α2 . Since β ∈ A, there exists β ′ ∈ A such that β′ ≥ β and x β′ ∉ U . Then β′ ∈ S.
Now we construct a subnet using the inclusion map ı ∶ S → A. Note that ı is increasing
as a consequence of the partial ordering on A. The map is cofinal since x α →
/ x. That is,
′
′
′
given α ∈ A, there exists α ∈ S such that ı(α ) = α ≥ α. Now consider the subnet ⟨x α ⟩α∈S .
By our definition of S, the open set U contains no tail of ⟨x α ⟩α∈S and thus ⟨x α ⟩α∈S has no
(sub-)subnet converging to x.
Proof of (a). Suppose f is continuous. Let ⟨(x α , f (x α ))⟩α∈A be a net in G f that converges to
(x, y) ∈ X × Y . Then x α → x in X and f (x α ) → y in Y by Problem 2. Since f is continuous,
we also have f (x α ) → f (x). Since Y is Hausdorff, by Problem 1, we know that f (x) = y.
Then (x, y) = (x, f (x)) ∈ G f . Hence G f contains its contact points and G f is closed.
Conversely, suppose that G f is closed. We will to show that f takes convergent nets to convergent nets. Let ⟨x α ⟩α∈ A be a net in X such that x α → x in X. Consider the net ⟨ f (x α )⟩α∈A
in Y and let ⟨ f (x α β )⟩β∈B be a subnet. Since Y is compact, every net in Y admits a convergent
subnet. Let ⟨ f (x α β γ )⟩γ∈C be a convergent (sub-)subnet of ⟨ f (x α β )⟩β∈B with limit y ∈ Y . Then
⟨(x α βγ , f (x α β γ ))⟩γ∈C is a net in G f that converges to (x, y) in X × Y . Since G f is closed, we
must have (x, y) ∈ G f . Then we have y = f (x). Since every subnet of the net ⟨ f (x α )⟩α∈A has
a (sub-)subnet that converges to f (x), by Lemma ??, f (x α ) → f (x). Hence f is continuous.
Solution, part b:
Suppose G f is closed in X × Y . Let V ⊆ Y be closed and consider f −1 (V ). Since Y is
compact, we know that π X ∶ X × Y → X is closed (P9 Midterm). We claim that f −1 (V ) =
π X ((X × V ) ∩ G f ). Observe that x ∈ f −1 (V ) if and only if f (x) ∈ V , which holds if and
only if (x, f (x)) ∈ X × V . Now by definition, (x, f (x)) ∈ G f so we have (x, f (x)) ∈
(X × V ) ∩ G f which holds if and only if x ∈ π X ((X × V ) ∩ G f ). Note that X × V is closed
in X × Y and thus the intersection of closed sets (X × V ) ∩ G f is closed. Since π X is a
closed map, then π X ((X × V ) ∩ G f ) = f −1 (V ) is closed in X. Hence f is continuous.
6. (Solution by David Maxwell)
Show that the homeomorphism group of a connected manifold acts transitively. In other
words, show that if M is a connected manifold, then for any two points p and q in M
there is a homeomorphism ψ ∶ M → M such that ψ(p) = q.
Solution:
Let G be the homeomorphism group of M. For each x ∈ M, let O x = {y ∶ y = f (x)for some f ∈ G},
so O x is the orbit of x.
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Math F651: Homework 8
Due: March 29, 2017
Lemma 5A: Let M be a topological n-manifold. Given any x ∈ M there is an open set V
containing x such that V ⊆ O x .
Proof. Let U be an open set containing x that is homeomorphic to Rn , and let Φ ∶ U → Rn
be a homeomorphism taking x to 0. Let V = Φ−1 (B1 (0)) and let W = Φ−1 (Bn ). Since V is
open in U and since U is open in M, we see that V is open in M. Since Bn is compact in Rn ,
W is compact in U, and by inclusion also in M. Since M is Hausdorff we conclude that W
is closed in M.
Let z ∈ V , and let f ∶ Bn → Bn be a homeomorphism that fixes the boundary and takes 0 to
Φ(z). Define F ∶ W → W by F = Φ−1 ○ f ○ Φ. Then F(x) = Φ−1 ( f (0)) = Φ−1 (Φ(z)) = z.
Let G be the identity map of the closed set M ∖ V . Notice that (M ∖ V ) ∩ W = W ∖ V
and that F is the identity on W ∖ V . Since M = W ∪ (M ∖ V ), the pasting lemma then
applies to construct a continuous map G ∶ M → M that is the identity on M ∖ V and takes
x to z. Moreover, G has a continuous inverse that is constructed in the same way, using f −1
rather than f . So G is a homeomorphism taking x to z. Since z ∈ V is arbitrary, the result is
proved.
Corollary 5B: For any x ∈ M, O x is open.
Proof. Let y ∈ O x , so O y = O x . By the previous lemma, there is an open set V with y ∈ V ⊆
O y = O x . Since y ∈ O x is arbitrary, O x is open.
We now turn to the main solution. Fix p ∈ M. To show that G acts transitively, it suffices
to show that the orbit O p = X. Since X is connected, to show that O p = X, it suffices to
show that O p is both open and closed. We have just shown that O p is open. Moreover,
O cp is a union of (open) orbits and is hence open. So O p is closed.
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