TOPOLOGY ASSIGNMENT 7
PRODUCT SPACES. METRIC TOPOLOGIES.
IGOR WIGMAN
(1) (a) The only nontrivial bit is proving the triangle inequality. Let x, y, z ∈ X, and
˜ y) + d(y,
˜ z) ≥ x,˜z. This follows from the fact that d is a
we are to prove d(x,
metric, unless one of the numbers d(x, y), d(y, z), d(x, z) is > 1. If d(x, z) > 1, it
˜ y) + d(y,
˜ z) ≥ 1 = d(x,
˜ z).
forces d(x, y) + d(y, z) ≥ d(x, z) > 1, and then also d(x,
˜
˜
˜
Otherwise, if d(x, z) ≤ 1 and d(x, y) > 1 (say), then d(x, y) + d(y, z) ≥ 1 ≥ d(x, z).
(b) Let B 0 = {B(x, r) : x ∈ X, 0 < r ≤ 1} ⊆ B {B(x, r) : x ∈ X}. We claim that B
is a basis for the topology on X. To see that it is sufficient to show that the
topology generated by B 0 is finer than the topology generated by B, i.e. for every
x, r > 0, there exists r0 < 1, so that B(x, r0 ) ⊆ B(x, r), which is obvious by
choosing r0 = min(r, 1/2) (say). Now that we know that, we notice that Bd˜(x, r) =
Bd (x, min(r, 1)) ∈ B 0 . (2) (a) It is evident that B 0 = Bd (x, n1 ) : x ∈ X, n ∈ N is a topology basis indeed; since
B 0 is a subset of the standard basis B of all balls for the topology on X, it is
sufficient to show that the topology generated by B 0 is finer than T ; to this end,
given x ∈ X, r > 0, it is sufficient to choose n > 1r , so that B(x, n1 ) ⊆ B(x, r).
For each x ∈ X the countable family Bd (x, n1 ) : n ∈ N satisfies the postulated
property a first countable topology: every neigbourhood U of x necessarily contains
the balls Bd (x, n1 ) for n sufficiently large. Hence it is satisfied by an arbitrary metric
space.
(b) If A is a set and {xn 6= x} is a sequence convergent to x, then x ∈ A0 always
(regardless on whether X is, in fact, metric). Conversely, let x ∈ A0 and Un a
sequence of neighbourhoods of x satisfying the defining property of a first countable
topology. With no loss of generality we may assume that {Un } is decreasing sequence
of neighbourhoods in the sense that for every n, Un+1 ⊆ Un (by passing to Un0 =
n
T
Uk ) We then choose xn ∈ Un \ {x} arbitrary. Let U be any neighbourhood of
k=1
x, then, since Un is decreasing, necessarily Un ⊆ U provided that n is sufficiently
large. Since U is arbitrary, it means that xn → x as required.
(c) This follows by choosing the neighbourhoods of the postulated limit x to be of the
form B(x, ) for > 0.
1
(d) The choice B = {B(q, m
) : q ∈ Qn , m ≥ 1} will do the job.
(3) Assume otherwise, i.e. that there exists such a metric d. By Q2, any neighbourhood of
x0 = (0)i∈N contains some ball Bn = {x : d(x, x0 ) < n1 } for Q
n sufficiently large. The ball
0
0
0
Bn in its turn contains a basic open neighbourhood U =
Un;m
⊆ B, where Un;m
is
m∈N
0
a neighbourhood of 0 in R. Let 0 ∈QUn;n ( Un;n
be a neighbourhood of 0 in R, properly
0
contained in Un;n , and define U =
Un;n , a neighbourhood (in the box topology of x0 .
n∈N
0
Then for each n, U cannot contain Bn , since πn U = Un;n ( Un;n
⊆ πn Bn .
Our proof is a version of Cantor’s diagonal method; in fact, it gives the stronger
statement that Rω is not first countable.
(4) First we need to prove that di are metrics on X. While for d1 it follows directly from di
t
being metrics, it is less clear for d0 . To see that we work with the function f (x) = 1+t
for
t = di (x, y) ≥ 0; f (x) is continuous, convex and increasing on this domain (by evaluating
f 0 and f 00 , say), thus it is a bijection, satisfying f (t + u) ≤ f (t) + f (u). It then follows
that for every i ∈ N , f (di (x, y)) + f (di (y, z)) ≥ f (di (x, y) + di (y, z)) ≥ f (di (x, z)), and
hence each of the summands in the series satisfies the triangle inequality, the other metric
properties being trivial.
1
2
ASSIGNMENT 7
Now to see that d0 generates the product topology (say), let x0 = (x0i ) ∈ X and
k
∞
Q
Q
R be a basic open neighbourhood of x0 . Let i > 0, i = 1 . . . k be
U =
Ui ×
i=1
i=k+1
numbers satisfying (x0i − i , x0i + i ) ⊆ Ui . Since f (t) is continuous, we may choose
i)
. If x satisfies d0 (x, x0 ) < δ, then for all i ≤ k, it implies
δ > 0 so that f (δ) < min f (
2i
i≤k
f (di (xi ,x0i ))
2i
< δ, so that f (di (xi , x0i )) < 2i δ, and thus di (xi , x0i ) < i by the construction
of δ. Therefore, Bd0 (x0 , δ) ⊆ U , and since U is arbitrary, we conclude that the metric
topology on X generated by d0 is finer than the product topology on X.
Conversely, let x0 ∈ X, r > 0 and B = B(x0 , r). We are to squeeze a basic neighbourhood of x0 inside B. We proceed like in the proof of the corresponding theorem in the lecK
Q
Q
tures. Assume 0 < r < 1 and set K = −blog( 4r )c. We let U =
B(x0i , f −1 (r))×
Xi ,
i=1
i>K
a basic open neighbourhood of x0 . Evaluating d(x, x0 ) for x ∈ U using similar analysis
as presented in the lectures, we obtain d(x, x0 ) < r, so that necessarily U ⊆ B.
The analysis for d1 is similar (but somewhat easier).
(5) We claim that T2 is strictly finer than T3 , which is strictly finer than T1 . In fact, T1 is
the trivial topology on Y , since if f differs from g in finitely many points, then f ∼ g,
i.e. f represent the same point as g, so that if
U=
Y
R×
i6=i1 ,...,in
n
Y
Uij ,
j=1
then every f ∈ L2 ([0, 1]) satisfies that f ∼ g for some g ∈ U and then U = Y ; it then
only remains to compare the other two topologies.
Now we show that T2 is finer than T3 . Let f0 ∈ L2 ([0, 1]), and let B = Bd (f, r) be a
basic open set in Y ; we are to squeeze
a basic open set in the box topology inside B.
Q
(f0 (x) − r/2, f0 (x) + r/2) ∩ L2 ([0, 1]); for f ∈ U
We may take, for example U =
x∈[0,1]
we have d(f, f0 ) ≤
1
R
0
r2
4 dx
1/2
=
r
2
< r, so that indeed U ⊆ B. On the other hand,
T2 and T3 are not equal, since the set {f ∈ Y : ∀x.|f (x)| < 21 } ∈ T2 \ T3 is open in the
box topology but not in the L2 one, since for every > 0 we may take a (continuous)
function f ∈ Y , so that d(f, 0) < , but f (0) > 21 (say).
(6) (a) Requires a picture.
(b) ((0, 1] ∪ [2, 3)/(1 ∼ 2).
It is homeomorphic to the open interval (0, 2).
(c) This is homeomorphic to a sphere.
(d) Homeomorphic to a sphere.
(e) *
See http://en.wikipedia.org/wiki/Klein_bottle
(7) (a) Choose a space as hinted (or otherwise): X = R × {0, 1} ⊆ R2 , where R is equipped
with the standard topology, and {0, 1} is discrete; X may be viewed as two nonintersecting lines in R2 . The space X is Hausdorff as a subspace of a Hausdorff space.
Let ∼ be the equivalence relation: ∀x 6= 0, (x, 0) ∼ (x, 1) (making it reflexive and
symmetric in the usual way). Let Y = X/ ∼ and consider the two distinct points
z1 = [(0, 0)], z2 = [(0, 1)] (where [P ] is the equivalence class of P ∈ X under ∼, i.e.
the image of P under the quotient map). Then any neighbourhoods of z1 , z2 in Y
will necessarily intersect, so that Y is not Hausdorff.
(b) Assume first that Y = X/ ∼ is Hausdorff. We are then to prove that the complement
U = {(x, y) ∈ X × X : x y} is open in X × X. Let (x, y) ∈ U , i.e. x y.
Since p(x) = [x] 6= p(y) = [y] and Y is Hausdorff, we may separate [x], [y] in Y ,
i.e. there exist disjoint open neighbourhoods Ux0 , Uy0 ⊆ Y of x, y respectively. Let
Ux = p−1 (Ux0 ), Uy = p−1 (Uy0 ) ⊆ X be open neighbourhoods of x, y; they are disjoint
(recall that p is continuous by the construction of quotient topology). Moreover,
(x, y) ∈ Ux × Uy ⊆ U since otherwise its image under p would contain the diagonal
TOPOLOGY ASSIGNMENT 7
PRODUCT SPACES. METRIC TOPOLOGIES.
3
which would contradict Ux0 , Uy0 disjoint. Note that we didn’t use the fact that p is
open for this direction.
Conversely, assume that the set {(x, y) ∈ X × X : x ∼ y} is closed, i.e. its complement U is open. Since p is open, (p × p)(U ) := (p(x), p(y) : (x, y) ∈ U ) is open
in Y × Y (to see that p × p is open, it is sufficient to see that on the basic open
sets U1 × U2 in the product topology on X × X, which is then clear); note that its
complement V = Y × Y \ p(U ) = {(y, y) : y ∈ Y } is the diagonal; it is closed in
Y ×Y (as a complement of an open). Thus by Assignment 5, Q3(c), Y is necessarily
Hausdorff.
(c) Define f in the following way: for P = [(x, y)] = p(x, y) let f (P ) := F (x, y) =
x2 + y 2 ; f is well defined since F is constant along the equivalent classes of ∼, i.e. if
(x1 , y1 ) ∼ (x2 , y2 , then necessarily F (x1 , y1 ) = F (x2 , y2 ); by construction F = f ◦ p.
It remains to show that f is a homeomorphism. First, it is clear that f is bijective
(by interpreting Y = X/ ∼ as the radius of the centred circle containing (x, y) ∈ X).
An open basic set of [0, ∞) is either of the form (a, b) with 0 < a < b < ∞ or [0, b)
with b > 0. The corresponding inverse image is f −1 ((a, b)) or f −1 ([0, b)); to check
that the latter is open, by the definition of the quotient topology we are to check
that P −1 (f −1 ((a, b))) = F −1 ((a, b)) or p−1 (f −1 ([0, b))) = F −1 ([0, b)), which is an
open annulus for the former or an open disc for the latter case.
Finally, to show that f is open, we note that if U ⊆ Y is open, then, by the definition,
p−1 (U ) ⊆ X is necessarily open, and rotation invariant; it is then a collection of open
annuli and perhaps a centred discs, with no loss of generality, either a single open
annulus or an open disc, whence (as both p and F are surjective), f (U ) = F (p−1 (U ))
is of either form (a, b) or [0, b), open in [0, ∞).