Math 131 2.

Second, if x = a × 0, but a 6= 0, then {(a − 1/n ×

Solution Set 4 0, a × 1/n) | n > 1a } forms a countable basis. Since
Dustin Cartwright b = 0, c must be less than a. Thus there is some n
such that a − 1/n > c. And as in the first case, either
21 2 Let X and Y be metric spaces with metrics dX and dY , e > a and any basis element works or e = a and there
respectively. Let f : X → Y have the property that for is some n such that 1/n < f . Take the larger of the
every pair of points x1 , x2 of X, two ns.
dY ( f (x1 ), f (x2 )) = dX (x1 , x2 ) 3. Third, if x = 0 × 0, then {[0 × 0, 0 × 1/n) | n > 1}
forms a basis. In this case the basis element around
Show that f is an imbedding. It is called an isometric
x is B = [0 × 0, e × f ). If e = 0, then there is n big
imbedding of X in Y . (5 points)
enough such that 1/n < f . If e 6= 0, then any n gives
First, f is injective, because if f (x1 ) = f (x2 ), then a neighborhood contained in B.
dX (x1 , x2 ) = dY ( f (x1 ), f (x2 )) = 0. Thus, by the proper- 4. The final two cases are when x = a × 1 and either
ties of the metric, x1 must be the same as x2 . This shows a 6= 1 or a = 1. These are similar to the third and four
that f is a bijective map onto its image f (X). cases respectively.
Second, this bijection is in fact a homeomorphism. Let
U be an open set in X, and y = f (x) any point in f (U) ⊂ 21 8 Let X be a topological space and let Y be a metric
f (X) ⊂ Y . Then there exists an ε such that x ∈ BX (x, ε) ⊂ space. Let fn : X → Y be a sequence of continuous
U. I claim that B f (X) (y, ε) ⊂ U. For any y0 = f (x0 ) ∈ functions. Let xn be a sequence of points of X con-
B f (X) (y, ε), verging to x. Show that if the sequence ( fn ) converges
uniformly to f , then ( fn (xn )) converges to f (x). (10
dY (y, y0 ) = dY ( f (x), f (x0 )) = dX (x, x0 ) < ε points)
so x0 is in the ε-ball around x, and thus it is in U. Thus, Let ε be any positive real number. Since ( fn ) con-
f (x0 ) = y0 is in f (U), so B f (X) (y, ε). This means that f (U) verges uniformly, f is continuous. So we can pick a
is open in f (X). δ > 0 such that for all x0 ∈ B(x, δ ), f (x0 ) ∈ B(x, ε/2). Fur-
Note that f −1 : f (X) → X is also an isometric imbed- thermore, since (xn ) converges to x, we can pick an N1
ding, so by the same reasoning f −1 takes open sets to open such that xn ∈ B(x, δ ) for all n > N1 . On the other hand,
sets. Therefore f is an imbedding. since fn converges uniformly, we can find an N2 such that
d( fn (x), f (x)) < ε/2 for all x ∈ X, and n > N2 . Now let N
21 4 Show that R` and the ordered square satisfy the first be the larger of N1 , N2 . For all n > N,
countability axiom. (10 points)
d( fn (xn ), f (xn )) < ε/2 because n > N2
Let x be a point in R` . A countable basis for x is {[x, x + d( f (xn ), f (x)) < ε/2 because n > N1
1/n) | n ∈ Z+ }. If U is any neighborhood of R` , then there
is a basis element [a, b) contained in U such that a ≤ x, So, by the triangle inequality, d( fn (xn ), f (x)) ≤ ε. There-
b > x. Pick n large enough such that 1/n < b, and then fore ( fn (xn )) converges to f (x).
x ∈ [x, 1/n) ⊂ [a, b) ⊂ U. 23 1 Let T and T 0 be two topologies on X. if T 0 ⊃ T ,
Let x = a × b be a point in the ordered square I02 , and what does connectedness of X in one topology imply
B = (c × d, e × f ) a basis element around x. There are five about connectedness in the other? (5 points)
cases to consider:
If X is connected in T 0 , then X is connected in T .
1. First, if b does not equal 0 or 1, then the set I’ll prove the contrapositive. Suppose X is disconnected
1 1 with topology T . Then there exist U, V open, non-empty,
{a × (b − 1/n, b + 1/n) | n > max( , )} which partition X in T . But U, V , are also open in T 0 , so
1−b b
they form a separation in T 0 , which is therefore discon-
forms a countable basis. If c < a and e > a, then any nected.
of these elements is contained in B. If c = a, then
d < b, so some n is big enough that d < b − 1/n. 23 5 A space is totally disconnected if its only connected
Similarly, for e = a, f > b, so there is some n such subspaces are one-point sets. Show that if X has
that f > b + 1/n. If both of these conditions apply, the discrete topology, then X is totally disconnected.
just take an n larger than either one. Does the converse hold? (5 points)

1
 Let A be any subset of X, which consists of more than 24 1(b) Suppose that there exist imbeddings f : X → Y and
one point. If x, y are any two distinct points in A, then {x} g : Y → X. Show by means of an example that X and
and A − {x} 3 y are open in both X and A. Since they are Y need not be homeomorphic. (5 points)
non-empty and partition A, A is not connected.
The converse does not hold. There are many examples: Let X = [0, 1] and Y = (0, 1). Let f be the map which
sends x to x/2 + 1/4, and thus imbeds X in Y as f (X) =
1. Q or any interval (a, b) or [a, b] in Q [ 41 , 34 ] ⊂ (0, 1). Let g be the the inclusion map. These are
2. {1/n | n ∈ Z+ } ∪ {0} each imbeddings, but X is not homeomorphic to Y from
part (a).
3. The Cantor set
24 1(c) Show Rn and R are not homeomorphic if n > 1. (5
All of these are completely disconnected, but have topolo- points)
gies distinct from the discrete topology.
Take any point in x ∈ R, then R − {x} = (−∞, x) ∪
23 7 Is the space R` connected? Justify your answer. (5 (x, ∞), which is disconnected. On the other hand, for
points) x = 0, Rn − {x} = Rn − {0}, and this is connected by Ex-
No. The sets U = (−∞, 0) = n∈Z+ [−n, −n + 1) and
S ample 4 on page 156.
V = [0, ∞) = n∈Z+ [n − 1, n) are open (since they are the
S
24 8(a) Is a product of path-connected spaces necessarily
union of basis elements) and partition R` .
path connected? (5 points)
23 8 Determine whether or not Rω is connected in the uni-
Yes. Suppose X = ∏α∈J Xα , which each Xα is path con-
form topology. (5 points)
nected. Let x = (xα )α∈J , y = (yα )α∈J be two points in X.
It is not connected. I claim the sets B, the set of bounded By the path connectedness of the Xα , there exist continu-
sequences and U, the set of unbounded sequences are both ous functions pα : [0, 1] → Xα such that pα (0) = xα and
open and thus form a separation. Suppose x = (x1 , x2 , . . .) pα (1) = yα . Define p : [0, 1] → X to be the product of
is a bounded sequence and y = (y1 , y2 , . . .) is an un- all this functions, which is continuous by Theorem 19.6,
bounded sequence. Then for all n, |xn | < R for some bound and by construction, p(0) = x, p(1) = y. Thus, X is path
R. On the other hand, y is unbounded, so there must be connected.
some yn such that |yn | > R + 1. Thus, d(xn , yn ) > 1, so
24 8(b) If A ⊂ X and A is path connected, is A necessarily
d(x, y) = 1.
path connected? (5 points)
Thus, for any point in either set, the 12 -ball around that
point doesn’t contain any points in the other set. Therefore No. For example take the set:
both B and U are open, and so Rω is not connected.
S = {x × sin(1/x) | 0 < x ≤ 1}
24 1(a) Show that no two of the spaces (0, 1), (0, 1], and [0, 1]
are homeomorphic. (5 points) which is path connected because it is the image of (0, 1]
under a continuous map (part c). However it’s closure is
(0, 1) is distinct from the others because if you remove not path connected, as demonstrated in Example 7, pages
a point, x ∈ (0, 1), then (0, 1) − {x} = (0, x) ∪ (x, 1). How- 156-157.
ever, with the other two sets, you can remove x = 1, and
the sets will remain connected: 24 8(c) If f : X → Y is continuous and X is path-connected,
is f (X) necessarily path connected? (5 points)
(0, 1] − {1} = (0, 1)
[0, 1] − {1} = [0, 1) Yes. Take any two points y, y0 in f (X). We can find
x, x0 ∈ X such that y = f (x) and y0 = f (x0 ). Then there
The other two sets are distinct because you can remove exists a path p : [0, 1] → X such that p(0) = x and p(1) =
a point from (0, 1], and it will be homeomorphic to (0, 1), x0 . Then f ◦ p : [0, 1] → f (X) ⊂ Y is a continuous function
as shown above. However, if you suppose you remove such that f ◦ p(0) = y and f ◦ p(1) = y0 . Thus, f (X) is
a point x from [0, 1]. If 0 < x < 1, then [0, 1] − {x} = path connected.
[0, x) ∪ (x, 1] which is disconnected. On the other hand, if
x = 0 or x = 1, then [0, 1] − {x} = (0, 1] or = [0, 1), both of 24 8(d) If {Aα } is a collection of path-connected subspaces
of X and if Aα 6= 0,
T S
which are homeomorphic to (0, 1], and thus not to (0, 1), / is Aα necessarily path con-
by above. Thus, these two sets are distinct. nected? (5 points)

2
 Let f : Rω → Rω be the product of these functions for
S
Yes. Suppose x1 and x2 are two points in Aα , say
x1 ∈ Aα1 and x2 ∈ Aα2 . Let b be a point in Aα . Because n ∈ Z+ . This is a continuous function since each compo-
T

Aα1 is path connected, there exists a path p1 from x1 to b. nent is continuous. Thus, f −1 (U1 ×U2 × . . .) = f1−1 (U1 ) ×
Because Aα2 is path connected, there exists a path p2 from f2−1 (U2 ) × . . ., and each of the fn is continuous.
b to x2 . Thus the function p : [0, 2] → Aα :
T
Note that since x is not eventually zero, there are in-
 finitely many xn such that fn (xn ) = n, and so f (x) is un-
p1 (x) if x ∈ [0, 1]
p(x) = bounded. Of course f (y) = 0, which is bounded. Thus by
p2 (x − 1) if x ∈ [1, 2]
the separation of Rω , f (x) and f (y) must lie in separate
This is a continuous by the pasting lemma, so it is a path components, so x and y must lie in separate components.
Conversely, if x is eventually zero, say, xn = 0 for all
S
from x1 to x2 . Thus, Aα is path connected.
n > N. Then x lies in RN × {0} × {0} × . . . ⊂ Rω , which
25 2(a) What are the components and path components of Rω is homeomorphic to RN , which is known to be connected.
(in the product topology)? (5 points) Thus x and y = 0 are in the same component.
By problem 24 8(a), Rω is path connected. Thus, it is
also connected, so the only path component is all of Rω
and the only component is Rω .

25 2(b) Consider Rω in the uniform topology. Show that x

and y lie in the same component of Rω if and only if
the sequence
x − y = (x1 − y1 , x2 − y2 , . . .)
is bounded. (10 points)

As the hint states, we can assume that y is 0. This is

because translation by y does not change the topology of
Rω . Thus, x and y are in the same component iff x − y and
0 are. For the rest of the problem, assume that y = 0.
Suppose the sequence x = (x1 , x2 , . . .) is bounded, and
then define the straight line path f : [0, 1] → Rω by:
f (t) = (x1t, x2t, . . .)
I claim this is continuous. Suppose we have some ε-ball
around the point f (t). Then the inverse image of this ball
ε
contains the max{x n}
-ball around t. (Note that this makes
sense only because x is bounded.) Thus, f is a path con-
necting y = 0 to x, so the two points lie in the same com-
ponent.
Conversely, if x is unbounded, then it must be in a dif-
ferent component from y by the separation exhibited in
23.8.

25 2(c) Give Rω the box topology. Show that x and y lie in

the same component of Rω if and only if the sequence
x − y is ”eventually zero.” (10 points)

As in the previous problem, assume that y = 0. Sup-

pose x = (x1 , x2 , . . .) is not ”eventually zero”. Define the
functions fn : R → R as:
 n
an when xn 6= 0
fn (an ) |xn |
an when xn = 0