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TOPOLOGY HW 2
CLAY SHONKWILER
20.9
Show that the euclidean metric d on Rn is a metric, as follows: If x, y ∈ Rn
and c ∈ R, define
x + y = (x1 + y1 , . . . , xn + yn ),
cx = (cx1 , . . . , cxn ),
x · y = x1 y1 + . . . xn yn .
(a) Show that x · (y + z) = (x · y) + (x · z).
Proof. Let x, y, z ∈ Rn . Then
x · (y + z) = x · (y1 + z1 , . . . , yn , zn )
= x1 (y1 + z1 ) + . . . xn (yn + zn )
= x1 y1 + x1 z1 + . . . xn yn + xn zn
=x·y+x·z
(b) Show that |x · y| ≤ ||x||||y||.
Proof. Let x, y ∈ Rn . Then
||
y
x
±
|| ≥ 0
||x|| ||y||
. Then
n
||
x
y 2 X
±
|| =
||x|| ||y||
i=1
xi
yi
±
||x|| ||y||
2
=
n X
xi ||y|| ± yi ||x|| 2
i=1
||x||||y||
Then, since ||x||||y|| ≥ 0,
n
X
(xi ||y|| ± yi ||x||)2 ≥ 0.
i=1
This in turn implies that
P
2 ||y||2 + y 2 ||x||2 ± 2x y ||x||||y||
0 ≤ ni=1
x
i
i
i
i
P
P
P
= ||y|| ni=1 x2i + ||x|| ni=1 yi2 ± 2||x||||y|| ni=1 xi yi
= ||y||2 ||x||2 + ||x||2 ||y||2 ± 2||x||||y||(x · y)
= 2||x||2 ||y||2 ± 2||x||||y||(x · y).
In other words,
||x||2 ||y||2 ± ||x||||y||(x · y) ≥ 0,
1
≥ 0.
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CLAY SHONKWILER
or
±(x · y) ≥
−||x||2 ||y||2
= −||x||||y||.
||x||||y||
Multiplying both sides by −1, we see that
∓(x · y) ≤ ||x||||y||,
which means that
|x · y| ≤ ||x||||y||.
(c) Show that ||x + y|| ≤ ||x|| + ||y||.
Proof. To see this, we note that, for any z ∈ Rn , ||z||2 = z · z. Hence,
||x + y||2 = (x + y) · (x + y)
= (x + y) · x + (x + y) · y
=x·x+y·x+x·y+y·y
= ||x||2 + 2(x · y) + ||y||2
≤ ||x||2 + 2||x||||y|| + ||y||2
= (||x|| + ||y||)2
since x · y ≤ ||x||||y|| by part (b) above. Hence, we can conclude that
||x + y|| ≤ ||x|| + ||y||.
(d) Verify that d is a metric.
Proof. Let x, y ∈ Rn . Then
d(x, y) = ||x − y|| = ((x1 − y1 )2 + . . . + (xn − yn )2 )1/2 ≥ 0
and d(x, y) = ||x − y|| = 0 if and only if xi = yi for all i = 1, . . . n, which is
to say precisely when x = y. Now,
d(x, y) = ||x − y|| = ((x1 − y1 )2 + . . . + (xn − yn )2 )1/2
= ((y1 − x1 )2 + . . . + (yn − xn )2 )1/2 = ||y − x|| = d(y, x).
Finally, if z ∈ Rn as well, then
d(x, z) = ||x−z|| = ||(x−y)+(y −z)|| ≤ ||x−y||+||y −z|| = d(x, y)+d(y, z)
by part (c) above. Hence, d is a metric on Rn .
TOPOLOGY HW 2
3
21.12
Prove continuity of the algebraic operations on R as follows: Use the
metric d(a, b) = |a − b| on R and the metric on R2 given by the equation
ρ((x, y), (x0 , y0 )) = max{|x − x0 |, |y − y0 |}.
(a) Show that addition is continuous.
Proof. Let > 0 and let δ < /2. If
ρ((x, y), (x0 , y0 )) = max{|x − x0 |, |y − y0 |} < δ,
then,
d(x + y, x0 + y0 ) = |(x + y) − (x0 + y0 )||(x − x0 ) + (y − y0 )|
≤ |x − x0 | + |y − y0 |
<δ+δ
< 2(/2)
= ,
so addition is continuous.
(b) Show that multiplication is continuous.
Proof. Given (x0 , y0 ) ∈ R2 , let 1 > > 0 and let δ <
|x0 |+|yo |+1 .
If
ρ((x, y), (x0 , y0 )) = max{|x − x0 |, |y − y0 |} < δ
then
d(xy, x0 y0 ) = |xy − x0 y0 |
= |xy + (xy0 − xy0 ) + (yx0 − yx0 ) + (x0 y0 − x0 y0 ) − x0 y0 |
= |x0 (y − y0 ) + y0 (x − x0 ) + (x − x0 )(y − y0 )|
≤ |x0 (y − y0 )| + |y0 (x − x0 )| + |(x − x0 )(y − y0 )|
= |x0 ||y − y0 | + |y0 ||x − x0 | + |x − x0 ||y − y0 |
< |x0 |δ + |y0 |δ + δ 2
= δ(|x0 | + |y0 | + δ)
< δ(|x0 | + |y0 | + 1)
< (|x0 | + |y0 | + 1) |x0 |+|y
0 |+1
< .
Hence, multiplication is continuous.
(c) Show that the operation of taking reciprocals is a continuous map
from R − {0} to R.
Proof. Let r : R − {0} → R be the reciprocal map. Let (a, b) be an open
interval in R. If both a and b are positive, then,
r−1 (a, b) = (1/b, 1/a)
which is open in R − {0}. If both a and b are negative, then
r−1 (a, b) = (1/b, 1/a),
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CLAY SHONKWILER
which is again open in R − {0}. If a is negative and b is positive, then
r−1 (a, b) = (1/a, 1/b),
which is open in R − {0}. If a is zero, then
r−1 (a, b) = (1/b, ∞)
which is open in R − {0}. Finally, if b is zero, then
r−1 (a, b) = (−∞, 1/a)
which is open in R − {0}. Hence, we can conclude that taking reciprocals is
a continuous map.
(d) Show that the subtraction and quotient operations are continuous.
Proof. To show that subtraction is continuous, let > 0 and let δ < /2. If
ρ((x, y), (x0 , y0 )) = max{|x − x0 |, |y − y0 |} < δ,
then
d(x − y, x0 − y0 ) = |(x − y) − (x0 − y0 )|
= |(x − x0 ) + (y0 − y)|
≤ |x − x0 | + |y0 − y|
<δ+δ
< 2(/2)
= ,
so subtraction is continuous. Finally, to show that the quotient is continuous
from R × R − {0} to R, we note that, for (x, y) ∈ R × R − {0},
x
1
=x
.
y
y
Thus, if q is the quotient map, p the product map, r the reciprocal map,
and r0 : R × R → R × R − {0} is defined by
r0 (x, y) = (x, r(y))
then
q = p ◦ r0 .
Note that r0 is continuous, since its coordinate functions (the identity and
the reciprocal maps) are, so q is continuous, since it is the composition of
two continuous functions.
22.1
Check the details of Example 3
Check: We want to show that, if p is the map from R to A = {a, b, c}
where

 a if x > 0
b if x < 0
p(x) =

c ifx = 0
the topology on A induced by p consists of the following open sets: {a}, {b}, {a, b}, {a, b, c}.
TOPOLOGY HW 2
5
In the quotient topology, we know that {a} is open, since
p−1 ({a}) = (0, ∞),
which is open in R. Similarly, {b} is open, since
p−1 ({b}) = (−∞, 0),
which is open in R. Also,
p−1 ({a, b}) = (−∞, 0) ∪ (0, ∞),
which is open in R, so {a, b} is open. Finally, {a, b, c} is open, since
p−1 ({a, b, c}) = R
which is open in R.
Now, we want to show that none of the other subsets of A (other than
the empty set, which is obviously open) are open in the quotient topology.
p−1 ({c}) = {0}
which is not open in R, so {c} is not open. Also, {a, c} and {b, c} are not
open, since
p−1 ({a, c}) = [0, ∞)
and
p−1 ({b, c}) = (−∞, 0],
neither of which is open in R. We’ve exhausted all subsets of A, so we see
that, indeed, the topology on A is precisely that depicted in Example 3.
♣
22.2
(a) Let p : X → Y be a continuous map. Show that if there is a continuous
map f : Y → X such that p ◦ f equals the identity map of Y , then p is a
quotient map.
Proof. If there exists a continuous map f : Y → X such that p ◦ f ≡ idY ,
then we want to show that p is a quotient map. p is clearly surjective since,
if it were not, p ◦ f could not be equal to the identity map. Now, let U ⊂ Y .
If p−1 (U ) is open in X, then
U = (p ◦ f )−1 (U ) = f −1 (p−1 (U ))
is open in Y since f is continuous. Hence, p is a surjective, continuous open
map, so it is necessarily a quotient map.
(b) If A ⊂ X, a retraction of X onto A is a continuous map r : X → A
such that r(a) = a for each a ∈ A. Show that a retraction is a quotient map.
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CLAY SHONKWILER
Proof. Let i : A → X be the inclusion map. Then i is continuous and, for
all a ∈ A,
(r ◦ i)(a) = r(i(a)) = r(a) = a
so r ◦ i ≡ idA . Hence, since r is continuous, we can conclude, using part (a)
above, that r is a quotient map.
22.4
(a) Define an equivalence relation on the plane X = R2 as follows:
x0 × y0 ∼ x1 × y1 if x0 + y02 = x1 + y12 .
Let X ∗ be the corresponding quotient space. It is homeomorphic to a familiar space; what is it?
Answer: X ∗ is homeomorphic to R. We see this as follows. Define
g : R2 → R by
g(x × y) = x + y 2 .
Then g is surjective since, for any r ∈ R, g(r × 0) = r + 02 = r. Also, if
A = (a, b) × (c, d) ⊂ R2 , then A = [a, b] × [c, d]. Now
g(A) = (a + inf x2 , b + sup y 2 )
x∈(c,d)
y∈(c,d)
and
g(A) = [a + inf x2 , b + sup y 2 ] = g(A),
x∈(c,d)
y∈(c,d)
so g is continuous. Also, we see that g(A) is open, so g is a surjective,
continuous, open map. In other words, a quotient map.
Now, for each z ∈ R,
g −1 (z) = {x × y ∈ R2 |x + y 2 = z}
Hence, X ∗ = {g −1 (z)|z ∈ R}. Thus, by Corollary 22.3, g induces a bijective
continuous map f : R2 → R which is a homeomorphism.
♣
(b) Repeat (a) for the equivalence relation
x0 × y0 ∼ x1 × y1 if x20 + y02 = x21 + y12 .
Answer: X ∗ is homeomorphic to R+ ∪{0}. We see this as follows. Define
g : R2 → R by
g(x × y) = x2 + y 2 .
√
Then g is surjective since, for any r ∈ R+ ∪ {0}, g( r × 0) = r2 + 02 = r.
Also, if A = (a, b) × (c, d) ⊂ R2 , a basis element of the topology on R2 , then
A = [a, b] × [c, d]. Now
g(A) = ( inf
x0 ∈(a,b)
x20 +
inf
y0 ∈(c,d)
y02 , sup x21 + sup y12 ),
x1 ∈(a,b)
y1 ∈(c,d)
and
g(A) = [ inf
x0 ∈(a,b)
x20 +
inf
y0 ∈(c,d)
y02 , sup x21 + sup y12 ] = g(A),
x1 ∈(a,b)
y1 ∈(c,d)
TOPOLOGY HW 2
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so g is continuous. Also, we see that g(A) is open, so g is a surjective,
continuous, open map. In other words, a quotient map.
Now, for each z ∈ R,
g −1 (z) = {x × y ∈ R2 |x2 + y 2 = z}
Hence, X ∗ = {g −1 (z)|z ∈ R}. Thus, by Corollary 22.3, g induces a bijective
continuous map f : R2 → R which is a homeomorphism.
♣
23.1
Let T and T 0 be two topologies on X. If T 0 ⊃ T , what does connnectedness of X in one topology imply about connectedness in the other?
Answer: If X is connected in T 0 , then there exist no two elements of T 0
that separate X. Since every element of T is an element of T 0 , this means
no two elements of T that separate X. In other words, if X is connected in
T 0 , then X is connected in T .
On the other hand, if X is connected in T , then it is not necessarily
connected in T 0 . For example, if we consider the three point set A from 22.1
above under the quotient topology described above and under the discrete
topology, then the discrete topology is finer than the quotient topology, A
is connected under the quotient topology, but A is not connected in the
discrete topology. This last is clearly true, since {a, b} and {c} are disjoint
open sets in the discrete topology that cover A. To see that A is connected
in the quotient topology, we note that the only open set containing the point
c is A, so it will be impossible to separate A by open sets.
♣
23.2
Let {An } be a sequence of S
connected subspaces of X, such that An ∩
An+1 6= ∅ for all n. Show that An is connected.
S
Proof. Suppose
An is not connected.
S
S Then there exist disjoint open sets C
and D in An such that C ∪ D = An . Then, to arrive at a contradiction,
we aim to demonstrate that each An lies in C or D. Since A1 is connected,
it must be contained entirely in either C or D, by Lemma 23.2. Suppose,
without loss of generality, that it lies in C. Fix k ∈ Z+ and suppose that
Ak ⊂ C. By Lemma 23.2, since Ak+1 is connected, it must lie entirely within
either C or D. Furthermore,
∅=
6 Ak+1 ∩ Ak ⊂ Ak ⊂ C,
so
Ak+1 ∩ C 6= ∅.
Therefore, it must be true that Ak+1 ⊂ C.
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CLAY SHONKWILER
S
Hence, by induction, every AS
An ⊂ C, contradicting the
n ⊂ C, so
assertion thatSC and D separate An . From this contradiction we conclude
that, in fact, An is connected.
23.6
Let A ⊂ X. Show that if C is a connected subspace of X that intersects
both A and X − A, then C intersects Bd A.
Proof. By contrapositive. Suppose C is connected and does not intersect
Bd A. Then
(1)
∅ = C ∩ (A ∩ X − A) = (C ∩ A) ∩ (C ∩ X − A).
Now, by definition of the subspace topology, C ∩A and C ∩X − A are closed
in C, so C − (C ∩ A) and C − (C ∩ X − A) are open in C. Now,
C − (C ∩ A) ∩ (C ∩ X − A) = C − ∅ = C.
However, by DeMorgan’s,
C = C − (C ∩ A) ∩ (C ∩ X − A) = C − (C ∩ A) ∪ C − (C ∩ X − A) .
Since C is connected, it must be true that either C − (C ∩ A) or C − (C ∩ X − A)
is empty.
If C − (C ∩ A) = ∅, then C ∩ A = C. By (1), then,
C ∩ X − A = ∅.
Since X − A ⊂ X − A, we can conclude that
C ∩ (X − A) = ∅.
On the other hand, if C − (C ∩ X − A) = ∅, then C ∩ X − A = C. By
(1),
C ∩ A = ∅.
Since A ⊂ A, we see that C ∩ A = ∅.
In either case, we see that, if C ∩ Bd A = ∅, then it is not the case that
C intersects both A and X − A. Hence, by contrapositive, we conclude that
if C intersects both A and X − A, then C intersects Bd A.
A
Let X and Y be connected topological spaces and endow X × Y with the
product topology. Now suppose that A ( X and B ( Y are proper subsets.
Show that (X × Y ) − (A × B) is connected. Give an example showing that
this fails if one of A and B is not a proper subset.
Proof. Fix a point a×b ∈ (X ×Y )−(A×B). Then the set X ×b is connected,
as it is homeomorphic to the connected set X and, for all x ∈ X − A, x × Y
is connected, since it is homeomorphic to the connected set Y . Now, X × b
and x × Y have the point a × b in common, so
Tx = (x × Y ) ∪ (X × b)
TOPOLOGY HW 2
9
is connected for all x ∈ X − A. Furthermore, the union
[
Cx =
Tx
x∈X−A
is connected, since each Tx is connected and contains the point a × b. Now,
also, a×Y is connected, since it is homeomorphic to Y and, for all y ∈ Y −B,
X ×y is connected, since it is homeomorphic to X, and contains a×b. Hence,
Ty = (a × Y ) ∪ (X × y)
is connected. Furthermore
Cy =
[
Ty
y∈Y −B
is connected, since each Ty is connected and contains the point a × b. Now,
finally, we can conclude that
Cx ∪ Cy
is connected, since both Cx and Cy are connected and contain a × b.
To see that Cx ∪ Cy = (X × Y ) − (A × B), let c × d ∈ (X × Y ) − (A × B).
Then, either c ∈ X − A or d ∈ Y − B (or both). In the first case, c × d ∈ Cx
and in the second, c × d ∈ Cy . Either way, c × d ∈ Cx ∪ Cy , so
(X × Y ) − (A × B) ⊆ Cx ∪ Cy .
Clearly,
Cx ∪ Cy ⊆ (X × Y ) − (A × B),
so, in fact, Cx ∪ Cy = (X × Y ) − (A × B), meaning (X × Y ) − (A × B) is
connected.
Counter-Example: Let X = Y = A = R and let B = [0, 1]. Then
(X × Y ) − (A × B) = (R × (−∞, 0)) ∪ (R × (1, ∞)) .
Now, R × (−∞, 0) and R × (1, ∞) are both clearly open in (R × (−∞, 0)) ∪
(R × (1, ∞)) and disjoint, so they form a separation. Hence, we see an
example where A is not a proper subset of X and, thereby, (X ×Y )−(A×B)
is not connected, even though X and Y are.
♣
B
Let X = [0, 1] ⊂ R have the product topology. Now suppose f : X → X
is continuous. Prove (without using the intermediate value theorem) that
f has a fixed point. That is show that there exists an x0 ∈ X such that
f (x0 ) = x0 . For the purposes of this exercise you may assume that X is
connected.
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CLAY SHONKWILER
Proof. Define g : [0, 1] → [−1, 1] by
g(x) = x − f (x).
Then g is continuous, g(0) = −f (0) ≤ 0 and g(1) = 1 − f (b) ≥ 0. Let
A = g([0, 1]) ∩ (−∞, 0)
and
B = g([0, 1]) ∩ (0, ∞).
Both A and B are open in g([0, 1]). Furthermore, if there is no point c ∈ [0, 1]
such that g(c) = 0, then f ([0, 1]) = A ∪ B. Since A and B are disjoint, this
constitutes a separation of g([0, 1]). However, the image of a connected set
like [0, 1] under a continuous map like g is connected by Thm. 23.5. Hence,
there must be some c ∈ [0, 1] such that g(c) = 0. Now,
f (c) = c − g(c) = c − 0 = c,
so f has a fixed point.
DRL 3E3A, University of Pennsylvania
E-mail address: [email protected]