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You got trolled by the underscore.
Not real.
Y! Sources: Andrew Bynum en route to Orlando as part of 3 team Dwight Howard blockbuster. Will take some time to sort out all the specifics. -- Adrian Wojnarowski (@WojYahooNBA_)
If this happens I'm gonna be really pissed.
You got trolled by the underscore.
Not real.
I hope its not real. Would be nice if we were the third team and got either Dwight or Andrew. Wishful thinking.
Dwight Howard to Houston, Bynum and a 1st to Orlando, Scola, Lowry, Parsons and 3 1st round picks to LA, according to sources. -- Adrian Wojnarowski
@WojYahooNBA
Hope this is fake cause it actually makes Houston a threat. Lakers getting 3 first round picks I'm not sure about that.
Lol I'm not a troll. I just saw this on his twitter and I thought it was interesting. Sorry if its not true but I think this guys legit.
Never mind it was a fake account my bad.
yep the trend during FA, draft, trade deadline etc, is create fake accounts close to those of legit nba sources.
1.1. Prove that |x| ≤
n
i=1
xi .
Answer. Let e1 , . . . , en be the standard basis of Rn and let x = (x1 , . . . , xn ) ∈ Rn . Then we have
x = x1 e1 + · · · + xn en . Since xj ej = xj for each index j, applying the triangle inequality shows
that
|x| ≤ x1 + · · · + |xn | .
1.2. When does equality hold in Theorem 1-1(3)?
Answer. Theorem 1-1(3) states that |x + y| ≤ |x| + |y|. We wish to find a condition that is
both necessary and sufficient to guarantee that |x + y| = |x| + |y|. Note this holds if and only if
|x + y|2 = (|x| + |y|)2 . Expanding the left-hand side gives
|x + y|2 = |x|2 + 2 x, y + |y|2 .
Expanding the right-hand side gives
(|x| + |y|)2 = |x|2 + 2 |x| |y| + |y|2 .
These two expressions are equal if and only if x, y = |x| |y| holds. Theorem 1-1(2) shows that x
and y must then be linearly dependent, so that one of the vectors is a multiple of the other. If
either vector is zero the result is obvious; assume neither is the zero vector. Then there is a real
number a such that y = ax holds. Then x, y = a |x|2 and |x| |y| = |a| |x|2 , so that the equality of
|x + y|2 and (|x| + |y|)2 is equivalent to the condition that a ≥ 0. Hence, equality holds in Theorem
1-1(3) precisely when one vector is a nonnegative real multiple of the other.
1.3. Prove that |x − y| ≤ |x| + |y|. When does equality hold?
Answer. Let z = −y so that x + z = x − y. Then we have
|x − y| = |x + z| ≤ |x| + |z| = |x| + |y| .
Note equality can only hold if x and z are linearly dependent. Note the equality is trivial when
either x or y is the zero vector. Assume neither is; then z cannot be the zero vector either. From
Problem 1.2 we know there must be a positive real number a such that z = ax then holds. Since
y = −z, we then have y = (−a)x. Hence, equality holds if and only if one of the vectors x and y is
a nonpositive multiple of the other.
1 Functions on Euclidean Space
2
1.4. Prove that ||x| − |y|| ≤ |x − y|.
Answer. Since x = (x − y) + y and y = (y − x) + x, the triangle inequality shows that
|x| ≤ |x − y| + |y|
|y| ≤ |x − y| + |x| .
Subtracting |y| from both sides of the first equation and |x| from both sides of the second then
gives
|x| − |y| ≤ |x − y|
|y| − |x| ≤ |x − y| .
These two conditions together just say that |x − y| ≥ ||x| − |y||.
1.5. The quan y |y − x| is called the distance between x and y. Prove and interpret geometrically
the “triangle inequality”: |z − x| ≤ |z − y| + |y − x|.
Answer. Note that z − x = (z − y) + (y − x), so Theorem 1-1(3) shows that that |z − x| ≤
|z − y| + |y − x| holds. In R2 and R3 this inequality states that in the triangle formed by the points
x, y, z the edge connecting points x and z cannot be longer than the sum of the lengths of the edges
connecting the points y, z and x, y, respectively. In other words, it shows the length of one side of
a triangle cannot be more than the sum of lengths of the other two sides.
1.6. Let f, g be integrable on [a, b].
(a) Prove that
b
a f
·g ≤(
b 2 1/2
b
( a g 2 )1/2 .
a f )
(b) If equality holds, must f = λg for some λ ∈ R? What if f and g are continuous?
(c) Show that Theorem 1-1(2) is a special case of (a).
Answer. We can emulate Spivak’s derivation of the Schwarz inequality by defining the quadratic
b
polynomial p(λ) = a (f − λg)2 dμ, where μ is the Lesbesgue measure on R1 . Note that p ≥ 0 holds
for all real numbers λ since the integrand (f − λg)2 is nonnegative and b ≥ a holds.
Consider first the case that f = λg a.e. on [a, b]. Then there can be no real root λ of the polynomial
p, since (f − λg)2 > 0 a.e. on [a, b] in this case. Expand this polynomial as
b
p(λ) = λ2
a
g 2 dμ − 2λ
b
b
f g dμ +
f 2 dμ.
a
a
Then the discriminant
2
b
∆ = 4
f g dμ
a
must be negative, so that (
b
a f g dμ
<(
b 2
1/2 (
a f dμ)
b
− 4
a
b
g 2 dμ
f 2 dμ
a
b
b 2
b 2
2
a f g dμ) < ( a f dμ)( a g dμ). Taking
b 2
1/2 when f = λg a.e. on [a, b].
a g dμ)
square roots then shows that
1 Functions on Euclidean Space
3
Now consider the case that f = λg a.e. on [a, b]. Then we have f g = λg 2 a.e. Thus we have
b
b
2
f dμ
2
g dμ
a
= λ
2
b
2
2
g dμ
a
a
b
=
2
λg 2 dμ
a
2
b
f g dμ
=
.
a
Taking square roots then shows that
b
b
f g dμ =
1/2
b
2
f dμ
a
1/2
2
g dμ
a
a
when f = λg a.e. on [a, b]. In any case, we have shown the main result
b
b
a
f g dμ ≤
1/2
b
f 2 dμ
1/2
g 2 dμ
a
a
holds for all integrable maps f, g : [a, b] → R.
Note the result proved in the text is just a special case of this: define the step functions f, g by the
rules
n
n
, g(t) = y k
,
f (t) = xk
b−a
b−a
where k is the unique integer such that a + (k − 1)(b − a)/n ≤ t < a + k(b − a)/n. Note that we
then have f (t)2 = (xk )2 n/(b − a), g(t)2 = (y k )2 n/(b − a), and f (t)g(t) = xk y k n/(b − a). We can do
the integral
b
a+(b−a)/n
f 2 dμ =
a+2(b−a)/n
f 2 dμ +
a+(b−a)/n
k 2
a
1 2
a
= (x ) + (x2 )2 + · · · + (x )
b
f 2 dμ + · · · +
b
f 2 dμ
a+(n−1)(b−a)/n
= |x|2 .
b
Similarly, we have a g 2 dμ = |y|2 and a f g dμ = n xi y i . Thus from our integral inequality we
i=1
then get the result of Theorem 1-1(2): namely, that
n
i=1
Rn
1.7. A linear transform T :
→
preserving if T x, T y = x, y .
Rn
xi y i ≤ |x| |y| .
is norm preserving if |T (x)| = |x|, and inner product
(a) Prove that T is norm preserving if and only if T is inner-product preserving.
(b) Prove that such a linear transformation T is 1–1 and T −1 is of the same sort.
Answer. Let T : Rn → Rn inner product preserving. Then |T x|2 = T x, T x = x, x = |x|2 holds.
Since norms are nonnegative, we then have |T x| = |x| so that T is norm preserving.
Assume that T : Rn → Rn is not norm preserving. Then there is some x ∈ Rn such that |T x| = |x|,
so that |T x|2 = |x|2 also holds. But |T x|2 = T x, T x and |x|2 = x, x then show that T x, T x =
1 Functions on Euclidean Space
4
x, x , and thus T is not inner product preserving. Hence, a linear transform T : Rn → Rn is inner
product preserving if and only if it is norm preserving.
Note that Rn is a vector space, so it is also an abelian group under addition. Since a linear
transformation T : Rn → Rn is a morphism, we only need to show it has trivial kernel to
establish that it is injective. Note that 0 is in the kernel of T since
T (0) = T (0 + 0) = T (0) + T (0)
shows that T (0) = 0 must hold. Let x ∈ ker T so that T (x) = 0. Since T is a norm-preserving map,
we have |x|2 = |T (x)|2 = 0 so that x = 0 must hold. Hence, T has trivial kernel and is therefore
injective. Since a linear map between a finite-dimensional vector space and itself is bijective if and
only if it is injective (see Theorem 3.21, [1]), we therefore see T gives an automorphism Rn → Rn .
Thus there is an inverse linear map T −1 : Rn → Rn that sends T (x) → x.
Let y ∈ Rn . Then there is some x ∈ Rn such that y = T (x). Thus T −1 (y) = x holds, and we have
T −1 (y) = |x|2 = |T (x)|2 = |y|2
since T is norm-preserving. This shows the inverse map T −1 : Rn → Rn is also norm-preserving.
1.8. If x, y ∈ Rn are non-zero, the angle between x and y, denoted ∠(x, y), is defined as
arccos ( x, y / |x| · |y|), which makes sense by Theorem 1-1(2). The linear transformation T is angle
preserving if T is 1-1, and for x, y = 0 we have ∠(T x, T y) = ∠(x, y).
(a) Prove that if T is norm preserving, then T is angle preserving.
(b) If there is an orthogonal basis x1 , . . ., xn of Rn and numbers λ1 , . . ., λn such that T xi = λi xi ,
prove that T is angle preserving if and only if all |λi | are equal.
(c) What are all angle preserving T : Rn → Rn .
Answer. From Problem 1.7 we know a norm preserving linear map is bijective and is also inner
product preserving. Thus, when x, y = 0 we have
T (x), T (y)
x, y
=
.
|T (x)| |T (y)|
|x| |y|
Taking inverse cosines of both sides then shows that ∠(T (x), T (y)) = ∠(x, y), which then establishes
that a norm preserving linear map is angle preserving.
Let x1 , x2 , . . . , xn be an orthogonal basis of Rn and let x = a1 x1 +· · ·+an xn , y = b1 x1 +· · ·+bn xn ∈
Rn . Since T xi = λi xi , we then have
T x, T y
=
ai T xi ,
i
bj T x j
j
a i λi x i ,
=
i
=
bj λ j x j
j
a i bj λ i λ j x i , x j
i,j
a i bi λ 2 x i , x i .
i
=
i
1 Functions on Euclidean Space
5
Furthermore, we also have
x, y
=
ai xi ,
bj xj
i
=
j
a i bj x i , x j
i,j
=
a i bi x i , x i .
i
Assume first that |λi | = λ for each index. Then we have T x, T y = λ2
Then we have
T x, T y
=
|T x| |T y|
λ2 x, y
λ2
λ2
x, x
x, y
=
y, y
x, x
y, y
=
i ai bi
xi , xi = λ2 x, y .
x, y
.
|x| |y|
Taking inverse cosines then shows that ∠(T (x), T (y)) = ∠(x, y) for all x, y ∈ Rn so that T is
angle-preserving.
Conversely, assume not all |λi | not equal and assume without loss of generality that |λ1 | = |λ2 |,
with λ1 , λ2 = 0. Take x = x1 , y = x1 + x2 , and λ2 = λ2 . We then have
2
1
T x, T y
= λ2 |x1 |2
1
|T x| = |λ1 | |x1 |
|T y| =
Thus we have
T x, T y
=
|T x| |T y|
x, y
λ2 |x1 |2 + λ2 |x2 |2 .
1
2
|λ1 | |x1 |
λ2 |x1 |2 + λ2 |x2 |2
1
2
= |x1 |2
|x| = |x1 |
|y| =
This shows we have
x, y
=
|x| |y|
|x1 |2 + |x2 |2 .
|x1 |
|x1 |2 + |x2 |2
.
Taking α = |x2 | / |x1 |, we then get
T x, T y
=
|T x| |T y|
1
1 + (λ2 /λ1 )2 α2
1
x, y
= √
|x| |y|
1 + α2
Since λ1 , λ2 = 0 and |λ1 | = |λ2 | shows that (λ2 /λ1 )2 = 1, we see that
T x, T y
x, y
=
|T x| |T y|
|x| |y|
1 Functions on Euclidean Space
6
for x = x1 , y = x1 + x2 . In other words, we have cos ∠(T x, T y) = cos ∠(x, y).
Note since −1 ≤ u, v /(|u| |v|) ≤ 1 for u, v = 0 and since the inverse cosine map [−1, 1] → [0, π] is
bijective, we therefore see that ∠(T x, T y) = ∠(x, y) in this case. Thus the map T : Rn → Rn isn’t
angle-preserving when |λ1 | = |λ2 | (λ1 , λ2 = 0). Thus we see that T is angle preserving if and only
if |λ1 | = |λ2 | = · · · = |λn |.
Before moving onto part (c), let us show a couple of preliminary results: (1) the composition of
angle-preserving maps is angle preserving and (2) if x1 , . . . , xn is an orthogonal basis of Rn and
T : Rn → Rn is a 1-1 linear map, then setting yj = T (xj ) for each index j gives us an orthogonal
basis y1 , . . . , yn of Rn . The first is easy to prove: letting S, T : Rn → Rn be angle-preserving means
that S ◦ T is 1-1 and for any x, y ∈ Rn we have
S(T (x)), S(T (y))
=
|S(T (x))| |S(T (y))|
S(T (x)), S(T (y))
|T (x)| |T (y)|
=
x, y
.
|x| |y|
Thus the composition map S ◦ T : Rn → Rn is angle-preserving. To prove (2), note that the list
T (x1 ), . . . , T (xn ) is linearly independent. Assume that
0 = c1 T (x1 ) + · · · + cn T (xn ) = T (c1 x1 + · · · + cn xn ).
Then c1 x1 + · · · + cn xn = 0 must hold (recall that T is 1-1). But c1 x1 + · · · + cn xn = 0 holds if and
only if c1 = · · · = cn = 0, since x1 , . . . , xn form a basis of Rn . Thus T (x1 ), . . . , T (xn ) is a linearly
independent list of length n in Rn , so it must be a basis of Rn . Note if we have i = j, we then have
xi , xj
T (xi ), T (xj )
=
= 0
|T (xi )| |T (xj )|
|xi | |xj |
since T is angle-preserving and since x1 , . . . , xn forms an orthogonal basis of Rn . Thus we have
shown T (x1 ), . . . , T (xn ) gives an orthogonal basis of Rn .
We now move on to part (c). Let x1 , . . . , xn be an orthogonal basis of Rn . We will show a linear
map T : Rn → Rn is angle-preserving if and only if it factors into T = A ◦ N , where N : Rn → Rn
is a norm-preserving linear map and A : Rn → Rn is an angle-preserving linear map such that
A(yj ) = λj yj for all j, where y1 , . . . , yn is some orthogonal basis of Rn .
Assume first that T is an angle-preserving linear map and define the linear map N : Rn → Rn by
the rule
n
ai T (xi ) |xi |
N (x) =
,
|T (xi )|
i=1
where x =
a1 x
1
+ ··· +
an e
n.
Then we see that
n
N (x), N (x)
=
ai T (xi ) |xi |
,
|T (xi )|
i=1
n
n
ai aj
=
i=1 j=1
n
j=1
aj T (xj ) |xj |
|T (xj )|
|xi | |xj |
|T (xi )| |T (xj )|
T (xi ), T (xj ) .
Since T is angle-preserving, we then have
T (xi ), T (xj ) =
|T (xi )| |T (xj )|
|xi | |xj |
xi , xj .
1 Functions on Euclidean Space
7
Subs ution of this result into the equation above then shows that
n
N (x), N (x)
n
n
i j
a a xi , xj
=
(ai )2 xi , xi
=
=
x, x .
i=1
i=1 j=1
This then shows that |N (x)| = |x|, so that N : Rn → Rn is a norm-preserving linear map. By
the result of Exercise 1.7(b), we therefore see that N is bijective with norm-preserving inverse
N −1 : Rn → Rn .
Define the composition map A : Rn → Rn by the rule A = T ◦ N −1 . Then A is bijective since T and
N −1 are. Note N is angle-preserving, since a 1-1 norm-preserving map is angle-preserving. Thus
the composition map A = T ◦ N −1 is angle-preserving.
Note that A(T (xj )) = T (N −1 (T (xj ))). To find N −1 (T (xj )), let N (y) = T (xj ), with y = b1 x1 +
· · · + bn xn ∈ Rn . Then we have
n
T (xj ) =
i=1
bi |xi |
T (xi ).
|T (xi )|
By linear independence we must then have bj = |T (xj )| / |xj | and bi = 0 for all i = j. Thus we
have y = (|T (xj )| / |xj |)xj . This then shows that
A(T (xj )) = T (y) = λj T (xj ),
where λj = |T (xj )| / |xj |. Since we know this map is angle preserving and since T (x1 ), . . . , T (xn )
is an orthogonal basis of Rn , we therefore know from part (b) that |λ1 | = · · · = |λn |.
Thus we have shown the forward implication in our if and only if statement; the reverse direction
is far easier. Let A : Rn → Rn be an angle-preserving linear map, N : Rn → Rn a norm-preserving
linear map, and let T : Rn → Rn be the map T = A ◦ N . Since a norm-preserving linear map is
trivially seen to be angle-preserving, we then see the composition T = A◦N is also angle-preserving.
Thus we have shown that T : Rn → Rn is an angle-preserving linear map if and only if T = A ◦ N ,
where
• N : Rn → Rn is a norm-preserving linear map.
• A : Rn → Rn is an angle-preserving linear map of the form A(yj ) = λj yj for each y1 , . . . , yn ,
where this list is some orthonormal basis of Rn .
Note from part (b) that we must have |λ1 | = · · · = |λn |.
1.9. If 0 ≤ θ < π, let T : R2 → R2 have the matrix
preserving and if x = 0, then ∠(x, T x) = θ.
cos θ sin θ
. Show that T is angle
− sin θ cos θ
Answer. Let us first show that T : R2 → R2 is injective. Let x, y ∈ R2 with T (x) = T (y). If
y1
x1
, we then have
and y =
x=
y2
x2
x1 cos θ + x2 sin θ
−x1 sin θ + x2 cos θ
=
y 1 cos θ + y 2 sin θ
.
−y 1 sin θ + y 2 cos θ
1 Functions on Euclidean Space
8
This leads to the equations
x1 − y 1 cos θ + x2 − y 2 sin θ = 0
− x1 − y 1 sin θ + x2 − y 2 cos θ = 0.
The second equation implies that x1 − y 1 sin θ = x2 − y 2 cos θ. Multiplying the first equation
by sin θ and subs uting this result then gives
x2 − y 2 cos2 θ + x2 − y 2 sin2 θ = 0
Since sin2 θ + cos2 θ = 1, this just says that x2 = y 2 must hold. Thus we must also have (x1 −
y 1 ) cos θ = 0 and (x1 − y 1 ) sin θ = 0. Since the cosine and sine of an angle cannot simultaneously
both be zero, we therefore see that x1 − y 1 = 0 must hold, and thus, T is injective. As an aside, it
must therefore be bijective since it is a linear mapping of a finite-dimensional vector space R2 into
itself.
Let x =
x1
x2
∈ R2 . Then we have
|T (x)|2 = (x1 cos θ + x2 sin θ)2 + (−x1 sin θ + x2 cos θ)2
= x2 + x2 = |x|2 .
2
1
Hence, we see the automorphism T : R2 → R2 is norm preserving. By Problem 1.8, it must also
then be angle preserving.
Letting x = 0, we have
T (x), x
= x1 (x1 cos θ + x2 sin θ) + x2 (−x1 sin θ + x2 cos θ)
=
x2 + x2 cos θ.
1
2
Thus we then have
T (x), x
= cos θ.
|T (x)| |x|
Since the inverse cosine is 1-1 on the half-open interval [0, π), we must then have
θ = arccos
T (x), x
|T (x)| |x|
= ∠(T (x), x).
1.10. If T : Rm → Rn is a linear transformation, show that there is a number M such that |T (h)| ≤
M |h| for h ∈ Rm .
Proof. Let A be the real n × m matrix of the linear transform T ; e.g., for all h ∈ Rm we have
T (h) = Ah. Let A have entries aij and let K be the maximum of |aij |. Letting y = Ah have entries
yi = m aik hk we then see that
k=1
n
|yi | ≤ mK
Thus we then see that
Setting M =
|hk | ≤ mKn |h| .
n
|T (h)|
mKn3/2
k=1
2
=
i=1
2
yi ≤ m2 K 2 n3 |h|2 .
and taking square roots then shows that |T (h)| ≤ M |h| for all h ∈ Rm .
1 Functions on Euclidean Space
9
1.11. If x, y ∈ Rn and z, w ∈ Rm , show that (x, z), (y, w) = x, y + z, w and |(x, z)| =
|x|2 + |z|2 . Note that (x, z) and (y, w) denote points in Rm+n .
Proof. Let x = x1 e1 +· · ·+xn en , y = y 1 e1 +· · ·+y n en , w = w1 e1 +· · ·+wm em , z = z 1 e1 +· · ·+z m em ,
u = (x, z) = x1 e1 + · · · + xn en + z 1 en+1 + · · · + z m en+m , and v = (y, w) = y 1 e1 + · · · + y n en +
w1 en+1 + · · · + wm en+m . Then we have
(x, z), (y, w)
= x1 y 1 + · · · + xn y n + z 1 w 1 + · · · + z m w m =
x, y + z, w .
Taking y = x and w = z then shows that
|(x, z)| =
x, x + z, z
|x|2 + |z|2 .
=
1.12. Let (Rn )∗ denote the dual space of the vector space Rn . If x ∈ Rn , define φx ∈ (Rn )∗ by
φx (y) = x, y . Define T : Rn → (Rn )∗ by T (x) = φx . Show that T is a 1–1 linear transformation
and conclude that every φ ∈ (Rn )∗ is φx for a unique x ∈ Rn .
Proof. Let x, y, z ∈ Rn and a ∈ R. Then we have
T (ax + by)(z) =
=
ax + by, z
ax, z + by, z
= a x, z + b y, z
= a T (x)(z) + b T (y)(z).
Thus the map T : Rn → (Rn )∗ is linear.
Let x ∈ ker T so that T (x)(z) = 0 for all z ∈ Rn . Then T (x)(x) = φx (x) = x, x = 0 must hold,
which shows that x = 0. Thus T is injective.
Let φ ∈ (Rn )∗ so that φ : Rn → R is a linear map. Set λj = φ(ej ) for each index j. Let
z = z 1 e1 + · · · + z n en ∈ Rn . Then we must have
φ(z) = z 1 φ(e1 ) + · · · + z n φ(en )
= z 1 λ1 + · · · + z n λn
=
z, x
=
x, z
= φx (z),
where x = λ1 e1 + · · · + λn en . Thus any map φ ∈ (Rn )∗ must be of the form φ = φx for some
x ∈ Rn . Note such x is unique by the injectivity of T : Rn → (Rn )∗ .
1.13. If x, y ∈ Rn , then x and y are called perpendicular (or orthogonal) if x, y = 0. If x and
y are perpendicular, prove that |x + y|2 = |x|2 + |y|2 .
Proof. From Theorem 1-2(4) we know that |x + y|2 = x + y, x + y . Bilinearity shows that x +
y, x + y = x, x + x, y + y, x + y, y . Since x, y = y, x holds, we then have
|x + y|2 =
x, x + 2 x, y + y, y
= |x|2 + 2 x, y + |y|2 .
Since x and y are orthogonal, x, y = 0 holds and the main result thus follows.
1 Functions on Euclidean Space
10
1.14. Prove that the union of any (even infinite) number of open sets is open. Prove that the
intersection of two (and hence of finitely many) open sets is open. Give a counterexample for
infinitely many open sets.
Proof. Let (Uα ) be a collection of open subsets of a topological space and let U = α Uα . If x ∈ U
there is some index α such that x ∈ Uα . Since the set Uα is open, x must be an interior point of
it. Thus there is some open set V ⊂ Uα with x ∈ V . Since Uα ⊂ U , we see that V ⊂ U also holds
so that x is also an interior point of U . Thus U must be open. Since a Euclidean space Rn is a
topological space, this result holds in that special case.
Now let U, V be open subsets of a metric space X. If U , V are disjoint, then U ∩ V = ∅ is open.
Consider the case that there is some x ∈ U ∩ V . Since U, V are open, there must be some open
balls U ′ = {y ∈ X : d(y, x) < r1 } and V ′ = {y ∈ X : d(y, x) < r2 } such that U ′ ⊂ U and V ′ ⊂ V .
If we take r = min{r1 , r2 } and W = {y ∈ X : d(y, x) < r}, we then see that W ⊂ U ′ ∩ V ′ ⊂ U ∩ V
so that x is an interior point of U ∩ V . Thus U ∩ V is open. Taking X = Rn and d(y, x) = |y − x|
shows the result holds in Euclidean space.
1 1
Note this result need not hold for infinite intersections. Consider the open segments Un = − n , n
of R1 . Then we have ∞ Un = {0}, which is not open in the real line.
n=1
1.15. Prove that U = {x ∈ Rn : |x − a| < r} is open (see also Problem 1-27).
Proof. Choose x ∈ U and let V be the set of all y ∈ Rn with |y − x| < s, where we have set
s = r − |x − a|. Letting y ∈ V , we then have
|y − a| ≤ |y − x| + |x − a| < s + |x − a| = r
so that V ⊂ U . Thus we have shown that U is open in Rn .
1.16. Find the interior, exterior, and boundary of the sets
A = {x ∈ Rn : |x| ≤ 1}
B = {x ∈ Rn : |x| = 1}
C = {x ∈ Rn : each xi is rational}.
Answer. The answers are pretty easy to see for A and B: int(A) = {x ∈ Rn : |x| < 1}, int(B) = ∅,
∂A = ∂B = B, ext(A) = {x ∈ Rn : |x| > 1}, and ext(B) = Rn − B = {x ∈ Rn : |x| < 1} ∪ {x ∈
Rn : |x| > 1}.
Note int(C) = ∅ since the irrationals are dense in the reals (and thus, so are the n-tuples of
irrationals in Rn ). Since the rationals and irrationals are both dense in the reals, any real n-tuple
x = (x1 , . . . , xn ) is arbitrarily close to a rational n-tuple q = (q 1 , . . . , q n ) and also to an irrational
n-tuple y = (y 1 , . . . , y n ). Thus we have ∂C = Rn . Note the density of the rationals in the reals
shows that any nonempty open rectangle in Rn contains an n-tuple of rationals, so ext(C) = ∅
must hold.
^
My head hurts.
baseline bum with the goods
"bijective since it is a linear mapping of a finite-dimensional vector space R2 into
itself."
Baseline bum: "Does this mean what my daddy used to say when I asked to borrow his car"? GFY If so, then you premise is correct but the conclusion is messy.
sheesh baseline. My brain is spinning.
I hate to tell you this but there's an error in that equation
Yeah and the world will end this year because the Mayans are always right.
We're ed
Isn't it "Wojnakowski"?
Moot point since x = 1.
BB taking a giant dump on this thread tbh.
Should've used LaTeX.
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