Solutions to suggested problems for exam 3
8.2.2a: The hyperplane must be of the form b1 x1 + b2 x2 + b3 x3 + b4 x4 = d for
the correct bi s and d. Plugging in the given points, the coecients must
satisfy the system
b1
2b1 + b2
= d
= d
b2 + b3
4b2 + b4
=
Note: this example works very well in polar coordinates. x2xy 2
+y
turns into sin cos if you put it into polar coordinates. Since theres
no r dependence, you get dierent limits if you approach the origin
along dierent rays.
Beware! This idea of going in a
for every sequence in V cfw_a (V is an open ball centered at a) for
which limk xk = a.
Proof (should look familiar, its the same as in R:
": Suppose that limxa f (x) = L. Take xk approaching a,
and take > 0. There is a > 0 so that 0 < x a < implies f (x)
Heine-Borel Theorem (9.2)
One particularly important type of set in R was closed, bounded
intervals (e.g., a continuous function dened on a closed, bounded set
was bounded, attained a maximum and minimum). The generalization
to Rn is closed, bounded set,
Limits of functions (9.3)
First o, were thinking of vector-valued functions, i.e., f : U Rm ,
U Rn .
Example:
Let f (x, y) = x2xy 2 . This is dened for all (x, y) = (0, 0), so domain
+y
2
U of f is U = R (0, 0). (Note: I should probably be writing f x
y
i
cover H. This will be based on the set T of open balls with rational
radii and centers in Qn . Then well show that this countable subcover
actually admits a nite subcover. This will be based on the BolzanoWeierstrass Theorem.
Proof of Borel Covering Lemma
Then k N implies k Ni for all i. Hence
(i)
xk a n max xk a(i) < n = .
1in
n
Example:
k 1 k
lim
= (1, 0) ,
,e
k
k
since k1 1 and ek 0 as k .
k
Main dierence between sequences in R and those in Rn is that you
cant talk about monotonicity (e.g., "a bounded
(1) Take x E. Suppose BWOC that > 0 so that B (x) E =
. Then the complement of B (x) is a closed set (by denition of closed) which contains E. Hence [B (x)]c contains
E, a contradiction. Conversely, suppose that x Rn has that
property that > 0, B (x) E =
Interior, closure, and boundary (8.4)
More topological denitions: we want to make precise the ideas of
"insides" and "edges" of a set.
Denition: Given an arbitrary set E Rn , we dene the interior
of E, written E o , by
Eo =
cfw_V : V is open, and V E
and
Proposition: f : E R is continuous i for every open set , we
have that f 1 () is an open set intersected with E.
This leads to the idea of a relatively open set, which Ill dene for
sets in Rn .
Denition: Suppose that E R. A set U is relatively open in E
i
Anyway, back to the theorem:
Proof:
(1) Suppose that the collection of open sets is V , A. Take
x A V . Then x is in at least one of the V s, call it
V0 . Since V0 is open and contains x, there is an > 0 so
that B (x) V0 . But V0 A V , so B (x) A V
as wel
For Rn , its a little suspicious to just quote the above proof, since
we cant really visualize R17 for example. So, for non-zero vectors a,
b in Rn , we dene the angle between them to be arccos aabb .
Since, by the Cauchy-Schwartz inequality, the fraction
Algebraic structure of Rn (8.1)
Now were starting multivariable advanced calculus. We dene Rn
to be the set of n-tuples of reals. This is a vector space, in the sense
that you can add two elements of Rn in a sensible way to get an element
of Rn , and you
other side will be similar. Anyway, we consider
b
b
f (x, y0 + h) dx a f (x, y0 ) dx
a
lim
h0+
h
b
f (x, y0 + h) f (x, y0 )
dx,
= lim+
h0
h
a
where wed really like to be able to take the limit inside the integral.
Its not that easy, though. We can apply
Proof: Hold x and x0 xed, and dene a function g(z) by
(16)
n1 (k)
f (x0 )
g(z) = f (z)
(zx0 )k (zx0 )n = f (z)Pn1 (z)(zx0 )n ,
k!
k=0
where is chosen so that g(x) = 0. For this to occur, the value of
must satisfy
n1
0 = f (x)
k=0
f (k) (x0 )
(x x0 )k (x
Analytic Functions (7.4)
Weve seem that if f equals a power series, i.e., f (x) = ak (x
k=1
x0 )k with radius of convergence R > 0, then f has derivatives of all
orders on the interval (x0 R, x0 +R), which we denote by f C (x0
R, x0 + R). Two questions n
Power Series (7.3)
In this section we will look at a special type of series of functions.
Denition: A power series in x is a series of the form ak xk ,
k=0
where cfw_ak is a sequence of real constants. A power series in x x0 is
a series of the form ak (x
(x0 , x0 + ), but of course it converges at every point of (x0 , x0 + ).
It follows that f is dierentiable on (x0 , x0 + ), with
d
f (x) =
dx
ak (x x0 )k
=
k=0
(13)
k=0
d
ak (x x0 )k
dx
kak (x x0 )k1
=
k=1
for all x (x0 , x0 + ). (Weve dropped the k = 0 t
Recall that, back in section 8.3, I proved that f : R R is continuous i f 1 () for every open set . Its time to generalize. First we
need a lemma, though. This says that there is a statement about relatively open sets which is analogous to the denition of
Theorem: Continuous images of connected sets are connected, i.e.,
if E Rn with E connected, and if f : E Rm is continuous, then
f (E) is connected.
Proof: Ill prove the contrapositive, i.e., if f (E) is not connected,
then E is not connected. So, suppose
Solutions to Assignment # 8
9.4.3: From Theorem 9.26, f is continuous on A i f 1 (V ) is relatively open
in A for every open set V . But since A is open, f 1 (V ) is relatively open
in A i f 1 (V ) is an open subset of Rn , concluding the proof.
9.4.6: Il
Solutions to Assignment 9
11.2.5: Note that the problem doesnt actually ask you to show that the function is not dierentiable for other . Away from the origin,
(
) (
) (
)1
4x3 x2 + y 2 x4 + y 4 x2 + y 2
2x
,
fx =
2 + y 2 )2
(x
and
fy =
(
) (
) (
)1
4y 3
Solutions to assignment # 7
9.2.1: Suppose that E is closed. Suppose that V , A, is an open cover for
E. Then, since E c is open, adding E c to the collection V gives an open
cover for K. Since K is compact, there is a nite subcover of K: V1 ,
, VN , and
Math 465 Lecture Notes
Brendan Hassett April 14-19, 2004
9
9.1
Cohomology of projective varieties
Cohomology of twisting sheaves
Denition 9.1 Let F be a coherent OPn -module. For each integer d, the Serre twist F(d) is dened by the gluing data F(d)|Uj :=
Math 465 Lecture Notes
Brendan Hassett January 30-February 11, 2004
4
4.1
Modules
Basic denitions
In the linear algebra of arbitrary rings, modules play the rle of vector spaces: o Denition 4.1 Let A be a ring. An A-module M is an additive abelian group M
Math 465 Lecture Notes
Brendan Hassett March 31-April 7, 2004
7
7.1
Cech cohomology
An obstruction computation
Let X be a topological space and consider an exact sequence of sheaves of abelian groups on X 0 K F G 0. Taking global sections gives an exact
Math 465 Lecture Notes
Brendan Hassett April 7-12, 2004
8
Cohomology on ane varieties
In this section, we consider algebraic varieties with the Zariski topology. The main reference is [Se].
8.1
Covering lemmas
Lemma 8.1 (Quasi-compactness lemma) Let X be
Math 465 Lecture Notes
Brendan Hassett January 26-28, 2004
3
3.1
Localization
Fundamental open subsets as ane varieties
D(x1 ) = cfw_a1 A1 (k) : a1 = 0 A1 (k).
Example 3.1 Consider the fundamental open subset
We show how this can be given the structure of
Math 465 Lecture Notes
Brendan Hassett January 23, 2004
2
2.1
Zariski topology
Abstract topological spaces
Denition 2.1 A topology on a set X is a collection of subsets U X, called open subsets, with the following properties: 1. and X are open; 2. the int
Math 410.500
Exam 3
4/20/12
1. (10 pts.) Dene the term in italics:
(a) the norm of a linear transformation T : Rm Rn . (We saw several denitions;
any of these is okay.) Solution: T = supx=1 T (x), or T = supx=0 T (x) , or
x
T = inf cfw_C > 0 : T (x) C x a