Math 312 Spring08 Day 13
HW 3: Due Monday May 19th.
11. 1 # 2, 3, 6, 10
11. 2 # 1, 3, 6
11.1 Continuous Functions and Mappings.
1. Definition. Let A be a subset of Rn
i. A mapping F: A → Rn is said to be continuous at the
point u in A provided that whenever a sequence {uk} in A
converges to u, the image sequence {F (uk)} converges
to F(u).
ii. A mapping F: A → Rn is said to be continuous provided
that it is continuous at every point in its domain.
2. Proposition 11.1 For each index i with 1 ≤ i ≤ n, the ith
component projection function pi: Rn → R is continuous.
Proof: Follows directly from the Componentwise Convergence
Criterion (Do you see how?)
3. Theorem 11.3 Let A be a subset of Rn that contains the point u
and suppose that the functions h: A → R and g: A → R are
both continuous at u.
i. For any real numbers α and β, the function
α h + βg: A → R is continuous at u.
ii. Also the product function hg: A → R is continuous at u.
iii. Furthermore if g(v) ≠ 0 for all v in A, then the quotient
function h/g: A → R is continuous at u.
Proof: Just like in the old days, these follow directly from the
sum, product, and quotient properties for sequences of real
numbers.
4. Theorem 11.5 Let A be a subset of Rn that contains the point
u. Suppose that the mapping G: A → Rm is continuous at the
point u. Let B be a subset of Rm with G(A) ⊆ B and suppose
that the mapping H: B → Rk is continuous at the point G(u).
Then the composition H ! G: A → Rk is continuous at u.
Proof: Let {uk} be a sequence in A that converges to the point
u…
5. Definition. Given a mapping F: A → Rm, where A is a subset
of Rn, and an index i with 1 ≤ i ≤ m, we define the function
Fi: A → R, to be the composition of F with the ith component
projection function. We call the function Fi the ith component
function of F. Thus F(u) = (F1(u), …, Fm(u))
6. Theorem 11.9 (The Component Continuity Criterion) Let A
be a subset of Rn that contains the point u. The mapping F =
(F1, …, Fm) : A → Rm is continuous at u if and only if each of
its component functions Fi: A → R is continuous at u.
Proof: Follows directly from the Componentwise
Convergence Criterion
7. Corollary 11.10 Let A be a subset of Rn that contains the point
u and suppose that the mappings G: A → Rm and H: A → Rm
are both continuous at the point u. Then for any real numbers
α and β, the mapping
α H + βG: A → Rm is continuous at u.
Proof: Follows directly from (you guessed it!) the
Componentwise Convergence Criterion (and Theorem 11.3).
8. Theorem 11.11 Let A be a subset of Rn that contains the point
u. The following assertions about F: A → Rm are equivalent:
i. The mapping F: A → Rm is continuous at the point u;
that is, for a sequence {uk} in A,
lim dist(F(u k ), F(u)) = 0 if lim dist(u k , u) = 0 .
k!"
k!"
ii. For every ε > 0, there exists a δ > 0 such that for every v
in A, dist(F(u), F(v)) < ε if dist(u, v) < δ.
Proof: Let F: A → Rm. Suppose that ii) is not true. Then there
exists an ε > 0 for which no adequate δ > 0 exists.
Then in particular, for every k in N, there is a point vk in A with
the property that dist(u, vk) < 1/k but dist(F(u), F(vk)) ≥ ε.
First note that the sequence {vk} in A clearly converges to u, so
lim dist(v k , u) = 0 .
k!"
However, since for each index k, dist(F(vk), F(u)) ≥ ε, it is just
as clear that lim dist(F(v k ), F(u)) # 0 . Thus i) is not true.
k!"
So by contrapositive, we have shown that i) ⇒ ii).
Now suppose that ii) is true and let {uk} be a sequence in A for
which lim dist(u k , u) = 0 . Let ε > 0. By ii) there exists a δ > 0
k!"
such that for every v in A, dist(u, v) < δ ⇒ dist(F(u), F(v)) < ε.
Now choose K such that k≥K⇒ dist(u k , u) < ! . Then k≥K⇒
dist(F(uk), F(u)) < ε. Therefore lim dist(F(u k ), F(u)) = 0 . Thus
k!"
i) is true. So we have proven that i) and ii) are equivalent.
Theorem 11.12 Let O be an open subset Rn of and consider the
mapping F: O → Rm. The following are equivalent.
i. The mapping F: O → Rm is continuous.
ii. F -1(V) is an open subset of Rn whenever V is an open
subset of Rm.
Note: F -1(V) = {u in O: F(u) is in V}, there is NO assumption
that F is invertible.
Proof: Saved for Wednesday
11. 2 Sequential Compactness, Extreme Values, and Uniform
Continuity
1.
Definition: Let A be a subset of Rn. Then A is said to be
sequentially compact provided that every sequence in A has a
subsequence that converges to a point in A.
2.
Definition: A subset A of Rn is said to be bounded provided
that there is a number M such that u ! M for all points u in
A.
3.
Theorem (Balzano-Weierstrass Theorem) A subset of Rn is
sequentially compact if and only if it is closed and bounded in
Rn.
Proof: Suppose the set A of Rn is sequentially compact. Then A
is closed since given any sequence in A that converges, this
sequence must have a subsequence that converges to a point in A
(so the limit of the whole sequence is a point in A).
Now suppose that A is not bounded. Then for any natural
number, k, there is a point uk in A that has a norm bigger than k.
But then the sequence {uk}has a subsequence { u ki } that
converges to some point u in A. But then the sequence of real
numbers u ki converges to u . This is a contradiction since the
sequence u ki is unbounded.
Now suppose that the subset A of Rn is closed and bounded. Let
{uk} be a sequence in A. Since the sequence {uk} is bounded, it
has a convergent subsequence (generalization – using induction
– of the fact that a bounded sequence of real numbers has a
convergent subsequence). Now since A is closed, this
subsequence must converge to an element of A. Thus A is
sequentially compact.
4.
Theorem 11.20 Let A be a subset of Rn and suppose that the
mapping F: A → Rm is continuous. If the domain A is
sequentially compact, then the image F(A) is also sequentially
compact.
Proof: To do on Wednesday
5. Lemma 11.21 Every nonempty sequentially compact subset of
R has a smallest and largest member.
Proof: Let A be a nonempty sequentially compact subset of R.
Since A is bounded, it has a greatest lower bound (call it a) and a
least upper bound (call it b). Now for each k in N, there exists an
element ak in A such that a ≤ ak < a +1/k and an element bk in A
such that b -1/k < bk ≤ b. Note that the sequence {ak} converges
to a, and the sequence {bk} converges to b. These sequences are
both in A, so since A is closed, a and b are in A as well!
Theorem 11.22 (The Extreme Value Theorem) Let A be a
nonempty sequentially compact subset of Rn and suppose that
the function f:A → R is continuous. Then the function f:A → R
attains a smallest and largest value.
Proof. f(A) is sequentially compact by Theorem 11.20. But then
by Lemma 11.21, since f(A) lives in R, it has a smallest and
largest element (thus f attains a smallest and largest value).
6.
Definition. A nonempty subset A of Rn is said to have the
Extreme Value Property provided that every continuous
function f: A → R has a maximum and a minimum functional
value.
7.
Theorem 11.24 A nonempty subset of Rn has the Extreme
Value Property if and only if it is sequentially compact.