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Analysis in Several Variables
Math 140C—Fall 2006
Bernard Russo
December 2, 2006
Contents
1 Friday September 22—Course information; Schwarz inequality (As- signment 1) 1 1.1 Course Information............................ 1 1.2 Schwarz inequality............................ 2
2 Monday, September 25—The triangle inequality and open sets (As- signment 2) 4 2.1 The triangle inequality.......................... 4 2.2 Open sets................................. 4
3 Wednesday September 27—More on open sets; closed sets, bound- ary, and closure (Assignments 3,4,5,6) 5 3.1 The final word on open sets (just kidding!)............... 5 3.2 Closed sets................................. 6 3.3 Boundary and closure........................... 7
4 Friday, September 29—more on closed sets; cluster points (Assign- ment 7) 7 4.1 More on closed sets and boundary.................... 7 4.2 Cluster points............................... 8 4.3 Proof of Proposition 3.8 ((vii) on page 32 of Buck).......... 9
5 Monday October 2—Compactness I (Assignment 8) 10 5.1 Bolzano-Weierstrass and Heine-Borel properties............ 10 5.2 HB implies closed............................. 11
6 Wednesday October 4—Compact sets II 11 6.1 Equivalence of HB and BW....................... 11
7 Friday October 6—Compactness III 12 7.1 Corrected proof of Step 3 of Proposition 6.1.............. 12
i
11 Monday October 16—Differentiability implies continuity I (Assign-
19 Friday November 3—More on existence of differentials (Assignments
- 7.2 Closed and Bounded implies Compact
- 8 Monday October 9—Continuity I (Assignment 9)
- 8.1 Overview
- 8.2 Continuous functions—continuous image of a compact set
- 9 Wednesday October 11—Continuity II (Assignment 10)
- 9.1 Limits of sequences of points in Rn
- 9.2 Continuity and limits of sequences
- 10 Friday October 13—Continuity III (Assignment 11)
- 10.1 Uniform continuity
- ment 12)
- 11.1 Motivation—one variable
- 11.2 Partial Differentiation
- 11.3 Differentiability (+ continuity) implies continuity
- 12 Wednesday October 18—Proof of Theorem 11.3
- 13 Friday October 20—First Midterm
- 13,14) 14 Monday October 23—More on closed sets and closure (Assignments
- 14.1 A discussion of closed sets and closure
- 14.2 A characterization of closed sets in terms of convergent sequences
- case of functions) 15 Wednesday October 25—Differential as a Linear approximation (the
- 15.1 Higher order partial derivatives
- 15.2 Linear Approximation
- 16 Friday October 27—Transformations (Assignments 15,16)
- 16.1 Transformations
- 17 Monday October 30—Uniqueness of the differential (Assignment
- 17.1 The case of functions
- 17.2 Coordinate free definition of derivative
- 18 Wednesday November 1—Existence of the differential
- 18.1 The case of functions
- 18.2 Differential as a linear approximation—the case of transformations
- 18,19)
- 19.1 Proof of Theorem 18.5
- 19.2 Two questions on differentials
- 20 Monday November 6—Chain rule for transformations
- 20.1 Composition of transformations; statement of chain rule
- 20.2 Proof of the one-dimensional chain rule
- (Assignments 20, 21) 21 Wednesday November 8—Proof of the chain rule for transformations
- 21.1 Two lemmas
- 21.2 Proof of the chain rule
- 22 Friday November 10—holiday (Veteran’s Day)
- orems 23 Monday November 13—How to use the chain rule; mean value the-
- 23.1 An application of the chain rule
- 23.2 Mean Value Theorems
- rem; local invertibility (Assignment 23) 24 Wednesday November 15—Applications of Big Mean Value Theo-
- 24.1 Alternate proof of linear approximation for differentiable transformations
- 24.2 The local invertibility theorem
- 25 Friday November 17—Implicit Function Theorem I (Assignment 24)
- 25.1 Motivation
- 25.2 Implicit function theorems
- 26 Monday November 20—Implicit Function Theorem II (Assignment
- signment 26) 27 Wednesday November 22—Proof of Open Mapping Theorem (As-
- 28 Friday November 24—holiday (Thanksgiving)
- 29 Monday November 27—Proof of Inverse Mapping Theorem
- 29.1 Automatic continuity of the inverse
- 29.2 The inverse function theorem
- 30 Wednesday November 29—Mixed Partials Theorem
- 30.1 Mixed Partials Theorem—weak version
- 30.2 Mixed Partials Theorem—strong version
- Course summary
- 31.1 Motivation and statement of the problem
- 31.2 The extension theorem
- 31.3 Course summary—from Buck (and the Minutes)
1 Friday September 22—Course information; Schwarz
inequality (Assignment 1)
1.1 Course Information
- Course: Mathematics 140C MWF 1:00–1:50 ET 204 FALL 2006 Webpage for the course: www.math.uci.edu/∼brusso
- Prerequisite: Math 140AB. Rigorous study of differentiation and integration of real-valued functions of one real variable. All of this can be found in the six chapters of the recent text for 140AB, namely, Elementary Analysis:The Theory of the Calculus, by Kenneth A. Ross. This includes the set of real numbers and the completeness axiom; sequences of real numbers, continuity, uniform continuity, sequences and series of functions, differentiation and integration up to the fundamental theorem of calculus.
- Instructor: Bernard Russo MSTB 263 Office Hours M 2:30-3:30 W 10:30-11: and by appointment (a good time for short questions is right after class just out- side the classroom; appointments can be arranged by email—[email protected])
- Discussion section: TuTh 1:00–1:50 HICF 100M
- Teaching Assistant: TBA
- Homework: There will be approximately 35 to 40 assignments with about one week’s notice before the due date. Most, but not all of these assignments will be from the textbook (Buck).
- Grading: The in-class exams are “closed book and notes.” Homework and take home midterm are “open book and notes”. First midterm (in class) October 20 (Friday of week 4) 20 percent Second midterm (take home) November 17 (Friday of week 8) 20 percent Final Exam (in class) December 6 (Wednesday) 40 percent Homework approximately 35-40 assignments 20 percent
- Holidays: November 10, 23, and 24
- Text: R. C. Buck, Advanced Calculus
- Material to be Covered. (Page numbers refer to the text Buck)
Schwarz inequality Theorem 1, page 13 (1 lecture) topology §1.5 pp 28–33: open, closed, boundary, interior, exterior, closure, neighborhood, cluster point (5 lectures) compactness §1.8 pp 64–67: Heine-Borel and Bolzano-Weierstrass properties (Theorems 25,26,27, page 65) (3 lectures)
Proof: Let Q := αp − βq where α and β are unspecified real numbers. From |Q|^2 ≥ 0 we obtain α^2 |p|^2 + β^2 |q|^2 − 2 αβp · q ≥ 0 for all α, β ∈ R.
Choosing α = |q| and β‖p|, we have 2|p||q|p · q ≤ 2 |q|^2 |p|^2 from which the theorem follows. 2
Assignment 1 (Due September 29)
- Read sections 1.2,1.3,1.4 in Buck (The lectures will continue with section 1.5). Do not waste your time reading about the concepts angle, orthogonal, hyper- plane, normal vector, line, convexity, which are discussed in section 1.3 of Buck. We have no immediate use for them. Thus, you may skip pages 15-18 and 21- for now.
- • Buck [§1.2 page 10 #5,10,23]
- Buck [§1.3 page 18 #1,2,5,6]
THINKING OUTSIDE THE BOX
- If xj and yj are infinite sequences, then
∑^ ∞
j=
xj yj ≤
∑^ ∞
j=
x^2 j
1 / 2
∑^ ∞
j=
y j^2
1 / 2 ,
provided the series on the left converges.
- If f and g are continuous functions on a closed interval [a, b], then from the Schwarz inequality applied to Riemann sums
∑^ n
j=
f (tj )g(tj )(xj − xj− 1 ) =
∑^ n
j=
f (tj )(xj − xj− 1 )^1 /^2 g(tj )(xj − xj− 1 )1/ 2
∑^ n
j=
f (tj )^2 (xj − xj− 1 )
1 / 2
∑^ n
j=
g(tj )^2 (xj − xj− 1 )
1 / 2 ,
you get the Schwarz inequality for functions
∫ (^) b a f^ (x)g(x)^ dx^ ≤^
(∫ b a f^ (x) (^2) dx
) 1 / 2 (
∫ g(x)^2 dx)^1 /^2.
∫ (^) b a f^ (x)g(x)^ dx, then^ f^ ·^ g^ has the same properties as the scalar product p · q and the proof above of Theorem 1.1 applies word for word to give an alternate proof of the Schwarz inequality for functions.
2 Monday, September 25—The triangle inequality
and open sets (Assignment 2)
2.1 The triangle inequality
Here is an important consequences of the Schwarz inequality.
Corollary 2.1 (Triangle Inequality) For any two vectors p, q, |p + q| ≤ |p| + |q|
Proof: |p+q|^2 = (p+q)·(p+q) = p·p+p·q+q·p+q·q ≤ |p|^2 +2|p||q|+|q|^2 = (p|+|q|)^2. 2
2.2 Open sets
A very important type of subset of Rn^ is a ball. An open ball is defined, for a given point p ∈ Rn^ and r > 0 by
B(p, r) := {q ∈ Rn^ : |p − q| < r}.
The center of B(p, r) is p and the radius is r. Today we want to prove (the two statements):
Triangle inequality ⇒
{ open ball is open set
} ⇒
{ characterization of interior
}
Definition 2.2 Let S ⊂ Rn^ and q ∈ Rn. The point q is interior to S if there exists δ > 0 such that B(q, δ) ⊂ S. The interior of S is the set of all points which are interior to S, notation int S, that is
int S = {q ∈ Rn^ : ∃δ > 0 such that B(q, δ) ⊂ S}.
Finally, S is an open set if S = int S.
Proposition 2.3 Let p ∈ Rn^ and r > 0. Then the ball B(p, r) is an open set.
Proof: Let x ∈ B(p, r) so that |x − p| < r. Choose δ := r − |x − p|. Then the triangle inequality implies that B(x, δ) ⊂ B(p, r), showing that every point of B(p, r) is an interior point of B(p, r).
MIDTERM ALERT: It is very important that the 10 propositions (i)- (x) on page 32 of Buck be mastered before the first miderm. Here is one of them.
Proposition 2.4 ((vi) on p.32 of Buck) Let S be any non-empty subset of Rn. Then int S is the largest open subset of S; more precisely
(a) int S is an open set;
Proof: Take ϕ(u) = up−^1 in the theorem. 2
Corollary 3.3 (H¨older Inequality) Let x 1 ,... , xn and y 1 ,... , yn be real numbers and let p ∈ (1, ∞). Then with q := p/(p − 1),
∑^ n j=
|xj yj | ≤
∑^ n j=
|xj |p
1 /p ∑^ n j=
|yj |q
1 /q .
Proof: Take a = |xj |/‖x‖p and b = |yj |/‖y‖p in the corollary, where ‖x‖p denotes (∑ n j=1 |xj^ | p ) 1 /p
. 2
Assignment 4 (Due October 13) Give a rigorous proof of Theorem 3.1. More precisely,
Step 1 First establish, for c ∈ [0, ∞), the formula
∫ (^) c
0
ϕ(u) du +
∫ (^) ϕ(c)
0
ψ(v) dv = cϕ(c). (2)
Step 2 Use (2) to prove (1).
Step 3 Prove the “moreover” statement.
Remark 3.4 Every open set in Rn^ is the union of (not necessarily disjoint) open balls.
Assignment 5 (Due October 13)
- Show that in R^1 , the open balls can be assumed to be disjoint
- Show that every open set in Rn^ is the union of a countable collection of open balls. (Hint: The answer is somewhere in the minutes for my 140C class of Fall
Here are the first two propositions on page 32 of Buck. The proofs are written out in detail in Buck on pages 32–34. (i) If A and B are open sets, then so are A ∩ B and A ∪ B. (ii) If {Aα : α ∈ I} is an arbitrary family of open sets, then ∪α∈I Aα is an open set.
3.2 Closed sets
Definition 3.5 A subset S of Rn^ is said to be a closed set if its complement Rn^ \ S is an open set.
Remark 3.6 Assignment 3 shows that the set {q ∈ Rn^ : |q − p| ≤ r} is a closed set for any p ∈ Rn^ and r > 0. Needless to say, we call such a set a “closed ball”.
In order to facilitate the study of closed sets, we recall De Morgan’s laws. If {Aα : α ∈ I} is an arbitrary family of sets, then
Rn^ \ ∪α∈I Aα = ∩α∈I (Rn^ \ Aα)
and Rn^ \ ∩α∈I Aα = ∪α∈I (Rn^ \ Aα). Using De Morgan’s laws we obtain immediately from (i) and (ii) the following propositions ((iii) and (iv)) on page 32 of Buck. From the definition of closed set, (v) is obvious, and (vi) has already been proved in Proposition 2.4 above.
(iii) If A and B are closed sets, then so are A ∩ B and A ∪ B. (iv) If {Aα : α ∈ I} is an arbitrary family of closed sets, then ∩α∈I Aα is a closed set. (v) A set is open if and only if its complement is closed.
3.3 Boundary and closure
Definition 3.7 Let S ⊂ Rn^ and let p ∈ Rn. We say that p is a boundary point of S if every ball with center p meets both S and its complement Rn^ \ S, that is, for every δ > 0, B(p, δ) ∩ S 6 = ∅ and B(p, δ) ∩ (Rn^ \ S) 6 = ∅. The boundary of S, denoted by bdy S, is the set of all boundary points of S. The closure of S, notation S is defined to be S ∪ bdy S.
The following proposition is the analog for closed sets of (vi) on page 32 of Buck. It will be proved in the next lecture.
Proposition 3.8 ((vii) on p.32 of Buck) Let S be any subset of Rn. Then S is the smallest closed set containing S. (you know what this means.)
Assignment 6 (Due October 6) Prove the following assertions:
(a) int S = ∪{G : G is open , G ⊂ S}
(b) S = ∩{F : F is closed , S ⊂ F }
4 Friday, September 29—more on closed sets; clus-
ter points (Assignment 7)
4.1 More on closed sets and boundary
We already mentioned the next proposition last time.
Proposition 4.1 ((iii) and (iv) on p.32 of Buck)
(a) If A and B are closed subset of Rn, then so are A ∩ B and A ∪ B.
Proposition 4.6 ((ix) on p.32 of Buck) Let S be any subset of Rn. Then S is a closed set if and only if every cluster point of S belongs to S.
Proof: Step1: If S is a closed set, then every cluster point of S must belong to S.
Proof: Indirect. Suppose p is a cluster point of the closed set S. If p 6 ∈ S, then since Rn^ \ S is open, there exists a ball B(p, δ) ⊂ Rn^ \ S, that is, B(p, δ) ∩ S = ∅. But B(p, δ) ∩ S is an infinite set, contradiction, so step 1 is proved.
Step 2: If a set S contains all of its cluster points, then S is a closed set.
Proof: Let S be a set containing all of its cluster points. We shall show that Rn^ \S is open. Let p ∈ Rn^ \ S, that is, p 6 ∈ S. It follows from our assumption that p is not a cluster point of S. This means that for some δ > 0, the set B(p, δ) ∩ S consists of only finitely many points, say p 1 ,... , pm. Since these points are in S and p 6 ∈ S, if we set δ′^ = min{|p − pk| : 1 ≤ k ≤ m},
then δ′^ > 0. Moreover, B(p, δ′) ∩ S = ∅, that is, B(p, δ′) ⊂ Rn^ \ S. Thus Rn^ \ S is open, and S is closed. Step 2 is proved.
Steps 1 and 2 constitute a proof of Proposition 4.6. 2
Assignment 7 (Due October 6) [Buck §1.5 page 36 #2,6,10,11]
4.3 Proof of Proposition 3.8 ((vii) on page 32 of Buck)
Proof of Proposition 3.8^2 Step 1: S is a closed set. Proof: We have to prove that the complement Rn^ \ S is an open set, so let q ∈ Rn^ \ S. We must find a ball B(q, δ) ⊂ Rn^ \ S. Since q 6 ∈ S = S ∪ bdy S, q 6 ∈ S and q 6 ∈ bdy S. The latter implies that there is a δ > 0 such that either B(q, δ) ∩ S = ∅ or B(q, δ)∩(Rn^ \S) = ∅. The point q belongs to the latter set, so for sure B(q, δ)∩S = ∅, that is, B(q, δ) ⊂ Rn^ \ S. We complete the proof of Step 1 by showing that in fact B(q, δ) ⊂ Rn^ \ S. If this were not true, there would be a point q′^ ∈ B(q, δ) ∩ S. Since B(q, δ) ⊂ Rn^ \ S, in fact we have q′^ ∈ B(q, δ) ∩ bdy S. Since B(q, δ) is an open set, there is > 0 such that B(q′, ) ⊂ B(q, δ). Since q′^ is a boundary point of S, B(q′, ) ∩ S 6 = ∅, a contradiction. This proves that S is a closed set.
Step 2: If F is a closed set and S ⊂ F , then S ⊂ F.
Proof: Since S = S∪bdy S, and we are given that S ⊂ F , we have to show only that bdy S ⊂ F. Suppose that p ∈ bdy S and p 6 ∈ F. If we arrive at some contradiction, we will be done. Since F is closed, Rn^ \ F is open, so there exists δ > 0 such that B(p, δ) ⊂ Rn^ \ F , that is, B(p, δ) ∩ F = ∅. By the definition of boundary point, B(p, δ) ∩ S 6 = ∅. This is the desired contradiction, since B(p, δ) ∩ S ⊂ B(p, δ) ∩ F. Steps 1 and 2 constitute a proof of Proposition 3.8. 2 (^2) This proof was not done in class. Please make sure you read and understand it
5 Monday October 2—Compactness I (Assignment
5.1 Bolzano-Weierstrass and Heine-Borel properties
Definition 5.1 Let S be any subset of Rn.
BW S satisfies the Bolzano-Weierstrass property if every infinite sequence from S has a cluster point in S. In other words, if T = {p 1 , p 2 ,.. .} ⊂ S is infinite, then there exists a point p ∈ S such that for every δ > 0, B(p, δ) ∩ T is an infinite set.
HB S satisfies the Heine-Borel property if every open cover of S can be reduced to a finite subcover. In other words, if G is a collection of open sets and if S ⊂ ∪G∈G G, then there is a finite subset G 1 ,... , GN of G such that S ⊂ G 1 ∪ G 2 ∪ · · · ∪ GN.
EXAMPLES:
- (0, 1) does not satisfy BW or HB.
- [0, ∞) does not satisfy BW or HB.
- [0, 1] satisfies BW. This is the Bolzano-Weierstrass theorem, which you learned in Mathematics 140A or 140B. You can also find it in Buck [Theorem 21,p. 62].
- [0, 1] satisfies HB. This is [Theorem 24,p.65] in Buck..
We shall show that the two properties are equivalent, that is, an arbitrary set S ⊂ Rn either satisfies both properties or neither property. This will be stated in a proposition below.
Definition 5.2 Let S be any subset of Rn. We say S is compact if it satisfies HB.
Assignment 8 (Due October 13) Prove directly the following three assertions. The fourth assertion will be proved in class.
(a) If S satisfies BW, then S is a closed set.
(b) If S satisfies BW, then S is a bounded set.
(c) If S satisfies HB, then S is a bounded set.
(d) (This will be done in class, not part of the homework—it is included here for comparison purposes only) If S satisfies HB, then S is a closed set.
These assertions are stated in Buck as [§1.8 page 69 #1,2]
Again, since Q is dense in R, we can find a vector qp with rational coordinates such that qp ∈ B(p, rp/2). By the triangle inequality, B(qp, rp/2) ⊂ B(p, rp) (Check this!), so for each p ∈ S, we have p ∈ B(qp, rp/2) ⊂ Gp. The collection {B(qp, rp/2) : p ∈ S} is countable, so we can enumerate it as {B(qpj , rpj /2)}∞ j=1, where {pj } is a sequence of points in S. For each j = 1, 2 ,... pick the corresponding Gpj ∈ G. Then S ⊂ ∪∞ j=1Gpj.
Step 3: BW⇒ HB. Proof: Assume that S satisfies BW. By step 2, it suffices to prove that any count- able open cover of S can be reduced to a finite subcover. Let S ⊂ G 1 ∪ G 2 ∪ · · ·. We must find N such that S ⊂ G 1 ∪ G 2 ∪ · · · ∪ GN. If this is not true, then for every n = 1, 2 ,...
S 6 ⊂ G 1 ∪ · · · ∪ Gn.
For each n there is thus a point pn ∈ S such that^4 pn 6 ∈ {p 1 ,... , pn− 1 } and
pn 6 ∈ Gk for 1 ≤ k ≤ n. (3)
Because S satisfies BW, there is a cluster point, say p of the infinite sequence T = {p 1 , p 2 ,... , } and p ∈ S. Since p ∈ S, there is a k 0 such that p ∈ Gk 0. Since Gk 0 is an open set, there is a δ > 0 such that B(p, δ) ⊂ Gk 0. Since p is a cluster point of T , B(p, δ) ∩ T is infinite, therefore B(p, δ) ∩ T = {pn 1 , pn 2 ,... , } is a subsequence, so n 1 < n 2 < · · · → ∞. We now have a contradiction: take any nj > k 0. Then pnj ∈ Gk 0 , which contradicts (3). Step 3 is proved and this completes the proof of Proposition 6.1. 2
7 Friday October 6—Compactness III
7.1 Corrected proof of Step 3 of Proposition 6.
Let S ⊂ G 1 ∪ G 2 ∪ · · ·. Since S 6 ⊂ G 1 , choose p 1 ∈ S − G 1. Choose n 1 such that p 1 ∈ Gn 1 −(G 1 ∪· · ·∪Gn 1 − 1 ). Since S 6 ⊂ G 1 ∪· · ·∪Gn 1 , choose p 2 ∈ S −(G 1 ∪· · ·∪Gn 1 ) and choose n 2 such that p 2 ∈ Gn 2 −(G 1 ∪· · ·∪Gn 2 − 1 ). Continuing in this way we obtain a sequence of distinct points T := {pk}∞ k=1 ⊂ S and a subsequence n 1 < n 2 < · · · such that for each k ≥ 1,
pk ∈ [S − (G 1 ∪ · · · ∪ Gnk− 1 )] ∩ [Gnk − (G 1 ∪ · · · ∪ Gnk − 1 )]. (4)
By BW, there is a point p ∈ cl T ∩ S. Choose k 0 such that p ∈ Gk 0 and then choose δ > 0 such that B(p, δ) ⊂ Gk 0. Since B(p, δ) ∩ T is infinite, there exists m > k 0 such that pm ∈ B(p, δ) and thus pm ∈ Gk 0. But by (4), pm 6 ∈ G 1 ∪ · · · ∪ Gnm− 1. This contradicts the fact that pm ∈ Gk 0 , since k 0 < m ≤ nm − 1. 2
(^4) This does not seem to be true; however all you need to know is that the sequence {pk}∞ k=1 is actually infinite. See subsection7.
7.2 Closed and Bounded implies Compact
Theorem 7.1 Let S be any subset of Rn. If S is closed and bounded, then S is compact.
We shall prove this theorem by showing that a closed and bounded set satisfies BW. In this form, the theorem is known as the Bolzano-Weierstrass theorem (in Rn). Of course you may want to prove this theorem by showing that a closed and bounded set satisfies HB. In that form, the theorem is known as the Heine-Borel theorem (in Rn). You will find the Heine-Borel theorem in Buck as Theorem 24 on page 65 (for n = 1) and Theorem 25 on page 65 of Buck for arbitrary n. The following two lemmas, well known facts (by now) about subsequences of se- quences of real numbers are the main tools in the proof of Theorem 7.1.
Lemma 7.2 (Bolzano-Weierstrass theorem in R) Every bounded sequence of real numbers has a convergent subsequence.
Lemma 7.3 Every subsequence of a convergent sequence of real numbers converges to the same limit as the sequence.
Proof of Theorem 7.1: Since S is bounded, there is a ball B(0, M ) with S ⊂ B(0, M ). Obviously
B(0, M ) ⊂ ∩nj=1{p = (a 1 ,... , an) ∈ Rn^ : −M ≤ aj ≤ M }.
Now let T = {p 1 , p 2 ,.. .} ⊂ S be an infinite sequence. We must find a point p ∈ S which is a cluster point of T. Choose a subsequence T 1 = {q 1 , q 2 ,.. .} of T such that the sequence of first coor- dinates converges (you used Lemma 7.2 here since the first coordinates of T lie in the closed interval [−M, M ]). Call the limit of the sequence of first coordinates x 1. Now choose a subsequence T 2 = {r 1 , r 2 ,.. .} of T 1 such that the sequence of second coordinates converges (Lemma 7.2 again) and call this limit x 2. By Lemma 7.3, the first coordinates of T 2 also converge to the previous x 1. Continuing in this way, you obtain subsequences
Tn ⊂ Tn− 1 ⊂ · · · ⊂ T 1 ⊂ T
such that the n coordinate sequences of Tn each converge to some number. We have decided to call these numbers x 1 ,... , xn, and we have thus defined a point p = (x 1 ,... , xn) ∈ Rn. Our proof will be complete as soon as we show that p is a cluster point of T. For then, since T ⊂ S, p will be a cluster point of S, and since S is closed, p will belong to S. To help us prove that p is a cluster point of T , we need some notation. Let Tn = {s 1 , s 2 ,.. .} and let
sk = (x( 1 k ),... , x( nk )) k = 1, 2 ,... ,
- (Ross, Theorem 19.2, p. 103) A continuous real valued function on a closed and bounded interval in R is uniformly continuous on that interval. (Application: a continuous function on a closed interval in R is Riemann integrable)
Here is a description of the first five theorems of Chapter 2 of Buck
Theorems 1,2 page 73-74 These concern a characterization of continuity at a point in terms of convergence of sequences, and are extremely useful. At least one of these will be proved below.
Theorem 3 page 76 This is a global characterization of continuity. It becomes messy if the domain D is not an open set, and for this reason we shall not spend any time on it right now.
Theorem 4 page 77 This concerns the “algebra” of continuous functions, that is sums, products, quotients, and is familiar from elementary calculus. This is important to know but we shall not spend time on it. It is used in Buck to give a proof of the extreme value theorem ([Theorem 11,page 91] of Buck), but we shall give an independent proof of the extreme value theorem, using only compactness.
Theorem 5 page 78 This involves composite functions and we shall discuss it in connection with our study of the chain rule, later in this course.
In [Buck, Section 2.3] we will discuss Defintion 2 on page 82 and Theorem 6 on page 84. We will not have time for Definition 3 and Theorem 7, which can be ignored. In [Buck, Section 2.4] Theorems 10 and 11 follow easily from Theorem 13, as we will show. Before we do that, let us note that Theorems 8 (p. 89), 9 (p. 90) and 12 (p. 92) can be skipped (we need Theorem 8 later, but we can wait and prove it later). Theorems 14 (p. 93), 15 (p. 94), 16 (p. 95) involve connectedness and we shall skip them now.
8.2 Continuous functions—continuous image of a compact
set
Definition 8.1 Let f : D → R be a function, where D is any subset of Rn, and let p 0 ∈ D. We say that f is continuous at p 0 if
∀ > 0 , ∃δ > 0
such that^6 |f (p) − f (p 0 )| < for all p ∈ D with |p − p 0 | < δ. (^6) δ depends in general on p 0 as well as on
It is important to realize that this lengthy definition can be put in the compact 7 form ∀ > 0 , ∃δ > 0 such that f [D ∩ B(p 0 , δ)] ⊂ B(f (p 0 ), ).
Here, we are using the notation
f (A) := {f (p) : p ∈ A} if A ⊂ D.
We refer to f (A) as the image of A under f. Please note that the above definition is a “local” one, that is, concerns a single point p 0 , together with “neighboring” points. We say f is continuous on D if it is continuous at each point of D. This gives a “global” definition of continuity.
Assignment 9 (Due October 27) [Buck, §2.2 page 80 #1 or 2, #3 or 4, #7 or 8, #12 or 13, #14 or 17] You are to hand in 5 problems, one from each of these 5 pairs. You will of course be responsible for all of the problems.
Theorem 8.2 The continuous image of a compact set is compact. In other words, if f : D → R is a continuous function on D, and D is a compact subset of Rn, then f (D) is a compact subset of R.
Proof: We shallo show that f (D) satisfies the HB property. By Lemma ??, we only need to deal with countable open covers. We shall use the fact that D satisfies the HB property (for arbitrary covers!). Let f (D) ⊂ ∪∞ k=1Gk
be an open cover of f (D). For each p ∈ D, f (p) ∈ f (D) and so there is a member of the cover, say Gkp , with f (p) ∈ Gkp. Since the cover is an open cover, Gkp is an open set so there is p > 0 such that B(f (p), p) ⊂ Gkp. Since f is continuous at every point of D, there exists δp > 0 such that
f [B(p, δp) ∩ D] ⊂ B(f (p), p)
We can now cover D^8 : D ⊂ ∪p∈DB(p, δp).
Since D is compact, the HB property tells us there are a finite number of points p 1 ,... , pm say, such that D ⊂ ∪mj=1B(pj , δpj ).
It follows that D = ∪mj=1[B(pj , δpj ) ∩ D], and therefore that
f (D) = ∪mj=1f [B(pj , δpj ) ∩ D] ⊂ ∪mj=1B(f (pj ), pj ) ⊂ ∪mj=1Gpj.
We have reduced the given (countable) cover to a finite subcover, so the proof is complete. 2
(^7) no pun intended (^8) the redundant cover!
