Real Variable Theory Flashcards
Axiom of Completeness
Every nonempty set of real numbers that is bounded above has a least upper bound aka supremum
Metric space
Let X be a set. We say that the pair (X,d) is a metric space with metric d if 1. d(x,y) = d(y,x) 2. d(x,y) = 0 iff x = y 3. d(x,z) < d(x,y) + d(y,z)
Lemma (u- epsilon < a)
Let A be a nonempty subset of |R. Then a real number u∈|R is the sup(A) iff 1. u is an upper bound for A and 2. for each epsilon > 0, there exists a member a∈A s.t. u- epsilon < a
Absolute value/basic metric space
d(x,y) = |x - y|
Bounded metric space
A ⊆X in a metric space (X,d) is called bounded if there exists a point x∈X and a pos. real number r such that A ⊆ {p in X | d(p,x) < r} = B_r(x) (ball of radius r centered at x)
Archimedean Property of |R
Let x be any real number. 1. Then there exists an n∈|N such that n > x and 2. for any y > 0, there exists n∈|N such that 1/n < y
(Proof of 1. BWOC, n
Taxicab metric
d(x,y) = |x1 - y1| + |x2 - y2|
Discrete metric
d(x,y) = {0 if x=y; 1 if x neq y}
Proof of triangle: 2 cases, x = z and x neq z
Convergent Seq (in metric space)
Let (a_n) be a sequence in (X,d). We say that a_n –> p in X if for all epsilon > 0 there exists N in |N such that d(a_n, p) < epsilon whenever n > N
Limit of a Convergent Seq is Unique (Prop)
Let (X,d) be a metric space and let a_n be a convergent seq in X. Then the limit of (a_n) is unique if a_n -> p and a_n -> q, then p = q.
(Proof: BWOC p neq q, use triangle inequality and let epsilon/2)
Monotone Convergence Theorem
Let (a_n) be a sequence in (|R , |.|) such that 1. a_n is bounded above and 2. a_n is increasing. Then a_n is convergent.
(Proof: Let A = {a_n}, use givens and property of sup so there exists an a_m in A s.t.
s - ep < a_m. WTS s - ep < a_n, let m = N.)Subsequence
Let a_n be any sequence. We say b_n is a subsequence of a_n if b_n = a_f(n) where f is an increasing function f: |N –> |N
Proposition of all subsequences of a convergent seq converge to same limit
If a_n is a convergent sequence, then any subsequence of a_n is convergent to the same limit as a_n
(Proof: suppose a_n -> p. WTS b_n –> p. use def of converg in met space, works cuz f is increasing f(n) > n)
Bolzano-Weierstrass
Let a_n be any bounded sequence in |R. Then a_n has a convergent subsequence
Nested Interval Property
Consider the closed (bounded) intervals
I_n = [a_n, b_n] for a_n, b_n in |R. Suppose that these intervals are nested in the sense that I1 >(ss) I2 > I3 >… Then the intersection of I_n is not empty.
(Proof: consider a_n of left endpoints of intervals; a_n is increasing and bounded, MCT; converges to a = sup…)
Open Set
A ⊆ X is open in X if for all a in A there exists an r > 0 such that B_r(a) ⊆ A
Closed Set
A subset X is closed if A^c = X - A is open in X A subset A is a closed subset of X if it contains all its limit points (A complement aka everything but A)
Limit Point
A point x in X is a limit point for A ⊆ X if there is a sequence a_n in A s.t. a_n→x and a_n neq x for all n in |N
Closure
Closure of A, denoted by A(bar) = A U A’
Isolated Point
a in A is an isolated point if a is not a limit point
Following are open in (X,d)…
X (metric space X), nonempty set, (0,1) in |R, 1st quadrant in |R^2
Following are closed in (X,d)…
X (b/c X^c = nonempty = open), nonempty set, [0,1] in |R
Sequentially Compact
Let (X,d) be a metric space. We say that A subset X is sequentially compact if any sequence in A has a convergent subsequence that converges to a point that is also in A
Heine-Borel Theorem
Consider (|R, |.|). Then A ⊆ |R is sequentially compact iff A is closed and bounded
Theorem, Any compact set is closed and bounded
Let (X,d) be a metric space and let A be a sequentially compact subset of X. Then A is bounded and closed in X.
(Boundedness proof: Assume A is seq compact but not bounded; use inverse of bounded metric space; convergent subseq.
Closeness proof: Assume A is seq compact but not closed; A^c is not open; use def of open; then find a_n in A with 1/n; convergent subseq)
Compact
A set K ⊆ R is compact if every sequence
in K has a subsequence that converges to a limit that is also in K.
An important property of sup
For each n>0 there exists a sequence a_n in A such that s - 1/n
Disconnected Set
Let (X,d) be a metric space and let A⊆X. Suppose there exists open sets U and V s.t.
(1) U n V = empty set (2) U n A neq empty set and V n A neq empty set (3) A⊆U u V
Connected Set
If A⊆X is not disconnected, then it is called connected
Theorem (on Compact)
A subset A of a metric space is compact if and only if it is sequentially compact
Theorem (also on compact)
Suppose A⊆X. If A is compact, then A is closed and bounded
Interior point
Suppose A⊆X. A point a in A is an interior point of A if there is an r>0 s.t. B_r(a) ⊆ A. The set of all interior points is denoted by A^o
Dirichlet Function
f(x) = {1 if x in Q; 0 if x not in Q
epsilon − δ Functional Limit
Let A be a subset of (|R, |.|), and let a be a limit point of A. Suppose f: A→|R. We say
lim x→a f(x) = L for some L in |R if for all epsilon >0 there exists some δ>0 such that |f(x) - L| < epsilon whenever |x-a|< δ
Continuous function
Let A⊆|R and let f: A → |R be a function. We say that f is continuous at x=a if for all epsilon > 0 there exists δ>0 such that |f(x) - f(a)| < epsilon whenever |x-a| < δ.
Proposition (continuity/limit point)
Let f: A→|R and let a be a limit point of A. Then the following are equivalent: (i) f is continuous at x=a (ii) lim x→a f(x) = f(a)
(Proof: => use def of continuity; limit point
<=use given to get def of continuity)
Proposition (continuity/sequence/limit)
Let f: A→|R and let a in A. TFAE: (i) f is continuous at x=a (ii) For each sequence (a_n) in A such that a_n → a we have lim f(a_n) = f(a)
Proposition (continuity/sequence)
Let f: A→|R and let a in A. TFAE: (i) f is continuous at x=a (ii) For each sequence (a_n) in A such that a_n → a we have
f(a_n) → f(a)
Algebraic Continuity Theorem
Assume f :A→R and g : A → R are continuous at point c ∈ A. Then,
(i) kf(x) is continuous at c for all k ∈ R;
(ii) f(x) + g(x) is continuous at c;
(iii) f(x)g(x) is continuous at c; and
(iv) f(x)/g(x) is continuous at c, provided the quotient is defined.
Algebraic Limit Theorem for Functional Limits
Let f and g be functions defined on a domain A ⊆ R, and assume limx→c f(x) = L
and limx→c g(x) = M for some limit point c of A. Then, (i) limx→c kf(x) = kL for all k ∈ R,
(ii) limx→c [f(x) + g(x)] = L + M,
(iii) limx→c [f(x)g(x)] = LM, and
(iv) limx→c f(x)/g(x) = L/M, provided M = 0.
Grand Theorem (dont need to know name)
Let A⊆|R and let f: A→|R be a function. Then f is continuous on A iff for any open set U⊆|R in the domain, f^-1(U) is open in the domain.
Corollary (compact)
f(compact) = compact implies that f: [0,1]→|R then f([0,1]) is compact in |R, f([0,1]) has a max/min
Theorem (closed set)
f: A→|R is continuous iff f^-1(K) is closed for any closed set K in |R.
Theorem (continuous image of compact set)
If K⊆|R is compact and f: K→|R is continuous, then f(K) is compact
Extreme Value Theorem
Suppose f: |R→|R is continuous and let K be a compact subset of |R. Then f:K-|R attains its max and min on K. In other words, there exists x_0, x_1 in K such that f(x_0)
Notes on [0,1]..
f is continuous on [0,1] and [0,1] is compact
Theorem (continuous image)
“The continuous of image of a connected set is connected”
Let f:|R→|R be continuous and suppose C is a connected subset of |R. Then f(C) is connected
Intermediate Value Theorem
Let a and b be real numbers and a
Fixed Point
A point x ∈ X is called a fixed point of a function f : X → X if f(x) = x
Uniformly continuous on A
f: A→|R is uniformly continuous on A is for all epsilon > 0, for all x,a in A there exists a
delta > 0 such that |f(x) - f(a)| < epsilon whenever |x-a| < delta
Criterion for lack of Uniform Continuity
f: A→|R is NOT uniformly continuous on A iff there exists sequences (x_n) and (a_n) in A such that (x_n - a_n) → 0 but there exists an epsilon > 0 for which | f(x_n) - f(a_n) | > epsilon for sufficiently large n
Two observations of uniform contin.
(1) Uniform continuity only implies continuity (not the other way)
(2) If f is uniformly continuous on A, then f is uniformly continuous on any subset B⊆A
Theorem (uniform contin/compact)
Let K be compact and f: K→|R continuous. Then f is uniformly continuous on K
Thomae’s Function
t(x) = { 1 if x=0; 1/n if x = m/n in Q;
0 if x not in Q}
Fact on continuity
Any polynomial is continuous
Limit definition of derivative
Let A be any interval. We say f: A →|R is differentiable at a in A if
f’(a) = limx→a f(x) − f(a) / x − a
Proposition (differentiable/continuous)
If f: A→|R is differentiable at x=a, then f is indeed continuous at x=a
Interior Extremum Value Theorem
If f is differentiable and it has a local max (or a local min) at x = a, then f’(a) = 0
Rolle’s Theorem
If f: [a,b] →|R is continuous on [a,b] and f is differentiable on (a,b) and f(a) = f(b), then
f’(c) = 0 for some c in (a,b)
Mean Value Theorem
Suppose f: [a,b] is continuous on [a,b] and f is differentiable on (a,b). Then there exists some c in (a,b) such that f’(c) = f(b) − f(a) / b − a
Consequence of MVT
If f: (a,b)→|R is a function such that f’(x) = 0 for all x in (a,b). Then f(x) = c (constant)
Product Rule for Differentiation
Suppose f and g are differentiable at some point a in A. Then f.g defined by f.g(x) = f(x)g(x) is also differentiable at x = a and moreover (f.g)’(a) = f’(a)g(a) + f(a)g’(a)
Cauchy Completeness
Let (X,d) be a metric space. We say that (X,d) is Cauchy Complete if any Cauchy sequence in X is convergent in X
Cauchy Sequence
Let (X,d) be a metric space and let (x_k) be a sequence in X. We say (x_k) is a Cauchy sequence if for all epsilon > 0 there exists N in |N such that whenever m and n satisfy m>N and n>N we have d(x_m, x_n) < epsilon
Big picture (Cauchy)
(1) Any convergent sequence is a Cauchy sequence in any metric space (2) Any Cauchy sequence is a convergent sequence in Cauchy complete spaces (3) (|R, |.|) is Cauchy complete so in |R Cauchy seq = convergent seq
Important Lemma (Cauchy/convergent)
Let (X,d) be a metric space and let (x_k) be a Cauchy sequence in X. Suppose (x_k) has a convergent subsequence (x_ki) and that x_ki→x in X. Then x_k is also convergent to x.
Theorem
(|R, |.|) is Cauchy complete
Theorem (Cauchy seq/bounded)
Let (X,d) be a metric space and let (a_n) be a Cauchy sequence in X. Then {a_n | n in |N} is a bounded subset of X
(Proof: def of Cauchy seq; let ep = 1 and n = N+1; use A = BUC, bounded}Partition
A partition P for the interval [a,b] is a finite set, P = {x_0,…,x_n} such that
x_0 = a < x1 < x2
Upper/Lower Sum
Let f be a (bounded) function on [a,b] and let P be a partition of [a,b]. We define upper/lower sums of f with respect to P as follows, P = {x_0,…,x_n}:
U(f,P) = sumM_k(x_k-x_k-1) where M_k = {f(x)|x in [x_k-1, x_k]}
L(f,P) = summ_k(x_k - x_k-1) where m_k = {f(x)|x in [x_k-1, x_k]}
Proposition (upper/lower sum)
For any partition P, L(f,P)
Refinement
Let f be bounded on [a,b] and let P and Q be partitions of f on [a,b]. We say Q is a refinement of P if P is a subset of Q
Integrable
Let f be bounded on [a,b]. We say that f is integrable on [a,b] if U(f) = L(f) where U(f) = inf{U(f,P) | P partition} and L(f) = sup{L(f,P) | P partition}
Lemma (2 paritions)
If P1 and P2 are any two partitions of [a,b], then L(f, P1)
Prop (refinement/parition)
If Q is a refinement of P, then L(f,P) eq U(f,Q)
Darboux’s Criterion for Riemann Integrability
Let f be a bounded function on [a,b]. Then f is integrable on [a,b] iff for all epsilon > 0, there exists some partition P_ep such that U(f,P_ep) - L(f,P_ep) < epsilon
Theorem (contin/integrable)
Any continuous function f on [a,b] is integrable
Fundamental Theorem of Calculus
Let f be a differentiable function. Assume f’ is integrable on [a,b]. Then integral from a to b f’(x)dx = f(b) - f(a)
