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