Analysis II 27/04/2024

• Metric space


  A nonempty set X together with a nonnegative function d: X x X -> [0, ∞), called the distance (or metric) on X, which satisfies:

  1. For all x, y in X, d(x, y) = 0 if and only if x = y (Definiteness)
  2. For all x, y in X, d(x, y) = d(y, x) (Symmetry)
  3. For all x, y, z in X, d(x, z) <= d(x, y) + d(y, z) (Triangle inequality)
• The Euclidean space (Rn, d), with d(x, y) := ||x - y||, is a metric space
• R, equipped with the absolute value distance |x - y| is a metric space
• Discrete metric


  d(x, y) = 1 if x != y, 0 if x = y
• Manhattan metric


  dNY(x, y) = |x1 - y1| + |x2 - y2|
• (X, d1) and (X, d2) are metric spaces, where X is the set of all continuous real-valued functions defined on [0, 1], d1(f, g) := max{|f(x) - g(x)| | x in [0, 1]}, and d2(f, g) := integral from 0 to 1 of |f(x) - g(x)| dx
• If (X, d) is a metric space and X0 is a subset of X, then (X0, d0), where d0 is the restriction of d to X0 x X0, is also a metric space
• Subsequence


  A sequence of the form (xf(k))k=0 to infinity, where f: N -> N is a strictly increasing function
• Accumulation point


  Given a set Y in a metric space X, x in X is an accumulation point of Y if there exists a sequence (yn)n>=0 in Y converging to x

  Given a sequence (xn)n>=0 in X, x in X is an accumulation point of (xn)n>=0 if some subsequence converges to x
• A stronger version: a sequence (xn)n in a metric space converges to x if and only if every partial sequence (xf(n))n (f increasing) has a sub-subsequence (xg(f(n)))n (g increasing) converging to x
• Hence, d(xf(n), x) < ε for all n ≥ N, as f(n) ≥ n
• We now prove the "if" part we can simply use that (xn)n≥0 is a subsequence (i.e, f(n) = n) and hence it converges to x
• A stronger version of the previous Lemma that is useful in some contexts asserts the following: a sequence (xn)n∈N in a metric space converges to x if and only if every partial sequence (xf(n))n∈N (f increasing) has a sub-subsequence (xg(f(n)))n∈N (g increasing) converging to x
• The "if" part is less trivial than in the previous lemma. One can argue by contraposition: If xn does not converge to x then we want to find a subsequence such that we cannot extract a sub-subsequence converging to x
• To do so we start by the negation of "xn converges to x


  1. Recall the negation of xn → x

  2. ∃ε > 0

  3. ∀N ∈ N

  4. ∃n ≥ N

  5. d(xn, x) ≥ ε
• The set of {n ∈ N | d(xn, x) ≥ ε} and hence there is an increasing sequence (nk)k≥0 such that d(xnk, x) ≥ ε. Notice that any sub-subsequence will still remain at distance ≥ ε from x and hence will not converge to x
• This stronger version can be used, for example, to prove that a continuous function f : [0, 1] → R has a unique minimum point if and only if all sequences (xn)n≥0 ⊂ [0, 1] such that f(xn) → min[0,1] converge to the same limit point
• A sequence in Rn converges (in the Euclidean distance) if and only if it converges coordinate-wise
• Proof of Lemma 9.24: Convergence in Rn versus coordinate-wise

  1. Assume xk → x ∈ Rn

  2. |xk,j - zj| ≤ √Σi(xk,i - zi)2 ≤ ε eventually

  3. This proves xk,j → xj for each j = 1,...,n

  4. Assume xk,j → xj for each j = 1,...,n

  5. Given ε > 0, for each j ∃Nj s.t. |xk,j - xj| < ε/√n for k ≥ Nj

  6. This implies ||xk - x|| < ε
• The space X of all bounded sequences (xn)n≥0 of real numbers, equipped with the distance d(x,y) = supn|xn-yn|, is a complete metric space
• If (X, d) is a metric space with completion (X, d), and (Y, dY) is a complete metric space, and f: X → Y satisfies d(x,y) = dY(f(x),f(y)) for all x,y ∈ X, then there exists a unique f: X → Y such that f = f ◦ ιX and d(x,y) = dY(f(x),f(y)) for all x,y ∈ X
• The completion of Q serves as a model for a field of real numbers R
• Construction of the reals R as the completion of the rationals Q

  1. C = set of Cauchy sequences in Q

  2. N = {(qn)∈C | lim qn = 0}

  3. R = C/∼ = C/N

  4. Define multiplication on R by [(pn)]*[(qn)] = [(pnqn)]

  5. Define order on R by x ≤ y if ∃(rn)∈N s.t. pn-rn ≤ qn∀n

  6. R is a complete ordered field
• R satisfies the Archimedean Principle: For x > 0 in R, there exists n ∈ N s.t. 1/n < x
• If X, Y ⊂ R are non-empty with x ≤ y for all x ∈ X, y ∈ Y, then there exists z ∈ R s.t. x ≤ z ≤ y for all x ∈ X, y ∈ Y
• ∅ and X are always both open and closed
• In general, a set U needs not be neither open nor closed
• The intervals (0,1) and (2,3) in X = (0,1) ∪ (2,3) are clopen (both open and closed)
• The set [0,1) is open in X = [0,2] but not in the whole R
• The open ball B(x,r) is an open set
• Every finite subset of X is closed
• Open set


  A set U is open if for every x in U, there exists an ε > 0 such that the open ball B(x, ε) is contained in U
• Closed set


  A set A is closed if its complement X\A is open
• The open ball B(x, r) is an open set
• The intersection of infinitely many open sets may not be open
• Interior


  The largest open set contained in a set E
• Closure


  The smallest closed set containing a set E
• Boundary


  The set E \ E°
• The interior E° is always open, while the closure E and boundary ∂E are always closed
• For balls B(x, r) in Rn, B(x, r) = {y ∈ Rn | d(x, y) ≤ r} and ∂B(x, r) = {y ∈ Rn | d(x, y) = r}