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Monday, May 26, 2014

Namesake Property of Completions (3.24(c))

Walter Rudin Principles of Mathematical Analysis, chapter 3, exercise 24(c):

MathJax TeX Test Page Let X be a metric space, and let X be the metric space of equivalence classes of Cauchy sequences under the relation {pn}~{qn} if limnd(pn,qn)=0 and under the distance metric Δ(P,Q)=\lim_{n→∞} d(p_n,q_n) for any Cauchy sequences \{p_n\}, \{q_n\} in P,Q respectively. Prove X^* is complete.

Proof: Lemma: Let s be a Cauchy sequence in some metric space. Say s is a quick Cauchy sequence if n,m≤N implies d(s(n),s(m)) < \dfrac{1}{N}. Then a subsequence of s is a quick Cauchy sequence, so particularly the class S of s under the above relation contains a representative that is a quick Cauchy sequence. Proof: Since s is Cauchy, for all n∈ℕ^+ let N_n be such that a,b≥N_n implies d(s(a),s(b)) < \dfrac{1}{n}. Let s' be the subsequence such that s'(n)=s(N_n), and it's seen that n,m≥N implies d(s'(n),s'(m))=d(s(N_n),s(N_m)) < \text{max}(\dfrac{1}{n},\dfrac{1}{m}) ≤ \dfrac{1}{N}.

Now, let \{S_n\} be a Cauchy sequence in X^*, with s_n a quick Cauchy sequence of S_n for all n. Let s be the sequence with s(n)=s_n(n) for all n.

We prove s is Cauchy: Let ε > 0. Let α∈ℕ be such that \dfrac{2}{α} < ε, and let N' be such that n,m ≥ N' implies Δ(S_n,S_m) = \lim_{z→∞} d(s_n(z),s_m(z)) < ε - \dfrac{2}{α}. Then if n,m ≥ \text{max}(α,N') we see d(s(n),s(m)) = d(s_n(n),s_m(m))≤ d(s_n(n),s_n(z))+d(s_n(z),s_m(z))+d(s_m(z),s_m(m)) < \dfrac{1}{α}+ε-\dfrac{2}{α}+\dfrac{1}{α} = ε for z sufficiently large.

We prove S, the class of s, is the point of convergence of \{S_n\}: Let ε > 0. Let α be such that \dfrac{1}{α} < ε, and let N' be such that n,m≥N' implies d(s(n),s(m)) < ε-\dfrac{1}{α} since s is Cauchy, then choose N ≥ \text{max}(α,N'). Given n≥N, observe Δ(S,S_n) = \lim_{z→∞} d(s(z),s_n(z)). For z≥N we bound d(s(z),s_n(z))=d(s_z(z),s_n(z))≤d(s_z(z),s_n(n))+d(s_n(n),s_n(z)) < ε-\dfrac{1}{α}+\dfrac{1}{α} = ε so Δ(S,S_n)≤ε when n≥N, and we are done.~\square

Sunday, May 18, 2014

Convergence of a Related Series (3.7)

Walter Rudin Principles of Mathematical Analysis, chapter 3, exercise 7:

MathJax TeX Test Page Let a_n≥0. Show that the convergence of ∑a_n implies the convergence of \sum \dfrac{\sqrt{a_n}}{n}
Proof:

Lemma: If a_n≥a_{n+1}≥0 and ∑a_n converges, then ∑\sqrt{a_{2^n}} converges. Proof: By theorem 3.25, ∑2^na_{2^n} converges, so by the root test \text{lim sup }\sqrt[n]{2^na_{2^n}}=2\text{lim sup }\sqrt[n]{a_{2^n}}≤1 so \text{lim sup }\sqrt[n]{a_{2^n}}≤1/2 implying ∑\sqrt{a_{2^n}} converges by the root test by \text{lim sup }\sqrt[n]{\sqrt{a_{2^n}}}≤\sqrt{1/2} < 1 (since one may show without advancing in theory that \text{lim sup }\sqrt{x_n}≤\sqrt{\text{lim sup }x_n} for any positive sequence x_n).

Now, every nonempty subset of a convergent series' terms \{a_n\} has an absolute greatest element, since by the epsilon-delta method there are only finitely many elements absolutely greater than any chosen term. Since ∑a_n=∑|a_n|, use this to inductively construct a rearrangement ∑b_n such that b_n≥b_{n+1} without losing convergence.

If we prove ∑\dfrac{\sqrt{b_n}}{n} converges, then though ∑\dfrac{\sqrt{a_n}}{n} is not generally a rearrangement, we may observe the effect of transpositions on partial sums to show the former bounds the latter, demonstrating the latter's convergence. So by 3.25, the former is equivalent to ∑\sqrt{b_{2^n}} converging, which is established by the lemma.~\square

Saturday, May 17, 2014

Behavior of a Complex-Valued Series (3.6(d))

Walter Rudin Principles of Mathematical Analysis, chapter 3, exercise 6(d):

MathJax TeX Test Page Investigate the behavior of ∑a_n when a_n=\dfrac{1}{1+z^n} for complex values of z.

Proof: Assume |z| > 1. Then ||\dfrac{a_{n+1}}{a_n}|-\dfrac{1}{|z|}|=||\dfrac{\dfrac{1}{1+z^{n+1}}}{\dfrac{1}{1+z^n}}|-|\dfrac{1}{z}||=||\dfrac{1+z^n}{1+z^{n+1}}|-|\dfrac{1}{z}||≤|\dfrac{1+z^n}{1+z^{n+1}}-\dfrac{1}{z}|=|\dfrac{z-1}{z+z^{n+2}}|=|\dfrac{z-1}{z}|·\dfrac{1}{|1+z^n|}≤|\dfrac{z-1}{z}|·\dfrac{1}{|z|^n-1}→0 as n→∞, so \text{lim }|\dfrac{a_{n+1}}{a_n}|=\dfrac{1}{|z|} < 1 and ∑a_n converges by the ratio test. Assume |z| ≤ 1. Then |\dfrac{1}{1+z^n}|≥\dfrac{1}{1+|z|^n}≥\dfrac{1}{2} and ∑a_n diverges seeing as a_n doesn't converge to 0.~\square

Friday, May 16, 2014

Rearrangement of a Different Convergence (Example 3.53)

Walter Rudin Principles of Mathematical Analysis, chapter 3, example 53:

MathJax TeX Test Page Consider the convergent series 1-1/2+1/3-1/4+... and one of its rearrangements 1+1/3-1/2+1/5+1/7-1/4+1/9+1/11-1/6+... As we have seen, the rearrangement s_n if it converges does so at a point larger than the point of convergence for the original series. Prove s_n converges.

Proof: For any sequence a_n, let f~:~ℕ→ℕ be such that a_{f(n)} and a_n-a_{f(n)} are convergent sequences. Then we see (\text{lim }a_n-a_{f(n)})+(\text{lim }a_{f(n)})=\text{lim }a_n so that a_n converges. As it applies to this problem, we will show the subsequence s_{3n} converges, and since \text{lim }s_n-s_{f(n)}=0 where f(n) rounds n to the next highest multiple of 3, s_n converges (technically s_{f(n)} has different terms from s_{3n}, but \text{lim }s_{f(n)}=\text{lim }s_{3n} since the former merely has finite copies of each of the latter's terms).

It suffices to compare the sum's terms a_n to \dfrac{4}{n^2}, which forms a convergent series: |a_n|=a_n=\dfrac{1}{4n-3}+\dfrac{1}{4n-1}-\dfrac{1}{2n}=\dfrac{8n-3}{2n(4n-3)(4n-1)}≤\dfrac{4}{(4n-3)(4n-1)}≤\dfrac{4}{n^2}~~\square

Sunday, May 11, 2014

Connected Base for Euclidean Space (2.29)

Walter Rudin Principles of Mathematical Analysis, chapter 2, exercise 29:

MathJax TeX Test Page Prove that every open set in ℝ^1 is the union of an at most countable collection of disjoint segments.

Proof: We shall prove something stronger—namely, that every open set in ℝ^k is the union of an at most countable collection of disjoint open connected spaces, which by Theorem 2.47 implies the proposition for k=1. This case could in fact be strengthened to path-connected spaces but is beyond the development of current theory.

Lemma 1 (Subspace Nonconnnectedness): Let E=A∪B for disjoint, nonempty, separated A,B. Every subset of E containing a point of A and a point of B is nonconnected. Proof: Let F⊆E be such that C=F∩A and D=F∩B are nonempty. Now, note that for spaces X,Y that\overline{X∩Y}=(X∩Y)∪(X∩Y)'⊆(X∩Y)∪(X'∩Y')=(X∪X')∩(Y∪Y')∩(X∪Y')∩(Y∪X')⊆\overline{X}∩\overline{Y}So observe C∪D=(A∪B)∩F=F and \overline{C}∩D=\overline{A∩F}∩D⊆\overline{A}∩\overline{F}∩D⊆\overline{A}∩B is empty, and similarly C∩\overline{D} is empty.

Lemma 2 (Upper Bounds of Chains of Connected Spaces): Let \{E_i\} be a collection of connected spaces totally ordered by inclusion. Then E=∪E_i is connected. Proof: Assume E=A∪B for disjoint, nonempty, separated A,B. Then some indexed summand E_α must contain a point of A, and similarly E_β for B. Then by hypothesis either E_α⊆E_β or E_β⊆E_α, in either case one of them being nonconnected by Lemma 1, a contradiction.

Lemma 3 (Union of Nondisjoint Connected Spaces): Let E and F be connected spaces with nonempty intersection. Then E∪F is connected. Proof: Assume E∪F=A∪B for disjoint, nonempty, separated A,B. We may assume some x∈E∩F is contained in A. As well, not all of E and all of F can be contained in A since B is nonempty, hence either E or F must not be connected by Lemma 1.

Now, let E⊆ℝ^k be an open set. Define a relation \text{~} on E by a \text{~} b if there exists a connected subspace of E containing both a and b. This is reflexive since E is an open set admitting a neighborhood of any point, and neighborhoods are convex hence connected (cf. Exercise 2.21 and the example below Definition 2.17); clearly symmetrical; and transitive by Lemma 3.

Choose a set of representatives in E inherited from this equivalence relation, and for any representative observe the collection C of connected subspaces of E containing the representative; by Lemma 2 and Zorn's Lemma there exists a maximal connected subspace under inclusion. This maximal subspace must in fact contain every point in the equivalence class, since we might otherwise construct a larger connected subspace by Lemma 3. Hence E is a union of disjoint open connected spaces, and it remains only to prove that there are a countable number of them. Generally, we will prove that there cannot be a disjoint collection of uncountably many nonempty open sets in ℝ^k.

Observe any collection of disjoint nonempty open sets in Euclidean space. Take a point from each set, and choose a neighborhood of each so that they are disjoint. However, as we've seen, within each of these neighborhoods we may find a member of the countable base for ℝ^k (cf. Exercises 2.22-23), implying the collection is at most countable.~\square

Sunday, May 4, 2014

Limit Points and Housekeeping (2.6-9)

Walter Rudin Principles of Mathematical Analysis, chapter 2, exercises 6-9:

MathJax TeX Test Page 6. Let E' be the set of all limit points of a set E. Prove that E' is closed. Prove that E'=\overline{E}'. Generally, is it true that E'=E''?
7. Let A_n be subsets of a metric space for n∈ℕ.
(a) If B_n=∪_{i=0}^n A_i prove that \overline{B_n}=∪_{i=0}^n \overline{A_i}.
(b) If B=∪A_n prove that \overline{B}⊇∪\overline{A_n}.
Show that this inclusion can be proper.
8. Is every point of every open set E⊆ℝ^2 a limit point of E? Answer the same question for closed sets in ℝ^2.
9. Let E ^\circ denote the set of all interior points of a set E, called the interior of E.
(a) Prove that E ^\circ is always open.
(b) Prove that E is open iff E ^\circ = E.
(c) If G ⊆ E and G is open, show G⊆E ^\circ.
(d) Prove {E^\circ}^c = \overline{E^c}.
(e) Does E ^\circ = \overline{E}^\circ?
(f) Does \overline{E}=\overline{E ^\circ}?

Proof: Lemma 1: Let ∪A_i be a union of subsets of a metric space. Then (∪A_i)'⊇∪A_i', and there is equality when the union is finite. Proof: The direction is clear since limit points of a subset are limit points of the whole set, so it remains to show when the union is finite. To wit, let p be a limit point of ∪A_i. For each i, let r_i=\text{inf }\{r∈ℝ~|~A_i∩N_r(p)-\{p\}≠∅\}. If r_i > 0 for all i then since the indices are finite choose some r such that 0 < r < \text{min}_i~r_i and we would have N_r(p)∩(∪A_i)=\{p\} and p is not a limit point, a contradiction. Hence r_i=0 for some i, therefore p is a limit point of A_i.

6. We now show E''⊆E' so that E' is closed. We have in fact already seen this, since a limit of limit points is a limit of the original points, since they lie arbitrarily close to these limit points. Now, by the lemma we see \overline{E}'=(E∪E')'=E'∪E''=E'. Finally, let E=(0,1). We see E'=\{0,1\} and E''=∅ so that generally E' need not equal E''.

7. (a) We see\overline{B_n}=B_n∪B_n'=(∪_{i=0}^n A_i)∪(∪_{i=0}^n A_i)'=(∪_{i=0}^n A_i)∪(∪_{i=0}^n A_i')=∪_{i=0}^n (A_i∪A_i')=∪_{i=0}^n \overline{A_i}with a containment between the appropriately modified terms three and four if the union is infinite. When A_i=\{1/(i+1)\} for all i, we see \overline{B} contains 0, whereas ∪\overline{A_i}=∪A_i does not.

8. ℝ^2 contains no isolated points, and every point of an open set is interior, and since interior and non-isolated implies limit point, every point of an open set in ℝ^2 is a limit point. Clearly, \{0\} is a closed set for which 0 is not a limit point.

9. Lemma 2: E^\circ = \overline{E^c}^c. Proof: () Suppose p∈E^\circ. Then p∈E so p∉E^c and also p∉{E^c}', hence p∉\overline{E^c} so p∈\overline{E^c}^c. () Suppose p∈\overline{E^c}^c, so p∉\overline{E^c}=E^c∪{E^c}' implying p∈E and p is not a limit point of E^c, i.e. p is an interior point of E and is contained in E.

(a) This follows from the lemma since \overline{E^c} is closed.

(b) This will follow from (c).

(c) Assume G⊆E is open and G⊈E^\circ. Then G'=G∪E^\circ is an open set contained in E with inclusions E^\circ = \overline{E^c}^c ⊂ G' ⊆ E. We then observe E^c ⊆ G'^c ⊂ \overline{E^c} are inclusions with G'^c closed containing E, yet \overline{E^c} is supposed to be the smallest closed set containing E^c, a contradiction.

(d) This was shown in the lemma.

(e) Let E=ℝ-\{0\}⊆ℝ^1. Then E^\circ=E yet \overline{E}^\circ=ℝ^\circ=ℝ. Hence, generally, E ^\circ ≠ \overline{E}^\circ.

(f) Let E=\{0\}. Then \overline{E}=E and \overline{E^\circ}=\overline{∅}=∅. Hence, generally, \overline{E}≠\overline{E ^\circ}.~\square