Real Analysis, Math 821.

Instructor: Dmitry Ryabogin

Assignment IV.

1. Problem 1.

Definition 1. A set M ⊂ R is called closed if it coincides with its closure. In other words, M ⊂ R is closed, if it contains all of its limit points.

Definition 2. A set A ⊂ R is called open if there is a closed set M ⊂ R such that A = R \ M. In other words, A is open if all of its points are inner points, (a point x is called the inner point of a set A, if there exists ϵ > 0, such that the ϵ-neighbourhood U_{ϵ}(x) of x, is contained in A, U_{ϵ}(x) ⊂ A).

Prove that any open set on the real line is a finite or infinite union of disjoint open intervals. (The sets (−∞, ∞), (α, ∞), (−∞, β) are intervals for us).

Hint. Let G be open on R. Introduce in G the equivalence relationship, by saying that x ∼ y if there exists an interval (α, β), such that x, y ∈ (α, β) ⊂ G.

a) Prove at first that this is indeed the equivalence relationship, i. e., that x ∼ x, x ∼ y implies y ∼ x, and x ∼ y, y ∼ z imply x ∼ z. Conclude that G can be written as a disjoint union G = ∪_{τ∈ℶ}I_{τ} of “classes” of points I_{τ}, I_{τ} = {x ∈ G : x ∼ τ}, ℶ is the set of different indices, i.e. τ ∈ ℶ if I_{τ} ∩ I_{x} = ∅, ∀x ∈ G.

b) Prove that each I_{τ} is an open interval.

c) Show that there are at most countably many I_{τ} (the set ℶ is countable).

2. Problem 2.

Definition 3. A set K ⊂ R is called nowhere dense in R, if for any open interval (α, β) there exists an open interval (γ, δ) ⊂ (α, β), such that K ∩ (γ, δ) = ∅.

a) Prove that a closed set K ⊂ R is nowhere dense, if and only if for any interval (α, β) there exists a point x ∈ (α, β) \ K.

b) Prove that the Cantor set is nowhere dense in R.

c) Let B be an open dense (the closure of B is R) subset of the real line. Is it true that R \ B is nowhere dense?

d) Let D be a dense subset of the real line. Is it true that R \ D is nowhere dense?

e) Prove that the closure of a nowhere dense set is nowhere dense.

3. Problem 3. a) Let (r_{k})_{k=1}^{∞} be a sequence of all rational numbers on the real line, and let v_{k} = (r_{k} − k^{−2}, r_{k} + k^{−2}). Prove that E := R \ (∪_{k=1}^{∞}v_{k}) is closed and nowhere dense.

Definition 3. A set K ⊂ R is called of first category if it is a countable union of nowhere dense sets. b) Prove that no interval in R is of the first category.

Hint. Our statement is an obvious consequence of the following

Lemma. The complement of any set A of the first category on the line is dense.

To prove the Lemma you could argue as follows. Let A = ∪_{i=1}^{∞}A_{i}, where A_{i} are nowhere dense. Let I be any interval on the real line, and let I_{1} be a closed subinterval of I \A_{1}, I_{2} be a closed subinterval of I_{1} \ A_{2}, and so on. Show that

∅ ≠ ∩_{n=1}^{∞} ⊂ (I \ A).

c)* Prove that R can be written as a disjoint union of A ∪ (R \ A), where |A| = 0, and R \ A is of first category.

Hint. Let (r_{k})_{k=1}^{∞} be a sequence of all rational numbers on the real line, and let I_{kj} = (r_{k} − 2^{−k−j}, r_{k} + 2^{−k−j}). Define

G_{j} := ∪_{k=1}^{∞}I_{kj}, A := ∩_{j=1}^{∞}G_{j}.

Show that |A| < ϵ, ∀ϵ > 0. On the other hand, G_{j} is open (why?) and dense (why?), hence R \ G_{j} is nowhere dense (Problem 2, c)).

4. Problem 4.

Definition 4. A set M ⊂ R is called perfect, if it is closed, bounded, and has no isolated points. Any closed interval is perfect. Cantor set is perfect (why?). A set containing two different points is not perfect.

a) Prove that M is perfect if and only if disjoint intervals from R \ M do not have “common ends” (intervals (a, b), (b, c) have common ends, intervals (a, b), (c, d), b < c, do not have).

b) Construct a nonempty perfect nowhere dense susbset of R \ Q.

Hint. Let (r_{k})_{k=1}^{∞} be a sequence of all rational numbers on the real line. Take r_{1}, and let ϵ_{1} be any irrational number. Define I_{1} := U_{ϵ_{1}}(r_{1}). Now take the first point from (r_{k})_{k=2}^{∞} \ I_{1}. Let it be r_{k_{2}}. Observe that r_{k_{2}} ≠ r_{1} ± ϵ_{1} (why?). Now take ϵ_{2} ∈ R \ Q such that U_{ϵ_{2}}(r_{k_{2}}) ∩ I_{1} = ∅, and such that U_{ϵ_{2}}(r_{k_{2}}), I_{1} have no common ends. Define I_{2} := U_{ϵ_{2}}(r_{k_{2}}). On the next step find among r_{k_{2}+1}, r_{k_{2}+2},..., the first point out of I_{1} ∪I_{2}. Let it be r_{k_{3}}. Observe that r_{k_{3}} ≠ r_{1} ± ϵ_{1}, r_{k_{3}} ≠ r_{k_{2}} ± ϵ_{2}. Now find ϵ_{3} ∈ R \ Q such that U_{ϵ_{3}}(r_{k_{3}}) ∩ I_{1}, I_{2} = ∅, and such that U_{ϵ_{3}}(r_{k_{3}}), I_{1}, I_{2} have no common ends. Define I_{3} := U_{ϵ_{3}}(r_{k_{3}}).

Continue this way to obtain a sequence (I_{k})_{k=1}^{∞}. Then E := R \ (∪_{k=1}^{∞}I_{k}) is the set you are looking for. Prove that E is closed. Prove that E ≠ ∅. Prove that E does not contain isolated points. Prove that E does not contain any rational point. Finally, prove that E is nowhere dense (use the closeness of E and Problem 2, a)).

5. Problem 5. a) Is it possible to construct a countable family (E_{i})_{i=1}^{∞} of perfect nowhere dense sets E_{i} ⊂ [a, b] such that | ∪_{i=1}^{∞} E_{i}| = b − a?

Hint. Take E_{i} ⊂ [a, b], such that |E_{i}| = b − a − 1/i, and E_{i} is perfect and nowhere dense (how to construct these sets?). To show that ∪_{i=1}^{∞}E_{i} satisfies the requirements of the problem define A_{i} := ∪_{k=1}^{i}E_{i}, and observe that ∪_{i=1}^{∞}E_{i} = ∪_{i=1}^{∞}A_{i}.

b)* Is it possible to construct a countable family (E_{i})_{i=1}^{∞} of pairwise disjoint perfect

nowhere dense sets E_{i} ⊂ [a, b] such that | ∪_{i=1}^{∞} E_{i}| ≡ ∑_{i=1}^{∞} |E_{i}| = b − a?

Hint. Consider the case [a, b] = [0, 1]. Construct a perfect nowhere dense set E_{1} ⊂ [0, 1], such that |E_{1}| = 1/2. Take any (a_{1,i}, b_{1,i}) ⊂ ([0, 1] \ E_{1}) and put there a perfect nowhere dense set E_{1,i} ⊂ (a_{1,i}, b_{1,i}), such that |E_{1,i}| = (b_{1,i} − a_{1,i})/2. Then | ∪_{i=1}^{∞} E_{1,i}| = 1/2^{2}. Observe that E_{2} := E_{1} ∪ (∪_{i=1}^{∞}E_{1,i}) is perfect and nowhere dense, moreover, |E_{2}| = 3/4.

Take any (a_{2,i}, b_{2,i}) ⊂ ([0, 1] \ E_{2}) and put there a perfect nowhere dense set E_{2,i} ⊂ (a_{2,i}, b_{2,i}), such that |E_{2,i}| = (b_{2,i} − a_{2,i})/2. Then | ∪_{i=1}^{∞} E_{2,i}| = 1/2^{3}. Observe that E_{3} := E_{2} ∪ (∪_{i=1}^{∞}E_{2,i}) is perfect and nowhere dense, moreover, |E_{3}| = 1 − 1/2^{3}.

Continue this way, and get a sequence of perfect nowhere dense sets

E_{1};  E_{1,1}, E_{1,2}, E_{1,3}, ..., E_{1,i}, ...;  E_{2,1}, E_{2,2}, E_{2,3}, ..., E_{2,i}, ...;
E_{k,1}, E_{k,2}, E_{k,3}, ..., E_{k,i}, ...; ...

Observe that all of them are disjoint, and make the desired conclusion.