Problem Set #3 Part C Official Solutions

Inna Zakharevich

October 16, 2003

(8) 1. Consider the sequence {x_{n}} where x_{n} ∈ F_{n}. Now, since E is compact, we know that this sequence has an accumulation point x. Let {y_{n}} be the subsequence that converges to x. Note that for each i, {y_{n}}_{n=i}^{∞} ⊂ F_{i}. In addition, since each of the F_{i} is closed, we know that each contains the limit of {y_{n}}, which is x. Thus x ∈ F_{i} for all i, so {x} ⊂ ∩_{n=1}^{∞}F_{n} so it is nonempty.

Now suppose that E is (0, 1] and let F_{n} = (0, 1/n]. Then for all x ∈ (0, 1], if we let N be the smallest integer greater than 1/x we see that x ∉ F_{N}. Thus ∩_{n=1}^{∞}F_{n} is empty.

(7) 2. Since E is compact, we know that for every r > 0 it can be covered with finitely many open balls of radius r. Let U_{n} be the set of open balls of radius 1/n that cover E. Let A_{n} be the set of centers of these balls. We know that A_{n} is finite, so it is countable. In addition, we know that every point in E is within 1/n of one of the points of A_{n}. Let A = ∪_{n=1}^{∞}A_{n}. This is a countable union of countable sets, so it is countable. Fix ϵ > 0, and let x ∈ E. Let N > 1/ϵ. Then we know that there is a point y ∈ A_{N} such that d(x, y) < ϵ, so there is a point y ∈ A such that d(x, y) < ϵ. Thus A is dense in E. So every compact space has a countable dense subset.

(25) 3.

(5) (1) Consider the complement of K. At the nth step, a union of open sets was taken out of K_{n}. Thus the complement of K, which is going to be the union of all of the intervals which was taken out, is going to be a union of open sets, which means that it will be open. Thus K is closed. We know that a closed subset of a compact space is compact. Therefore K is compact.

(5) (2) Note that at every step we take out a finite number of intervals. Since we do a countable number of steps, and at each point we take out a countable number of intervals, in the end we will have taken out a countable number of intervals. Note that at every step every interval present in K_{n} is reduced by 1/3 of its length. Thus at every step we take out 1/3 of the length that we have not yet taken out. So at the first step we will have taken out a total of 1/3. At the second step we take out a length of 2/3 · 1/3. At the nth step, we take out a length of (2/3)^{n−1} · 1/3. Thus total, we take out

∑_{n=1}^{∞} \frac{1}{3} (\frac{2}{3})^{n−1} = \frac{1}{3} ∑_{n=1}^{∞} (\frac{2}{3})^{n−1} = \frac{1}{3} · \frac{1}{1 − 2/3} = \frac{1}{3}3 = 1.

(5) (3) Suppose that the connected component of a point x is some interval in R. Then it will be of some length ϵ. Note, however, that that will mean that there is some interval of length ϵ that has not been taken out, which means that the total length taken out is at most 1 − ϵ ≤ 1. Since we know that 1 − ϵ = 0 (by part (2)) we see that ϵ = 0, which means that the connected component of the point is just the point itself.

(5) (4) We will prove this by induction. Write the real numbers in [0, 1] in base 3. Then K_{0} = [0, 1]. Note that when we take out the middle third of the interval we remove all real numbers such that a_{1} = 1, except for 1/3. However, 1/3 can be written in base 3 with a_{i} = 2 for i > 1 and a_{1} = 0. Thus after one step we have eliminated all points with a_{1} = 1. Now, suppose that after n steps we have removed all points with a_{i} = 1 for i ≤ n. Consider the n + 1-st step. Fix some interval. This will be the set of real numbers α where the first n a_{i} are a fixed sequence of 0s and 2s, and the rest are allowed to vary. When we remove the middle third of this we will remove all α in this interval such that a_{n+1} = 1 other than the one where a_{i} = 0 for all i > n + 1. However, this α can be written with a_{n+1} = 0 and a_{i} = 2 for all i > n + 1. Thus after n + 1 steps we will have removed all α with a_{n+1} = 1. In addition, we will not have removed any other numbers. Thus K will be the set of all α that can be written in the form ∑_{i=1}^{∞} a_{i}3^{−i} for a_{i} ∈ {0, 2}.

(5) (5) We will construct a bijection between the infinite sequences consisting of 0s and 1s, and the points of the Cantor set. Consider any point x ∈ K. Then we can write x = ∑_{i=1}^{∞} a_{i}3^{−i}. Map this to the sequence {a_{i}/2}, which will consist of 0s and 1s. Note that this is a bijection, since every sequence will yeild such a number, and every number yeilds a sequence. Thus we know that Card(K) = Card({0, 1}^{Z}), which we know is an uncountable set. Thus we see that K is uncountable.