Solution Set #1 - Microsoft Research

18.510 Introduction to Mathematical Logic and Set Theory
Solution Set #1
1.
(1) The function f : (0, 1) → R defined by f(x) = tan(π(x − 1/2)) is a bijection because it has the inverse
function f −1 (x) = 1/2 + arctan(x)/π.
(2) Choose distinct elements x 0 , x 1 , x 2 , . . . of (0, 1). Then we can define a bijection f : (0, 1) → [0, 1]
by f(x 0 ) = 0, f(x 1 ) = 1, f(x i+2 ) = x i for i ≥ 0, and f(y) = y for y ∈ (0, 1) \ {x 0 , x 1 , . . . }. This
amounts to a bijection between the countably infinite sets {x 0 , x 1 , . . . } and {0, 1, x 0 , x 1 , . . . }, while
doing nothing to the rest of the interval.
(3) Because we can compose with a bijection between R and [0, 1], all we need to construct is a bijection
between [0, 1] and {0, 1} N . Binary expansions almost work, but not quite, because two different binary
expansions can give the same number. Specifically, there’s an ambiguity between those ending in all
0’s and in all 1’s. (E.g., 1/2 = 0.1000 · · · = 0.0111 . . . in binary.) However, that’s the only ambiguity,
and it occurs only for rational numbers whose denominators are powers of 2. We can fix it as follows.
Let A be the set of rational numbers in [0, 1] whose denominators are powers of 2, and let B be
the subset of {0, 1} N consisting of sequences that are eventually all 0’s or all 1’s. Then taking binary
expansions gives a bijection between [0, 1] \ A and {0, 1} N \ B. Call it g : [0, 1] \ A → {0, 1} N \ B.
On the other hand, A and B are both countably infinite. Choose any bijection h: A → B between
them, and then
{
g(x) if x ∈ [0, 1] \ A, and
f(x) =
h(x) if x ∈ A.
defines a bijection from [0, 1] to {0, 1} N .
(4) Using the bijection from (3), all we need is a bijection between {0, 1} N and ( {0, 1} N) 2
. To do that,
we just send a sequence (a 0 , a 1 , . . . ) ∈ {0, 1} N to the pair ( (a 0 , a 2 , . . . ), (a 1 , a 3 , . . . ) ) .
(5) We want a bijection between R and R × {0, 1}. Using (3), that amounts to a bijection between {0, 1} N
and {0, 1} N × {0, 1}. We can send (a 0 , a 1 , . . . ) to ( (a 1 , a 2 , . . . ), a 0
)
.
2. Every set S of disjoint open intervals in R is countable. Specifically, define a function f : S → Q that maps
each element of S to a rational number contained in it (there is a rational number in each open interval; the
only reason why they have to be open is to avoid degenerate cases like [ √ 2, √ 2]). The disjointness of the
intervals in S shows that f is injective, and so it gives a bijection between S and a subset of Q. Because Q is
countable, so is every subset of Q and thus S is countable.
Alternatively, for all positive integers n and m, consider the subset S n,m of S consisting of the intervals of
length at least 1/n that are contained in [−m, m]. Then |S n,m | ≤ 2m/n, since the sum of the lengths of the
intervals in S n,m can be at most 2m. On the other hand,
S = ⋃ n,m
S n,m ,
since each open interval (a, b) in S is contained in S n,m when n > 1/(b − a) and m ≥ max(|a|, |b|). Thus, S
is a countable union of finite sets and hence countable.
3. There is an uncountable set of nested subsets of N. Because Q is in one-to-one correspondence with N,
it suffices to find subsets of Q with the same property. For each r ∈ R, let S r = {x ∈ Q : x ≤ r}. Then
these sets are distinct subsets of Q, and they are nested because S r1 ⊆ S r2 whenever r 1 ≤ r 2 . There are
uncountably many of them because R is uncountable.
4. Again, there is an uncountable set. As above, it suffices to construct subsets of Q instead of N. For each
real number r, pick a sequence s r 0, s r 1, . . . of rational numbers converging to r. Then the sets {s r i : i ∈ N}
are distinct for distinct r, because they have different limit points, and any two of these sets have finite
intersection: if there were an infinite intersection, it would give a subsequence of both sequences, which would
therefore converge to two different limits. Because R is uncountable, this gives an uncountable collection of
subsets of Q with finite intersections.
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Alternately, one can solve the problem as follows. Instead of considering subsets of N, we will consider
subsets of ⋃ n≥1 {0, 1}n ; since both sets are countable, this is equivalent. Given r ∈ [0, 1] write r = ∑ i≥1 r i2 −i
with r i ∈ {0, 1}; i.e., take the r i to be a binary expansion of r (if there are two, pick either). Define
S r = {(r 1 , . . . , r n ) : n ≥ 1}. Then (z 1 , . . . , z n ) ∈ S r ∩ S q only if the chosen binary expansions of r and q agree
to n digits and those digits are z 1 , . . . , z n ; if r ≠ q then the binary expansions differ at some finite point m,
so S r ∩ S q has size m − 1 (if m is the first place at which they differ).
5. There exists an uncountable set of inequivalent sequences. Consider the equivalence classes of sequences.
We’ll show that they are all countable, from which it follows that there must be an uncountable number of
equivalence classes. (If there were only countably many, then there would be only countably many sequences
in total.) Then taking one representative of each equivalence class gives an uncountable set of inequivalent
sequences.
Thus, all we need to do is to show that the equivalence class of a sequence (a 0 , a 1 , . . . ) is countable. For
each N, there are only countably many sequences (b 0 , b 1 , . . . ) such that a n = b n for all n ≥ N; the reason
is that such a sequence differs from the original only in (b 0 , b 1 , . . . , b N−1 ), which is an element of Q N and
thus has only countably many possibilities. Taking the union over all N gives a union of countably many
countable sets, which remains countable.
6. Suppose for all finite subsets T ⊆ S, we have ∑ s∈T f(s) ≤ B. Define S n = {s ∈ S : f(s) ≥ 1/n}. Then
for each n we must have |S n | ≤ Bn, for otherwise we could take a finite subset T ⊆ S n with |T | > Bn; then
∑
f(s) ≥ ∑ 1/n = |T |/n > B,
s∈T s∈T
contradicting our assumption. Then S ′ := ⋃ n∈N S n is a countable union of finite sets, and so is countable.
However, if f(s) > 0 then f(s) ≥ 1/n for some n, so f(s) > 0 ⇒ s ∈ S ′ , as desired.
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