# How to Prove It - Solutions

## Chapter - 7, Infinite Sets

### Summary

• Suppose A and B are countable sets. Then:
• $A \times B$ is countable.
• $A \cup B$ is countable.
• The union of countably many countable sets is countable. In other words, if $\mathcal F$ is a family of sets, $\mathcal F$ is countable, and also every element of $\mathcal F$ is countable, then $\cup \mathcal F$ is countable.
• Suppose $A$ is a set. A function $f : I_n \to A$, where $n$ is a natural number, is called a finite sequence of elements of $A$, and $n$ is called the length of the sequence.
• Suppose $A$ is a countable set. Then the set of all finite sequences of elements of $A$ is also countable.
• Cantor’s theorem: $\mathcal P(Z^+)$ is uncountable.
• $\mathbb R$ is uncountable.

Soln1

(a)

We know that from a theorem of previous section that $\mathbb Q$ is denumerable. Thus $\mathbb Q$ is countable. Suppose $\mathbb R \setminus \mathbb Q$ is countable. But by first theorem of this section, we know that if $A$ and $B$ are countable then $A \cup B$ is also countable. Thus in our case we can say that $(\mathbb R \setminus \mathbb Q) \cup \mathbb Q)$ is also countable. Thus $\mathbb R$ is countable. But $\mathbb R$ is uncountable. Thus we have a contradiction. Thus $\mathbb R \setminus \mathbb Q$ is not countable.

(b)

We need to prove that $\mathbb R \setminus \mathbb Q \; \sim \; \mathbb R$.

Suppose $A = \{ \sqrt 2 + x \; \vert \; x \in \mathbb Z^+ \}$. Clearly we can write a one-to-one and onto function from $\mathbb Z^+ \to A$. Thus $\mathbb Z^+ \sim A$. Thus $A$ is denumerable. Also we know that $\mathbb Q$ is denumerable(using a theorem of this section).
Since $A \sim \mathbb Z^+$ and $\mathbb Q \sim \mathbb Z^+$ and $A \cap \mathbb Q = \phi$, it follows that $A \cup \mathbb Q \sim \mathbb Z^+$. Since $\mathbb Z^+ \sim A$, it follows that $A \cup \mathbb Q \sim A$.

Now we shall try to use above theorems related to set operations of denumerable sets to prove the required theorem: $\mathbb R \setminus \mathbb Q \; \sim \; \mathbb R$. We can see that $\mathbb R = (\mathbb R \setminus (A \cup \mathbb Q)) \; \cup \; (A \cup \mathbb Q)$. Also $\mathbb R \setminus \mathbb Q = (\mathbb R \setminus (A \cup \mathbb Q)) \; \cup \; \mathbb Q$. Thus clearly by first theorem of this section, it follows that $(\mathbb R \setminus (A \cup \mathbb Q)) \, \cup \, (A \cup \mathbb Q) \; \sim \; (\mathbb R \setminus (A \cup \mathbb Q)) \, \cup \, \mathbb Q$. Thus $\mathbb R \setminus \mathbb Q \; \sim \; \mathbb R$.

Soln2

• $F$ is one-to-one:

Suppose $(f_1,a)$ and $(f_2,b)$ in $S_n \times A$ such that $F(f_1,a) = F(f_2,b)$. Thus $f_1 \cup \{(n+1,a)\} = f_2 \cup \{(n+1,b)\}$. Thus either $(n+1,a) \in f_2$ or $(n+1, a) = (n+1,b)$.
Since $f_2 \in S_n$, it follows that $f_2: I_n \to A$. Thus $(n+1,a) \notin f_2$. Thus $(n+1, a) = (n+1,b)$. Thus $f_1 = f_2$ and $a = b$. It follows that $(f_1,a) = (f_2,b)$. Thus $F$ is one-to-one.

• $F$ is onto:

Suppose $f \in S_{n+1}$. Thus $f$ is a $(n+1)$-length sequence. Thus $f: I_{n+1} \to A$. Suppose $a \in A$ such that $(n+1,a) \in f$. Thus we can write $f = (f \setminus \{(n+1,a)\}) \, \cup \, \{(n+1,a)\}$. Let $g = (f \setminus \{(n+1,a)\})$. Since $\vert f \vert = n+1$, it follows that $\vert g \vert = n$. Thus $g: I_n \to A$. Thus $g$ is a sequence of $n$ length. Thus $g \in S_n$. Thus $F(g,a) = g \cup \{(n+1,a)\} = f$. Thus $F$ is onto.

Soln3

Let $P = \{ X \in P(Z^+) \; \vert \; X \text{ is finite } \}$. Suppose $P_n = \{ X \subseteq Z^+ \, \vert \, \vert X \vert \le n \}$. Thus clearly we can check that $P = \cup_{i \in I_n} P_n$.

Suppose $X \in P_n$. Thus $X$ is finite and $\vert X \vert \le n$. Thus we can also write a finite sequence for $X$ such that $X = \{ a_1, a_2, ... a_k$ where $k \le n$. Thus $P = \cup_{i \in \mathbb N} P_n$ is a set of finite sequences of the countable set $Z^+$. It follows from theorem-7.2.4 that $P$ is countable.

Since $P$ is countable, it follows that either it is finite or denumerable. Suppose $P$ is finite. Thus we can choose some $n \in \mathbb N$ such that $f: P \to I_n$. Suppose $X \in P$ such that $\vert X \vert = n$. It follows from section-7.1, Ex-22 that there are $n!$ sequences for the set $\vert X \vert$. But all of these sequences are in set $P$. Thus for each of this sequence say $x$, there must be a unique $n$ in $I_n$. Clearly this is not possible since $n! > n$. Thus $P$ is not finite. Thus $P$ must be denumerable.

Soln4

Suppose $f: A \to P(A)$ is an arbitrary function. Let $D = \{a \; \vert \; a \notin f(a) \}$. Now suppose $a \in A$ is an arbitrary element. Thus we have following possible cases:

• Case $a \in f(a)$:

Thus $a \notin D$. It follows that $f(a) \ne D$.

• Case $a \notin f(a)$:

Thus $a \in D$. Since $a \notin f(a)$, it follows that $f(a) \ne D$.

Thus in both cases $f(a) \ne D$. Since $a$ is arbitrary, it follows that $\forall a \in A(f(a) \ne D)$. Thus $f$ is onto. Since $f$ is arbitrary it follows that there is no onto function such that $f: A \to P(A)$. Thus $A \nsim P(A)$.

Soln5

Note: I skipped proofs for proving the defined relations are functions.

(a)

Suppose $\mathcal H: {^A B} \times {^A C} \; \to \; {^{A} {(B \times C)} }$ such that $\mathcal H(f,g) = \{ h \in {^{A} {(B \times C)} } \; \vert \; if ((a_1, b) \in f \text{ and } (a_2, c) \in g) \text{ then } ( (a_1, (b,c)) \in h \text{ and } (a_2, (b,c)) \in h)$.

• $\mathcal H$ is one-to-one:

Suppose $f_1, f_2 \in {^A B}$ and $g_1,g_2 \in {^A C}$ such that $\mathcal H(f_1, g_1) = \mathcal H(f_2,g_2)$. Suppose $(a,b) \in f_1$ and $(a,c) \in g_1$. Thus $(a,(b,c)) \in \mathcal H(f_1,g_1)$. Since $\mathcal H(f_1, g_1) = \mathcal H(f_2,g_2)$, it follows that $(a,(b,c)) \in \mathcal H(f_2,g_2)$. It follows from the definition of $\mathcal H$ that $(a,b) \in f_2$ and $(a,c) \in g_2$. Thus $f_1 \subseteq f_2$ and $g_1 \subseteq g_2$. Similarly we can prove that $f_2 \subseteq f_1$ and $g_2 \subseteq g_1$. Thus $f_1 = g_1$ and $f_2 = g_2$.

• $\mathcal H$ is onto:

Suppose $h \in {^{A} {(B \times C)} }$. Thus $h: A \to (B \times C)$. Let $f = \{ (a,b) \, \vert \, (a,(b,c)) \in h \}$ and $g = \{ (a,c) \, \vert \, (a,(b,c)) \in h \}$. Clearly $\mathcal H(f,g) = h$. Thus $\mathcal H$ is onto.

(b)

Suppose $\mathcal H: ^A {(^B C)} \; \to \; ^{A \times B} C$ such that $\mathcal H(a,f) = \{ g \in ^{A \times B} C \; \vert \; if ((b,c) \in f) \text{ then } ((a,b),c) \in g \}$.

• $\mathcal H$ is one-to-one:

Suppose $f_1, f_2 \in {^B C}$ and $a_1,a_2 \in A$ such that $\mathcal H(a_1, f_1) = \mathcal H(a_2,f_2)$. Suppose $(b,c) \in f_1$. Thus $((a,b),c) \in \mathcal H(a_1,f_1)$. It follows that $((a_1,b),c) \in \mathcal H(a_2, f_2)$. Thus $a_1 = a_2$ and $(b,c) \in f_2$. Thus $f_1 \subseteq f_2$. Similarly it can be proved that $f_2 \subseteq f_1$. Thus $f_1 = f_2$. Thus $(a_1, f_1) = (a_2,f_2)$. Thus $\mathcal H$ is one-to-one.

• $\mathcal H$ is onto:

Suppose $g \in ^{A \times B} C$. Thus $g: (A \times B) \to C$. Let $a \in A$ and $f_a = \{ (b,c) \, \vert \, ((a,b),c) \in g \}$. Let $f = \cup_{a \in A} f_a$. Clearly $\mathcal H(f) = g$. Thus $\mathcal H$ is onto.

(c)

Suppose $\mathcal H : \mathcal P(A) \to ^A { \{ yes, no\} }$ such that:
$\mathcal H(X) = \{ f \in ^A { \{ yes, no\} } \, \vert \, a \in A, if(a \in X) \text{ then } (a,yes) \in f \text{ else } (a,no) \in f \}$.

• $\mathcal H$ is one-to-one:

Suppose $X_1, X_2 \in \mathcal P(A)$ such that $\mathcal H(X_1) = \mathcal H(X_2)$. Suppose $a \in X_1$. It follows that $(a,yes) \in \mathcal H(X_1)$. Thus $(a,yes) \in \mathcal H(X_2)$. Thus by the definition of $\mathcal H$, it follows $a \in X_2$. Thus $X_1 \subseteq X_2$. Similarly it can be proved that $X_2 \subseteq X_1$. Thus $X_1 = X_2$. Thus $\mathcal H$ is one-to-one.

• $\mathcal H$ is onto:

Suppose $f \in ^A { \{ yes, no\} }$. Thus $f: A \to \{ yes, no \}$. Suppose $X = \{ a \in A \, \vert \, (a,yes) \in f \}$. Now we can see that $\mathcal H (X) = f$. Thus $\mathcal H$ is onto.

(d)

We need to prove $^{Z^+} {\mathcal P(Z^+)} \, \sim \, \mathcal P(Z^+)$.

We know from part(c) $^{Z^+} { \{ yes, no\} } \sim \mathcal P(Z^+)$.

TODO.

Soln6

Suppose $A$ is denumerable. Thus we can represent elements of $A = \{a_1, a_2, a_3 ... \}$. Suppose $\mathcal F$ is a family of sets such that $\mathcal F = \{ X_p \; \vert \;$ p $\text{ is prime } \}$ where $X_p = \{ a_i \in A \; \vert \; \text{ the *smallest* prime divisor of } i+1 \text{ is } p \}$(note $i+1$ to take care for a_1).

Clearly $\mathcal F$ is a partition as it satisfies all the properties of partition:

• $\cup \mathcal F = A$ is satisfied.
• $\mathcal F$ is pairwise disjoint. Note that every $a_i$ exist in exactly one of the sets.
• $\forall X \in \mathcal F(X \ne \phi)$. Since $A$ is denumerable, This will be correct for all $X$.

Now we shall prove that $\mathcal F$ is denumerable.

Suppose $f: \mathcal F \to Z^+$ such that $f(X_p) = p$. Clearly $f$ is one-to-one function. Thus $\mathcal F$ is countable. Now if we prove that $\mathcal F$ is not finite, then it will complete the proof that $\mathcal F$ is denumerable. Suppose $\mathcal F \sim I_n$. Clearly the $n+1$-prime number in the list or ordered primes will contradict with $\mathcal F \sim I_n$.

Now suppose $X_p \in \mathcal F$. Suppose $f: X_p \to Z^+$ such that $f(a_i) = i$. Clearly $f$ is one-to-one. Thus $X_p$ is countable. Clearly there are infinite multiples of $p$ such that $p$ is their lowest common divisor. It follows that $X_p$ is not finite. Thus $X_p$ is denumerable.

Soln7

Suppose $f: (\mathcal P(A) \times \mathcal P(B)) \; \to \; \mathcal P(A \cup B)$ such that $f(X,Y) = X \cup Y$.

• $f$ is one-to-one.

Suppose $X_1, X_2 \in \mathcal P(A)$ and $Y_1, Y_2 \in \mathcal P(B)$ such that $f(X_1,Y_1) = f(X_2,Y_2)$. Thus $X_1 \cup Y_1 = X_2 \cup Y_2$. Since $X_1 \cap Y_2 = \phi$(given), it follows that $X_1 = X_2$. Similarly since $Y_1 \cap X_2 = \phi$, it follows $Y_1 = Y_2$. Thus $(X_1,Y_1) = (X_2,Y_2)$. Thus $f$ is one-to-one.

• $f$ is onto.

Suppose $Z \in A \cup B$. Suppose $X = \{ a \in A \; \vert \; a \in Z \}$ and $Y = \{ b \in B \; \vert \; b \in Z \}$. Clearly $f(X,Y) = Z$. Thus $f$ is onto.

Soln8

TODO

Soln9

Suppose $\mathcal F \subseteq \{ f \; \vert \; f: \mathbb Z^+ \to \mathbb R \}$ and $\mathcal F$ is countable. Since $\mathcal F$ is countable, we can represent it as $\{ f_1, f_2, f_3, .... \}$. Now consider function $g: Z^+ \to \mathbb R$ such that $g(n) = max( \vert f_1(n) \vert , \vert f_2(n) \vert , ... \vert f_n(n) \vert)$. Now we can easily check that $\mathcal F \subseteq \mathbb O(g)$.

Suppose $f_n \in \mathcal F$. Clearly $\forall x > n (\vert f_n(n) \vert \le \vert g(n)$. Thus $f_n \in \mathbb O(g)$. Since $f_n$ is arbitrary, it follows that $\mathcal F \subseteq \mathbb O(g)$.

Soln10

Suppose $A$ is the set of all words in english. Clearly $A$ is finite. Thus $A$ is countable. Consider $S =$ set of all finite sequences of $A$. It follows from theorem-7.2.4 that $S$ is countable.

Let $G$ is the set of all grammtical sentences. Since each grammatical sentence if a finite sequence of english words, it follows that $G \subseteq S$. Since $S$ is countable, it follows that $G$ is also countable.

Thus either $G$ is finite or denumerable. Suppose $G$ is not denumerable. Thus $G$ is must be finite. Thus for some $n \in \mathbb N$, $G \sim I_n$. It follows that total number of sentences in english is $n$. Clearly this is not correct as we can create a new grammatical sentence such that it is not in the $n$ sentences. Thus our assumption is not correct. Thus $G$ is denumerable.

Soln11

(a)

Suppose $S$ is the set of mathematical symbols defined by english sentence. Suppose $G$ is the set of english sentences. Lets define a function $f: S \to G$, such that $f(s) = g$ such that $g$ is the grammatical sentence for the symbol $s$. Clearly $f$ is one-to-one. Since $G$ is denumerable, we can choose a one-to-one function $h: G \to Z^+$. It follows that $h \circ g : S \to Z^+$ is a one-to-one function. Thus $S$ is denumerable.

(b)

We will prove this by contradiction. Suppose all real numbers can be defined by english sentences. Thus we can define a one-to-one function $f: \mathbb R \to G$. Since $G$ is denumerable, we can choose a one-to-one function $g: G \to Z^+$. Thus $g \circ f : \mathbb R \to G$ is a one-to-one function. Thus by the theorem 7.1.5, it follows that $\mathbb R$ is countable. But $\mathbb R$ is not countable. Thus we have a contradiction and our assumption is wrong. Thus there exist some real numbers that can not be defined by english sentences.