# How to Prove It - Solutions

## Chapter - 7, Infinite Sets

### Summary

• If $A$ and $B$ are sets, then we will say that $B$ dominates $A$, and write $A \precsim B$, if there is a function $f : A \to B$ that is one-to-one. If $A \precsim B$ and $A \nsim B$, then we say that $B$ strictly dominates $A$, and write $A \prec B$.
• (Cantor–Schroöder–Bernstein theorem): Suppose $A$ and $B$ are sets. If $A \precsim B$ and $B \precsim A$, then $A \sim B$.
• Using Cantor–Schroöder–Bernstein theorem, it is easier to prove that $\mathbb R \sim \mathcal P(Z^+)$. We need to define a one-to-one function $f: \mathbb R \to \mathcal P(Z^+)$ and other one-to-one function $g: \mathcal P(Z^+) \to \mathbb R$.

Soln1

(a)

Suppose $f: A \to A$ such that $f(a) = a$. Clearly $f$ is one-to-one and onto. It follows that $A \precsim A$. Thus $\precsim$ is reflexive.

(b)

Suppose $A \precsim B$ and $B \precsim C$. Thus we can choose one-to-one function $f: A \to B$ and another one-to-one function $g: B \to C$. Since $f,g$ are one-to-one, the function $g \circ f: A \to C$ is also one-to-one. Thus $A \precsim C$.

Soln2

(a)

Suppose $A \prec A$. It follows that $A \precsim A$ and $A \nsim A$. But $A \sim A$. Thus we have a contradiction. It follows that $A \nprec A$.

(b)

Suppose $A \prec B$ and $B \prec C$. Thus we can choose one-to-one function $f: A \to B$ and another one-to-one function $g: B \to C$. Since $f,g$ are one-to-one, the function $g \circ f: A \to C$ is also one-to-one. Thus $A \precsim C$.

Now suppose $A \sim C$, or we can also say $C \sim A$. Since $\sim$ is transitive and $A \sim B$, it follows that $C \sim B$, or $B \sim C$. But it contradicts with $B \prec C$. Thus our assumption is wrong. Thus $A \nsim C$.

Since $A \precsim C$ and $A \nsim C$, it follows that $A \prec C$.

Soln3

Suppose $A \subseteq B \subseteq C$ and $A \sim C$. Since $A \subseteq B$, it follows that $A \prec B$. Since $A \sim C$, it follows that $C \sim A$. Thus $C \prec A$. Since $\prec$ is transitive and $C \prec A \land A \prec B$, it follows that $C \prec B$.

Now since $B \subseteq C$, it follows that $B \prec C$. Thus we have $C \prec B$ and $B \prec C$. Thus by Cantor–Schroöder–Bernstein theorem, it follows that $B \sim C$.

Soln4

Suppose $A \precsim B$ and $C \precsim D$. Thus we can choose one-to-one function $f: A \to B$ and another one-to-one function $g: C \to D$.

(a)

Suppose $h: A \times C \to B \times D$ such that $h(a,c) = (f(a), g(c))$.

Suppose $(a_1, b_1), (a_2,b_2) \in A \times C$ such that $h(a_1,c_1) = h(a_2,c_2)$. Thus $(f(a_1), g(c_1)) = (f(a_2),g(c_2))$. Since $f,g$ are one-to-one, it follows that $a_1 = a_2$ and $c_1 = c_2$. Thus $(a_1,c_1) = (a_2,c_2)$. Thus $h$ is one-to-one. Thus we can conclude that $A \times C \precsim B \times D$.

(b)

Suppose $h: A \cup C \to B \cup D$ such that:

• if $a \in A$ then $h(a) = f(a)$.
• if $a \in C$ then $h(a) = g(a)$.

Suppose $a_1,a_2 \in A \cup C$ such that $h(a_1) = h(a_2)$. Since $A \cap C = \phi$. We have following possible cases:

• Case $a_1 \in A$.

Thus $a_1 \notin C$. Also by definition of $h$, $h(a_1) = f(a_1)$. Suppose $a_2 \in C$. Thus $h(a_2) = g(a_2)$. Since $h(a_1) = h(a_2)$, it follows $f(a_1) = g(a_2)$. Since $f(a_1) \in B$ and $g(a_2) \in D$, it follows that $B \cap D \ne \phi$. But $B \cap D = \phi$. Thus we have contradiction. Thus $a_2 \notin C$. It follows that $a_2 \in A$. Thus $h(a_2) = f(a_2)$. Since $h(a_1) = h(a_2)$, it follows $f(a_1) = f(a_2)$. Since $f$ is one-to-one, it follows that $a_1 = a_2$.

• Case $a_1 \in C$.

Similar to previous case, we can prove by contradiction that $a_2 \in C$. Thus $g(a_1) = g(a_2)$. Thus $a_1 = a_2$.

Thus by all possible cases, it follows that $a_1 = a_2$. Thus $h$ is one-to-one. Thus $A \cup C \precsim B \cup D$.

(c)

Suppose $h: \mathcal P(A) \to \mathcal P(B)$ such that $h(X) = \{ f(x) \; \vert \; x \in X \} = Y$.

• $Y \subseteq B$.

Suppose $y \in Y$. Clearly we can choose some $x \in A$ such that $f(x) = y$. Since $x \in A$, it follows that $f(x) \in B$, or $y \in B$. Since $y$ is arbitrary, $Y \subseteq B$. Thus $Y \in \mathcal P(B)$.

• $h$ is one-to-one.

Suppose $X_1$ and $X_2$ are in $\mathcal P(A)$ such that $h(X_1) = h(X_2)$. We can prove $X_1 = X_2$ as follows:

Suppose $x_1 \in X_1$ such that $y = f(x_1) \in h(X_1)$. Since $h(X_1) = h(X_2)$, it follows that $y \in g(X_2)$. Thus for some $x_2 \in X_2$, we have $y = f(x_2)$. Thus $f(x_1) = f(x_2)$. Since $f$ is one-to-one, it follows $x_1 = x_2$. Thus $x_1 \in X_2$.

Since $x_1$ is arbitrary, it follows that $X_1 \subseteq X_2$.

Similarly we can also prove $X_2 \subseteq X_1$.

Thus $X_1 = X_2$.

Thus $h$ is one-to-one. It follows that $\mathcal P(A) \precsim \mathcal P(B)$.

Soln5

Suppose $A \precsim B$ and $C \precsim D$. Thus we can choose one-to-one function $f: A \to B$ and another one-to-one function $g: C \to D$.

(a)

Since $A \ne \phi$, we can choose some element $a_0 \in A$. Lets define a function $f' : Ran(f) \to A$ such that:

• if $b \in Ran(f)$, then $f'(b) = f^{-1}(a)$. Clearly $f^{-1}$ is defined for all $b \in Ran(f)$.
• if $b \notin Ran(f)$, then $f'(b) = a_0$.

Clearly $f'$ is onto.

Now consider the function $h: ^A C \to ^B D$ such that $h(f) = g \circ f \circ f'$. Clearly $f \in ^A C$, thus $f: A \to C$. Thus $g \circ f \circ f'$ is a function from $B$ to $D$.

Now we will prove that $h$ is one-to-one. Suppose $f_1, f_2 \in ^A C$ such that $h(f_1) = h(f_2)$. Thus $g \circ f_1 \circ f' = g \circ f_2 \circ f'$. Suppose $a \in A$ is an arbitrary element. Since $f'$ is onto, it follows that we can choose $b \in B$ such that $f'(b) = a$. Since $b \in B$ and $g \circ f_1 \circ f' = g \circ f_2 \circ f'$, it follows that $g \circ f_1 \circ f'(b) = g \circ f_2 \circ f'(b)$.
Thus $\leftrightarrow g(f_1(f'(b))) = g( f_2 ( f'(b) ))$. Since $g$ is one=to-one, it follows $f_1(f'(b)) = f_2 ( f'(b) )$. Since $a = f'(b)$,
we get $f_1(a) = f_2(a)$. Since $a$ is arbitrary, it follows that $\forall a \in A f_1(a) = f_2(a)$. Thus $f_1 = f_2$. Thus $h$ is onto. Thus we can conclude that $^A C \precsim ^B D$.

(b)

Yes.

Suppose $A = \phi$. Since $A \precsim B$, it follows that $f: \phi \to B$ such that $f$ is one-to-one. Thus $f = \phi$.

Suppose $B = \{ b \}$, and $C = \{c\}$ and $D = \{d,e\}$.

Thus $^A C = \{ \phi \}$. And $^B D = \{ \{ (b,d) \}, \{ (b,e) \} \}$.

Thus the only function possible from $^A C$ to $^B D$ is $\phi$. But clearly $\phi$ is not one-to-one, because it is a part of every set. Thus there is no one-to-one function from $^A C$ to $^B D$.

Soln6

(a)

Suppose $\vert B \vert = n$. Since $B$ is finite, by countable set theorems of last section, we can choose one-to-one function $f: B \to Z^+$. Since $A \precsim B$, we can choose a one-to-one function $g: A \to B$. Thus $f \circ g: A \to Z^+$ is a one-to-one function. Thus $A$ is countable. Thus $A$ either finite or denumerable.

Now we will prove by contradiction that $A$ is finite. Suppose $A$ is infinite. Since $A$ is countable, it follows that $A$ must be denumerable. Thus we can represent elements of $A$ as $\{a_1, a_2, ... a_n, ... \}$. Let $b_i = f(a_i)$ where $i \in \mathbb N$. Thus $\{ b_i \; \vert \; b_i = f(a_i), i \in \mathbb N \} \subseteq B$. Since $g: A \to B$ is one-to-one, it follows that $b_i \ne b_j$, if $i \ne j$. Thus $\vert \{ b_1, b_2, .... b_n, b_{n+1} \} \vert = n+1$. Since $\{ b_1, b_2, .... b_n, b_{n+1} \} \subseteq B$, it follows $\vert \{ b_1, b_2, .... b_n, b_{n+1} \} \vert \le \vert B \vert$. Thus $n+1 \le \vert B \vert$. But $\vert B \vert = n$. Thus we have a contradiction. It follows that $A$ is finite.

Suppose $\vert A \vert$ contains $m$ elements. Suppose $m > n$. Thus $A = \{ a_1, a_2, ... a_m \}$. Similar to above step, lets define $b_i = f(a_i)$ where $0 \le i \le m$. Thus $\{ b_1, b_2, .... b_m \} \subseteq B$. Thus $\vert \{ b_1, b_2, .... b_m \} \vert \le \vert B \vert$. Since $g: A \to B$ is one-to-one, it follows that $\{ b_1, b_2, .... b_m \} = m$. Thus $m \le n$. This contradicts with our assumption that $m > n$. Thus $m \le n$. Thus $\vert A \vert \le \vert B \vert$.

(b)

Since $A \prec B$, it follows that there exist a one-to-one function $f: A \to B$. Thus $A \precsim B$. Since $B$ is finite it follows from part(a) that $A$ is finite and $\vert A \vert \le \vert B \vert$.

Now suppose $\vert A \vert = \vert B \vert$. Thus $A \sim B$. Since $A \prec B$, it follows that $A \nsim B$. Thus we have a contradiction. It follows that $\vert A \vert \ne \vert B \vert$.

Since $\vert A \vert \le \vert B \vert$ and $\vert A \vert \ne \vert B \vert$, it follows that $% $.

Soln7

Suppose $f: A \to \mathcal P(A)$ such that $f(a) = \{ X \in \mathcal P(A) \; \vert \; a \in X \}$.

Now we will show that $f$ is one-to-one. Suppose $a_1, a_2 \in A$ such that $f(a_1) = f(a_2)$. Clearly by definition of $f$, it follows $\{a_1\} \in f(a_1)$. Similarly $\{a_2\} \in f(a_2)$. Note that all the elements of $f(a_1)$ must contain $a_1$. Thus every set in $f(a_1)$ has more than one element except the set $\{a_1\}$. Similarly $\{a_2\}$ is the only set in $f(a_2)$ that contains one element. Since $f(a_1) = f(a_2)$, it follows that $\{a_1\} = \{a_2\}$. Thus $a_1 = a_2$. Thus $f$ is one-to-one.

Thus $A \precsim \mathcal P(A)$. From Section-7.2, Ex-4, we know that $A \nsim \mathcal P(A)$. Thus $A \prec \mathcal P(A)$.

Soln8

(a)

We will prove this by induction.

• Base Case:

For $m = n+1$, Thus $A_m = A_{n+1} = \mathcal P(A_n)$. Using Ex-7 we have $A_n \prec \mathcal P(A_n)$. Thus $A_n \prec A_m$.

• Induction Step:

Suppose for $m > n$, $A_n \prec A_m$.

We know from Ex-7 that $A_m \prec \mathcal P(A_m)$. Since $A_{m+1} = \mathcal P(A_m)$, it follows that $A_m \prec A_{m+1}$. Since $\prec$ is transitive and $A_n \prec A_m$ and $A_m \prec A_{m+1}$, it follows that $A_n \prec A_{m+1}$.

(b)

Suppose $S = \cup_{i \in \mathbb N} A_n$. Clearly this set is not equinumerous with any set $A_n$. Another example can be $\mathcal P(S)$. Thus $S_{n+1} = \mathcal P(S_n)$, where $n \in \mathbb N$. Thus there are infinite possible sets which are not equinumerous with $A_n$.

Soln9

Suppose $f: (0,1] \to (0,1)$ such that $f(x) = x/2$.

Suppose $g : (0,1) \to (0,1]$ such that $g(x) = x$.

Let $A = (0,1]$ and $B = (0,1)$. Also we can define $g^{-1}: Rang(g) \to A$ such that $g^{-1}(b) = a$ such that $g(a) = b$.

Clearly $f$ and $g$ are one-to-one. Thus by Cantor–Schröder–Bernstein Theorem, we can define a one-to-one and onto function $h: (0,1] \to (0,1)$ such that:

• if $x \in X$, then $h(x) = f(x) = x/2$.
• if $x \in Y$, then $h(x) = g^{-1}(x) = x$.

where $X = \cup_{i \in \mathbb N} A_i$ and $A_1 = A \setminus Rang(g)$ and $A_{n+1} = g(A_n)$.
$Y = A \setminus X$.

Lets compute $X$:

$A_1 = A \setminus Rang(g) = (0,1] \setminus (0,1) =  = \{1\}$.
$A_2 = g(f(A_1)) = g(\{1/2\}) = \{1/2\}$.
$A_3 = g(f(A_2)) = g(\{1/4\}) = \{1/4\}$.
….
$A_n = g(f(A_{n-1})) = \{1/2^{n-1}\}$.

Thus $X = \cup_{i \in \mathbb N} A_i = \{1, 1/2, 1/4, 1/8, .... 1/2^{n-1}, .... \} = \{ 1/2^i \; \vert \; 0 \le i \le 1 \}$.

Thus $Y = A \setminus X = (0,1] \setminus \{1\} = (0,1)$.

Thus we have $h: (0,1] \to (0,1)$ such that:

• if $x \in X$, then $h(x) = x/2$.
• if $x \in Y$, then $h(x) = x$.

Soln10

Let $\mathcal E = \{ R \; \vert \; R \text{ is an equivalence relation on } \mathbb Z^+ \}$.

(a)

Clearly $\mathcal E \subseteq \mathcal P(Z^+ \times Z^+)$. Thus $\mathcal E \prec \mathcal P(Z^+ \times Z^+)$. Since $Z^+ \sim Z^+ \times Z^+$, it follows from Section-7.1, Ex-5 that $\mathcal P(Z^+ \times Z^+) \sim \mathcal P(Z^+)$. Thus $\mathcal E \prec \mathcal P(Z^+)$.

(b)

Suppose $X_1, X_2 \in \mathcal P(A)$ such that $f(X_1) = f(X_2)$. Thus it follows that:
$\{ X_1 \cup \{1\} , (A \setminus X_1) \cup \{2\} \} = \{ X_2 \cup \{1\} , (A \setminus X_2) \cup \{2\} \}$.

Thus either $\{ X_1 \cup \{1\} = (A \setminus X_2) \cup \{2\}$, or $X_1 \cup \{1\} = X_2 \cup \{1\}$. Since $1 \notin (A \setminus X_2) \cup \{2\}$, it follows that $\{ X_1 \cup \{1\} \ne (A \setminus X_2) \cup \{2\}$. Thus the only possible case is $X_1 \cup \{1\} = X_2 \cup \{1\}$. Thus $X_1 = X_2$. It follows that $f$ is one-to-one.

(c)

We have $\mathcal P$ is the set of all partitions of $A$ and $\mathcal E$ is the set of all equivalence relation of $A$. As we saw in section-5.3, we can define a one-to-one and onto function $f: \mathcal P \to \mathcal E$ such that $f(X) = \cup_{X \in \mathcal P} (X \times X)$.

Thus $\mathcal P \sim \mathcal E$.

Now from part(b) we have $\mathcal P(A) \precsim \mathcal P$. Also since $A \subseteq Z^+$, it follows that $Z^+ \sim A$. Thus $\mathcal P(Z^+) \sim \mathcal P(A)$. Since $\mathcal P(Z^+) \sim \mathcal P(A)$ and $\mathcal P(A) \precsim \mathcal P$, it follows that $\mathcal P(Z^+) \precsim \mathcal P$. Since $P \sim \mathcal E$, it follows that $\mathcal P(Z^+) \precsim \mathcal E$.

Now from part(a) we have $\mathcal E \precsim \mathcal P(Z^+)$. Thus by applying Cantor–Schröder–Bernstein Theorem, it follows that $\mathcal E \sim \mathcal P(Z^+)$.

Soln11

TODO

Soln12

(a)

TODO[This is not complete.]

Suppose $a \in A$. Suppose $A_1 = \{a\}$ and $A_2 = A \setminus A_1$. Thus $A = A_1 \cup A_2$.

Since $A$ contains atleast two elements, by definition $A_1 \ne \phi$ and also $A_2 \ne \phi$.

Clearly $A_1 \cap A_2 = \phi$, thus it follows from Section-7.2, Ex-7, that $\mathcal P(A) = \mathcal P(A_1 \cup A_2) = \mathcal P(A_1) \times \mathcal P(A_2)$.

Since $A_1 \subseteq A$, it follows $\mathcal P(A_1) \subseteq \mathcal P(A)$. Similarly since $A_2 \subseteq A$, it follows $\mathcal P(A_2) \subseteq \mathcal P(A)$. Since $\mathcal P(A_1) \subseteq \mathcal P(A)$ and $\mathcal P(A_2) \subseteq \mathcal P(A)$, it follows:
$\mathcal P(A_1) \times \mathcal P(A_2) \; \subseteq \; \mathcal P(A) \times \mathcal P(A)$.

Since $\mathcal P(A_1) \times \mathcal P(A_2) \; \subseteq \; \mathcal P(A) \times \mathcal P(A)$, it follows $\mathcal P(A_1) \times \mathcal P(A_2) \; \prec \; \mathcal P(A) \times \mathcal P(A)$.

But we have already seen that $\mathcal P(A) \sim \mathcal P(A_1) \times \mathcal P(A_2)$. Thus it follows that $\mathcal P(A) \; \prec \; \mathcal P(A) \times \mathcal P(A)$.

Now if we also prove that $\mathcal P(A) \times \mathcal P(A) \; \prec \; \mathcal P(A)$, then applying Cantor–Schröder–Bernstein theorem will complete the proof.

I got stuck in proving this part: $\mathcal P(A) \times \mathcal P(A) \; \prec \; \mathcal P(A)$.

(b)

TODO.

Soln13

Suppose $\mathcal F$ is a set of nondegenerate intervals and $\mathcal F$ is pairwise disjoint.

Suppose $I \in \mathcal F$ is an arbitrary interval. Since it is nondegenerate interval, it must contains $\ge 2$ real numbers. We know from lemma 7.3.4, that it between every two real numbers, there exist a rational number. Thus we can define a non-empty set:
$% $.

Clearly $Q_I \ne \phi$ and $Q_I \subseteq \mathbb Q$.

Suppose $g: \mathcal F \; \to \; \mathcal P(\mathbb Q) \setminus \{ \phi \}$ such that:

$g(I) = Q_I$.

Now we will prove that $g$ is one-to-one:

Suppose $I_1, I_2 \in \mathcal F$ such that $g(I_1) = g(I_2)$. Now we will prove by contradiction that $I_1 = I_2$. Suppose $I_1 \ne I_2$. Since $\mathcal F$ is pairwise disjoint, it follows that $I_1 \cap I_2 = \phi$. Suppose $q \in g(I_1)$. It follows that for some $x_1, y_1 \in I_1$, we have $% $. Thus by definition of interval, $q \in I_1$. Similarly since $q \in g(I_2)$, we can prove that $q \in I_2$. Thus $q \in I_1 \cap I_2$. But $I_1 \cap I_2 = \phi$. Thus we have a contradiction. It follows that $I_1 = I_2$.

Thus $g$ is one-to-one.

Since $\mathbb Q$ is denumerable, we can define a one-to-one function $f: \mathbb Q \to Z^+$.

Thus we can define a funnction $h: \mathcal F \to Z^+$ as:

$h(I) = \text{ smallest element of } \{ f(q) \; \vert \; q \in g(I) \}$.

Since $g$ is one-to-one, it follows that $h$ is also one-to-one.

Thus $\mathcal F \sim Z^+$.

Soln14

TODO.