give answer in 2 step with explanation at the end of each step and final answer at the end:

Solving in all parts.Lemma 04.1. Let A be a set and S be a countable set. If there is a one-to-one functi

Question

give answer in 2 step with explanation at the end of each step and final answer at the end:

Solving in all parts.Lemma 04.1. Let A be a set and S be a countable set. If there is a one-to-one function
9: A — S mapping A into S (note that we are not assuming g to be onto), then A is
countable.
Proof. Note that f [A] is countable since it is a subset of a countable set. Now f: A —
J [A] is one-to-one onto. Thus A is countable. [mu]
Theorem 04.10. The product set N x N is countable.
Proof. Let f: Nx N — N be given by f (m,n) = 2™ 3". Then f is one-to-one as can
easily be checked. So in view of Lemma 0.4.1 the result follows. [m}
Let Xi, X>.... be nonempty sets. For each n € N let x, = {(z.,n) | z € Xa}. We
therefore have
X= {@)) |zeXi}
X= {@2) |zeXs), ec
Notice that for any k, m € N with k # m, x n Xi = () because the second coordinate
of each element of X, k is different from the second coordinate of any element of Xow: Thus
the sets X,, are mutually disjoint. We claim that card(J X,,) < card(U %). Indeed, for
0
every z € [J X,, let i(z) be the smallest positive integer such that z € Xj(.). Defining
n=1
oo CI
B:UXu— UXa by B(2) = (2i(2))
n=1 n=1
gives a one-to-

VIPSolutions · Accepted Answer

Let's break down the main result (Theorem 04.10: The product set $\mathbb{N} \times \mathbb{N}$ is countable) in 2 steps, with explanations for each step, and then provide the final answer.

---

### **Step 1: Constructing a one-to-one function from $\mathbb{N} \times \mathbb{N}$ to $\mathbb{N}$**

**Explanation:**  
The proof uses a specific function $ f: \mathbb{N} \times \mathbb{N} \to \mathbb{N} $ defined by $ f(m, n) = 2^m \cdot 3^n $.  
- This function is **one-to-one (injective)** because the Fundamental Theorem of Arithmetic guarantees that every natural number has a unique prime factorization. So, different pairs $(m, n)$ will map to different numbers in $\mathbb{N}$.

---

### **Step 2: Applying Lemma 04.1 to conclude countability**

**Explanation:**  
By Lemma 04.1, if there is a one-to-one function from a set $A$ to a countable set $S$, then $A$ is countable.  
- Here, $A = \mathbb{N} \times \mathbb{N}$ and $S = \mathbb{N}$ (which is countable).
- Since we have constructed a one-to-one function from $\mathbb{N} \times \mathbb{N}$ into $\mathbb{N}$, it follows that $\mathbb{N} \times \mathbb{N}$ is countable.

---

### **Final Answer:**

$\mathbb{N} \times \mathbb{N}$ is countable because there exists a one-to-one function from $\mathbb{N} \times \mathbb{N}$ into $\mathbb{N}$, and $\mathbb{N}$ is a countable set.

VIPSolutions · Answer

**Step 1: Construct a one-to-one function from $\mathbb{N} \times \mathbb{N}$ to $\mathbb{N}$.**

- *Explanation:*  
  Define the function $f : \mathbb{N} \times \mathbb{N} \to \mathbb{N}$ by  
  $$
  f(m, n) = 2^m \cdot 3^n.
  $$  
  This function is **injective** because of the Fundamental Theorem of Arithmetic, which states that every natural number has a unique prime factorization. Since different pairs $(m, n)$ produce distinct products $2^m 3^n$, the function $f$ is one-to-one.

---

**Step 2: Use Lemma 04.1 to conclude that $\mathbb{N} \times \mathbb{N}$ is countable.**

- *Explanation:*  
  Lemma 04.1 states that if there exists an injective function from a set $A$ into a countable set $S$, then $A$ is countable.  
  Here, $A = \mathbb{N} \times \mathbb{N}$ and $S = \mathbb{N}$. Since $f$ is a one-to-one function from $\mathbb{N} \times \mathbb{N}$ into $\mathbb{N}$, it follows that $\mathbb{N} \times \mathbb{N}$ must be countable.

---

### **Final answer:**

$\boxed{\text{The set } \mathbb{N} \times \mathbb{N} \text{ is countable}}$ because there exists an injective function from $\mathbb{N} \times \mathbb{N}$ into $\mathbb{N}$.

VIPSolutions · Answer

Let's understand why the set $\mathbb{N} \times \mathbb{N}$ is countable, step by step, with clear explanations.

---

### **Step 1: Constructing an injective (one-to-one) function from $\mathbb{N} \times \mathbb{N}$ to $\mathbb{N}$**

**Explanation:**  
To show that $\mathbb{N} \times \mathbb{N}$ is countable, a common approach is to find a way to assign each pair $(m, n)$ of natural numbers to a unique natural number.

One effective method is to leverage prime factorization. Consider the function:
$$
f(m, n) = 2^m \times 3^n
$$
where $2$ and $3$ are prime numbers.

- Since every natural number has a unique prime factorization, each pair $(m, n)$ corresponds to a unique product $2^m \times 3^n$.  
- Different pairs produce different products because the exponents of primes are unique.  
- Therefore, this function $f$ is **injective**; no two different pairs $(m, n)$ map to the same number.

---

### **Step 2: Applying the lemma to conclude $\mathbb{N} \times \mathbb{N}$ is countable**

**Explanation:**  
Recall Lemma 04.1:  
*If there's an injective function from a set $A$ into a countable set $S$, then $A$ is countable.*

Here,  
- $A = \mathbb{N} \times \mathbb{N}$  
- $S = \mathbb{N}$, which is known to be countable

Since we've constructed the injective function $f$ from $\mathbb{N} \times \mathbb{N}$ into $\mathbb{N}$, it follows directly from the lemma that **$\mathbb{N} \times \mathbb{N}$ is countable**.

---

### **Final conclusion:**

Because we can assign each pair $(m, n)$ a unique natural number via $f(m, n) = 2^m \times 3^n$, and since $\mathbb{N}$ is countable, the set $\mathbb{N} \times \mathbb{N}$ must also be countable.

**Thus, the set of all ordered pairs of natural numbers is countable.**

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