(5) Show that the set $$ E $$ of even positive integers is countable.

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To show that the set $$ E $$ of even positive integers is countable, we need to demonstrate that there exists a bijection (one-to-one and onto function) between the set $$ E $$ and the set of positive integers $$ \mathbb{N} $$.

The set of even positive integers can be expressed as:
$$
E = \{2, 4, 6, 8, 10, \ldots\}
$$

We can define a function $$ f: \mathbb{N} \to E $$ by the rule:
$$
f(n) = 2n
$$
for each $$ n \in \mathbb{N} $$.

### Step 1: Show that $$ f $$ is well-defined
For every positive integer $$ n $$, $$ f(n) = 2n $$ is an even positive integer. Thus, $$ f(n) \in E $$ for all $$ n \in \mathbb{N} $$.

### Step 2: Show that $$ f $$ is injective (one-to-one)
To show that $$ f $$ is injective, we need to show that if $$ f(n_1) = f(n_2) $$, then $$ n_1 = n_2 $$.

Assume $$ f(n_1) = f(n_2) $$:
$$
2n_1 = 2n_2
$$
Dividing both sides by 2 gives:
$$
n_1 = n_2
$$
Thus, $$ f $$ is injective.

### Step 3: Show that $$ f $$ is surjective (onto)
To show that $$ f $$ is surjective, we need to show that for every $$ e \in E $$, there exists an $$ n \in \mathbb{N} $$ such that $$ f(n) = e $$.

Let $$ e $$ be any even positive integer. We can express $$ e $$ as $$ e = 2k $$ for some positive integer $$ k $$. We can choose $$ n = k $$. Then:
$$
f(n) = f(k) = 2k = e
$$
This shows that for every $$ e \in E $$, there exists an $$ n \in \mathbb{N} $$ such that $$ f(n) = e $$. Thus, $$ f $$ is surjective.

### Conclusion
Since $$ f $$ is both injective and surjective, it is a bijection between $$ \mathbb{N} $$ and $$ E $$. Therefore, the set $$ E $$ of even positive integers is countable.

Thus, we conclude that the set $$ E $$ is countable.
The set $$ E $$ of even positive integers is countable because there is a bijection between $$ E $$ and the set of positive integers $$ \mathbb{N} $$.

(5) Show that the set of even positive integers is countable.

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