Mathematics, 10.07.2019 04:00 live4dramaoy0yf9

Prove that the set of even integers is denumerable

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to show that the set of all even integers is denumerable, we prove that there is a bijection from $\mathbb{z}^+$ to $2\mathbb{z}$ , where $2\mathbb{z} = \{2k \mid k \in \mathbb{z}\}$ , the set of all even integers.

proof. note that the set of even integers can be put into a list, that is, $2\mathbb{z} = \{0,2,-2,4,-4,6,-6,\ldots\}$ . we can intuitively see that all the even integers can be listed, allowing for a way to count all the even integers (however, we will never finish counting). we can be more rigorous in our argument by finding an explicit bijection. we can think about how to recreate the list we have made.

let the function $f: \mathbb{z}^+\to 2\mathbb{z}$ be defined by

$f(x) = \begin{cases}x & \text{ if $x$ is even} \\-(x-1) & \text{ if $x$ is odd.}\end{cases}$

for example, $f(1) = -(1-1) = 0$ , $f(2) = 2$ , $f(3) = -(3-1) = -2$ , $f(4) = 4$ , and so on. clearly this function generates this list. now we show that $f$ is a bijection. to do this, we must show that $f$ is injective and surjective.

injectivity

to show that $f$ is injective, we must show that for any $a,b\in\mathbb{z}^+$ , $f(a) = f(b)$ implies $a=b$ .

suppose that $a,b\in\mathbb{z}^+$ such that $f(a) = f(b)$ . then we have four cases:

case 1: if $a,b$ are both odd, then

$f(a) = f(b) \implies -(a-1) = -(b-1) \implies a - 1 =b - 1\implies a = b.$

case 2: if $a,b$ are both even, then

$f(a) = f(b) \implies a=b.$

case 3: if $a$ is even and $b$ is odd, then

$f(a) = f(b) \implies a=-(b-1) \implies -a = b-1.$ .

but since $a$ is a positive even integer, we have $a \ge 2 \implies -a \le -2$ and since $b$ is a positive odd integer, we have $b \ge 1 \implies b - 1 \ge 0$ .

we inferred that $-a = b-1$ is true in this case, but the left- and right-hand inequalities actually do not agree. this is a contradiction, so case 3 cannot happen.

case 4: if $a$ is odd and $b$ is even, then by similar argument as in case 3, case 4 cannot happen.

hence $f$ is injective.

surjectivity

now we show that $f$ is a surjection. to do this, we must show that for any $b \in 2\mathbb{z}$ , there is $a \in \mathbb{z}^+$ such that $f(a) = b$ . we must assert that there is an element in the domain that has $b$ as its image. so suppose we have $b \in 2\mathbb{z}$ . here, $b$ is even.

case 1: if $b > 0$ , then $b\in\mathbb{z}^+$ and $f(b) = b$ , since $b$ is even.

case 2: if $b < 0$ , then $-b > 0 \implies -b + 1 > 1$ so $-b + 1\in\mathbb{z}^+$ and $f(-b + 1) = + 1 - 1) = ) = b$ , since $-b + 1$ is odd.

case 3: if $b = 0$ then $f(1) = -(1-1) = 0$ .

so $f$ is also a surjection.

therefore, $f: \mathbb{z}^+\to 2\mathbb{z}$ is a bijection because it is injective and surjective.

it follows that $2\mathbb{z}$ , the set of all even integers, is denumerable.

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