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To show that two sets are equipotent, we must find a bijection between the two.

Proof

$\mathbb{N}=\{1,2,3,4,...\}$

Let $X \subseteq \mathbb{N}$ with $|X|=\infty$.

So, for some $x \in \mathbb{N}, X = \{x,x+1,x+2,x+3,...\}$

Define $f: \mathbb{N} \to X$ with $f(1)=$ min$(X)$ and for each $n \geq 2 \in \mathbb{N},$ let

$f(n)=$ min$(X)+(n-1)$.

One-to-one: Suppose $f(m)=f(n)$

min$(X)+m - 1 =$ min$(X) +n - 1$

$m-1=n-1$

$m=n$

Onto: Again, $f(n)=$ min$(X) + n - 1$

$n=f(n) -$ min$(X) +1 \in \mathbb{N}$

Note: since $X\subseteq \mathbb{N},$ min$(X) \geq$ min$\mathbb{(N)}$ for any chosen subset $X$. This clears up any issues regarding closure under subtraction when testing for surjectivity.

Since $f$ is one-to-one and onto, $f$ is a bijeciton.

$|X| = |\mathbb{N}|$

$\square$

Aram Nazaryan
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    $X$ is not necessarily of the form ${x,x+1,x+2,\dots}$, consider for example the set of even numbers. – leoli1 Feb 02 '21 at 19:03
  • The definition of $|X| = \infty$ is that there exists an injective map $f: X \rightarrow X$ that is not surjective. – WhatsUp Feb 02 '21 at 19:04
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    Is that a question or an answer? – Shaq Feb 02 '21 at 19:05
  • Would the function I defined work for the specific set $X={x,x+1,x+2,...}$ that I constructed? – Aram Nazaryan Feb 02 '21 at 19:05
  • @WhatsUp That's one definition of infinite. There are others (there exists an injection $\Bbb N\to X$ is another more common definition), and they aren't necessarily equivalent. – Arthur Feb 02 '21 at 19:07
  • @Arthur I assume that we are talking about ZFC here. – WhatsUp Feb 02 '21 at 20:09
  • You can make your life easier here : A subset of a countable set (here $\mathbb N$) is of course countable and since it is also an infinite set, it must have the same cardinality as $\mathbb N$. – Peter Feb 03 '21 at 13:39

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