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so we want to prove the cardinality of $\mathbb{N} \sim \{n+q: n \in \mathbb{N}, q \in \mathbb{Q}\}$. So far, during our course, we have done this by either:

  1. finding a bijective function $A \to B$.
  2. Cantor-Schröder-Bernstein theorem, which states that if for the Sets A and B, and there are injective functions $f: A \to B$ and $g: B \to A$, then there is a bijective function is $h: A \to B$.

So I went with the latter option for this one. i proved that an injective function $f$ exists: $\mathbb{N} \sim \{n+q: n \in \mathbb{N}, q \in \mathbb{Q}\}$. by saying $\forall~ n \in \mathbb{N}, f(n) = n \in \mathbb{Q}$ thus $\mathbb{N} \to \mathbb{Q}$. but I had to rely on proving that $\{n+q: n \in\mathbb{N}, q \in \mathbb{Q}\} = \mathbb{Q}$. I can't find a function for the other way around (I don't know if saying calkin-wilf tree is enough). and am I wrong for my presumption about " $\{n+q: n \in\mathbb{N}, q \in \mathbb{Q}\} = \mathbb{Q}$ ".

Abdallh Bder
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1 Answers1

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$\{n+q: n \in\mathbb{N}, q \in \mathbb{Q}\} = \mathbb{Q}$, which is easy to see as $(\mathbb{Q},+)$ is a group and any $n \in \mathbb{N} \subseteq \mathbb{Q} \implies n+q \in \mathbb{Q} ~\forall q \in \mathbb{Q}$.

Let $x = r/s \in \mathbb{Q}$,where $\gcd(r,s)=1$ and $r \in \mathbb{Z}, s \in \mathbb{N}$ then Define a map $f: \mathbb{Q} \to \mathbb{N}$ by,

$f(x)=p^r\cdot q^s$ if $r\geq 0$ and $f(x)=2 \cdot p^{-r}\cdot q^s$ if $r<0$, where $p,q$ are any two distinct odd primes.

This map is an injection.

Afntu
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