To prove that $2\Bbb{N}$ is countable, is it sufficient to define a bijection $f:\Bbb{2N}\rightarrow \Bbb{N}$ given by $f(n)=\frac{n}{2}, \forall n\in \Bbb{2N}$?
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5Yes, it is sufficient. – Kavi Rama Murthy Mar 27 '19 at 05:58
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It is sufficient. Of course, you might want a formal proof that it is a bijection, which is quite easy: $n/2=x$ implies $n=2x$ (injectivity) and for any $x\in\mathbb N$ it suffices to take $n=2x\in 2\mathbb N$ (surjectivity). In general, the map $\mathbb N\to a\mathbb N$ given by $x\mapsto ax$ gives a bijection between these two sets, so $a\mathbb N$ is always countable (for any $a\in\mathbb C$).
YiFan Tey
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Note:
To prove that $S= 2\mathbb{N} $ countable it suffices to show :
There exists a surjection $f: \mathbb{N} \rightarrow S$.
Choose $f(n)=2n$.
Is a surjection from the natural numbers enough to show that a set is countable?
Peter Szilas
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The $n/2$ approach is backwards. This answer, the $2n$ approach is right. Think of it this way...you want to COUNT them. This means you answer the questions, which is the first, second, etc., and then you need to show you didn't miss any (surjectivity). – Dean C Wills Mar 27 '19 at 15:53
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