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Since $\Bbb N$ and ${\sqrt2}$ are each countable sets, I see that the union is also countable. From this and the fact that $\Bbb N$ union ${\sqrt2}$ is infinite we know there exists a bijection to $\Bbb N$

I understand why such a bijection exists, but i'm not sure how to create the actual bijection itself?

If we let an element in the union be the smallest element we can show that it is impossible for two elements to be the smallest element in that set so that it would be one to one. and then because it is infinite, it is also onto. Is this correct?

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    $\begingroup$ The two sets are indeed countably infinite. But I think you are being asked to give an explicit bijection. Suppose we decide to let $f(\sqrt{2})=1$. Then we could let $f(1)=2$, $f(2)=3$, and so on. $\endgroup$ – André Nicolas Apr 1 '16 at 15:49
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Define $$f:\Bbb N\cup\{\sqrt{2}\}\to\Bbb N$$ by $f(\sqrt{2})=1$ and $f(n)=n+1$ for all $n\in\Bbb N$.

Its inverse is the map $$g:\Bbb N\to\Bbb N\cup\{\sqrt{2}\}$$ defined by $g(n)=n-1$ for $n\geq 2$ and $g(1)=\sqrt{2}$.

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  • $\begingroup$ Okay, then if I had something similar but the union was between the set of natural numbers and the set containing $\sqrt2$ and$\pi$ would I let $f(\sqrt2)=1$ and $f(\pi)=2$ and $f(n)=n+2$ ? $\endgroup$ – user319635 Apr 1 '16 at 15:51
  • $\begingroup$ @user319635 Yes, that would work. $\endgroup$ – Spenser Apr 1 '16 at 15:52
  • $\begingroup$ Awesome, thank you. If it were the other way around and I was to go from the set of natural numbers to the set of natural numbers excluding some values how would I go about this? $\endgroup$ – user319635 Apr 1 '16 at 15:54
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    $\begingroup$ Define $f:\Bbb N\to\Bbb N-\{2\}$ by $f(1)=1$, $f(2)=3$, $f(3)=4$, $f(4)=5$, and so on. $\endgroup$ – Spenser Apr 1 '16 at 16:01
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    $\begingroup$ Define $f:\Bbb N\to\Bbb N-\Bbb N_{odd}$ by $f(n)=2n$. $\endgroup$ – Spenser Apr 1 '16 at 16:02

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