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Show that the set $C = N$ x {a,b} is countable by constructing a bijection between $N$ and $C$. Note: that it is the set of natural numbers Cartesian product with {a,b}.

I know I have to map all the elements so that each element goes to a natural number, so here is what I tried:

(1,a) , (1,b), (2,a), (2,b) .... (n,a) (n,b)

1 -------2-----3------4-------(n-1)----n

I am not sure if I understood the question correctly..

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You didn't. The set $C$ is a set of ordered pairs with first coordinate a natural number and second coordinate $a$ or $b$, that is $\{(n,a),(n,b):n=1,2,3,\ldots\}$. Want to try again or a hint?

What you wrote down is $\Bbb N\times \{\{a,b\}\}$, rather.

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  • $\begingroup$ I have changed it, would this suffice now? $\endgroup$ – Mac Dec 12 '13 at 2:26
  • $\begingroup$ @Mac The idea is fine, but note that your rule is not correctly written. You can be explicit: send even numbers $2n$ to $(n,b)$, so $2\to (1,b)$, $4\to (2,b)$, &c and send odd numbers $2n-1$ to $(n,a)$ so $1\to (1,a)$, $3\to (2,a)$, $5\to (3,a)$ and so on. $\endgroup$ – Pedro Tamaroff Dec 12 '13 at 2:30
  • $\begingroup$ Awesome thanks! The question asked for a visual so I didn't bother considering cases!...I have to wait a few minutes before I can accept answer $\endgroup$ – Mac Dec 12 '13 at 2:31

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