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This question already has an answer here:

Suppose that $S_1$ , $S_2$ are disjoint countable sets of T .Then their union is countable

ATTEMPT

Let $S_1$ = ${x_1 ,x_2,...}$

$S_2$ = ${y_1,y_2,...}$

I am thinking of making pairs by doing $S_1 \times S_2$ ,but i donot know what to do furthure .Need hints

Thanks

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marked as duplicate by Asaf Karagila set-theory Jul 9 '15 at 7:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Well I am writing $S_1,S_2$ as

$S_1=\{a_{11},a_{12},a_{13},...\}$

$S_2=\{a_{21},a_{22},a_{23},...\}$.Then Listing,

$S_1\cup S_2=\{a_{11};a_{12},a_{21};a_{13},a_{22};a_{23},...\}$

Now defining the map, $f:S_1\cup S_2\to\Bbb N$, by

$$f(a_{pq}) = \begin{cases} 1, & \text{if $p=1=q$ } \\ \{2(p+q)-5+p\}, & \text{otherwise} \end{cases}$$,Which is an enumeration ,Hence $S_1\cup S_2$ is countable

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Hint: Identify $S_1$ with the set of odd numbers and $S_2$ with the set of even numbers.

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  • $\begingroup$ yes i was thinking about that earlier ,but there i can see two mappings but here mappings are not given .I need to give mapping from my side .Can you guide bit furthure $\endgroup$ – Taylor Ted Jul 9 '15 at 5:32
  • $\begingroup$ Countable means we have a mapping (by which I mean a bijection) from $S_1$ to $\Bbb N$. Make a mapping from $\Bbb N$ to the odds, and combine the two to get what you want. $\endgroup$ – Omnomnomnom Jul 9 '15 at 5:35
  • $\begingroup$ I am thinking of this as a two rows and infinite columm matrix .A function which assigns $a_{11}$ to 1 and $a_{21}$ to 2 ,$a_{12}$ to 3 and so on i bijection .What you have to say about that ? $\endgroup$ – Taylor Ted Jul 9 '15 at 5:36
  • $\begingroup$ isn't it infinite countable if the mapping is a bijection? Vs. finite countable if injection only? $\endgroup$ – Fluffy12 Jul 9 '15 at 5:37
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    $\begingroup$ @Fluffy12 terminology varies. Rudin, for example, takes "at most countable" to mean "finite or countably infinite" an "countable" to mean "countably infinite." $\endgroup$ – Omnomnomnom Jul 9 '15 at 5:42

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