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Good night. I have a problem with this exercise:

Prove that the union of two denumerable sets is denumerable.

Proof:

Be $A,B\subset\mathbb{R}$ where A and B are numerable, in other words $f:\mathbb{N\rightarrow}A$ biyective and $g:\mathbb{N\rightarrow}B$. Be $h:\mathbb{N\rightarrow}A\cup B$ and suppose:

$A=\left\{ a_{1},a_{2},a_{3},...\right\} $

$B=\left\{ b_{1},b_{2},b_{3},...\right\} $

I construct a function biyective such that

$1\rightarrow a_{1}$

$2\rightarrow b_{1}$

$3\rightarrow a_{2}$

$4\rightarrow b_{2}$

$.$

$.$

$.$

Where $h(x)=\begin{cases} 2k+1\rightarrow a_{k}\:k\epsilon\mathbb{N}\\ 2k\rightarrow b_{k}\:k\epsilon\mathbb{N} \end{cases}$

If we see the function $h(x)$ she is $\mathbb{N}$ in other words, $A\cup B$=$\mathbb{N}$ then $h:\mathbb{N\rightarrow}A\cup B$ is biyective.

But i feel my proof is too bad, can someone help me?

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    $\begingroup$ Almost straight from the get-go, you're assuming $A$ and $B$ are disjoint. This isn't too difficult an issue to amend, but you may want to mention something about it. $\endgroup$ – Aweygan Aug 2 '16 at 22:02
  • $\begingroup$ Your function $h$ need not ba bijection; for instance, if $A \cap B \ne \emptyset$ then $h$ is not a bijection. But $h$ is surjective, and this should help you. $\endgroup$ – Lee Mosher Aug 2 '16 at 22:03
  • $\begingroup$ If you want to construct a bijection, you can let $C = B \setminus A$ and note that $A \cup B = A \cup C$, where $A \cup C$ is a disjoint union. Then you can apply your method to $A$ and $C$ (but be careful that $C$ may not be infinite and could even be empty). $\endgroup$ – Bungo Aug 2 '16 at 22:07
  • $\begingroup$ I edited the tags and clarified the title and question to reflect the apparent intent. If the result is not what you intended, please fix accordingly. :-) $\endgroup$ – Bungo Aug 2 '16 at 22:11
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The proof is almost correct, and is actually a good way of approaching the problem. One small error is that $h$ is not actually bijective, since you can have $a_i=b_j$ for some $i, j$. However, $h$ is still surjective, which suffices to show the union is countable.

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We describe one way to modify your argument.

We assume that both sets are countably infinite. Suppose that we have defined $h(k)$ for every $k\lt n$.

If $n$ is even, let $h(n)=b_j$, where $j$ is the smallest positive integer such that $b_j\ne h(k)$ for any $k\lt n$.

If $n$ is odd, $h(n)=a_j$, where $j$ is the smallest positive integer such that $a_j\ne h(k)$ for any $k\lt n$.

Now one can verify that $h$ is a bijection from $\mathbb{N}$ to $A\cup B$.

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