If Set A and B are denumerable, then A union B is also Denumerable. I know Set A is denumerable if there exists a bijection to the natural numbers f: A -> N. Same story for B.Then A union B has no repeats, and I know all elements in A and B go to unique Natural number, so there exists a bijection from A union B to the Natural numbers as well as it essentially has all elements of A and B with no repeats, each of those elements being mapped to a natural number. 
Is this the correct way to go about this? 
 A: Let us first define a set $C = B\backslash A$ so that $C\cap A = \emptyset$ and $A\cup C = A\cup B$. Since $A,B$ (and consequently $C$) are countable, there exists bijections from $\mathbb{N}$, namely $f:\mathbb{N} \rightarrow A$ and $g:\mathbb{N} \rightarrow C$. 
Define $h:\mathbb{N} \rightarrow A\cup B = \begin{cases}f(\frac{n+1}{2}) \text{ if } n \text{ is odd } \\ g(\frac{n}{2})\hspace{4mm}\text{ if } n \text{ is even }\\ \end{cases}$
Then $A\cup B$ is countable.
A: If $A,B$ are two disjoint countable sets, since the set of odd natural numbers $O$ and also the set of even natural numbers $E$ are countable there are two bijections $A\equiv O$, $B\equiv E$. Hence $A\cup B \equiv O \cup E= \mathbb N$.
A: Lemma: If there are $h,r$ functions such that $h:X\to Y$ and $r:Y\to X$ are injective, then there is a bijection between $X$ and  $Y$.
Having this lemma: Suppose $f:A\to \mathbb{N}$ and $g:B\to \mathbb{N}$ are bijections.
Then, $h:\mathbb{N}\to A\cup B $ defined by $h(n)=f^{-1}(n)$ is injective. 
Let $r:A\cup B\to\mathbb{N}$ defined by $r(x)=\max\{2f(x),2g(x)+1\}$, to be more precise:
$$r(x)=\left\{\begin{array}{cc}2f(x)&  x\in A\setminus B\\
2g(x)+1&  x\in B\setminus A\\
\max\{2f(x),2g(x)+1\}& x\in A\cap B
\end{array}\right.$$
Convince yourself that $r(x)$ is injective since $g,f$ are in injective and an odd number never equals an even number.
Since $h,r$ are injective, there should be a bijection between $A\cup B $ and $\mathbb{N}$
A: If $\emptyset\ne S\subset \Bbb N$ and $S$ has no largest member then there is a bijection $e:S\to \Bbb N.$ E.g. for $s\in S$ let $e(s)=n$ iff $\{t\in S: t\le s\}$ is bijective with $\{m\in \Bbb N:m\le n\}.$ (I.e. $e(s)=n$ iff $\{t\in S:t\le s\}$  has exactly $n$ members.)
Let $f:A\to \Bbb N$ and $g:B\to \Bbb N$ be bijections. Since $A\cup B=A\cup (B$ \ $A),$ and since $A,\, (B$ \ $A)$ are disjoint, we can define an injective $h:A\cup B\to \Bbb N$ by:
$h(a)=2^{f(a)}$ for $a\in A$ and $h(b)=3^{g(b)}$ for $b\in B$ \ $A.$
Now with $S=\{h(x):x\in A\cup B\}$ let $e:S\to \Bbb N$ be a bijection.
Finally let $i(x)=e(h(x))$ for $x\in A\cup B .$ Then $i:A\cup B\to \Bbb N$ is a bijection.
