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I need to prove:

If $A$ and $B$ are denumerable sets then so is their union $A\cup B$.

In this case, denumerable is defined as:

A set $X$ is said to be denumerable if there is a bijection $\mathbb{Z}^+\rightarrow X$.


My attempt:

$A \text{ denumerable } \implies f:\mathbb{Z^+} \rightarrow A$

$B \text{ denumerable } \implies g:\mathbb{Z^+} \rightarrow B$

I need to construct a bijective function, $h:\mathbb{Z}^+\rightarrow A\cup B$.

So, define $h$:

$h(i)=\left\{ \begin{align} & f(n+1), & \text{ if } i=2n+1, \text{ for } n \in \{0,1,2,\dots\} \\ & g(n), & \text{ if } i=2n, \text{ for } n \in \{ 1,2,3,\dots\} \end{align} \right.$

Then, $h$ is bijective because both $f$ and $g$ are bijective by definition.

Hence, $A \cup B$ is denumerable, as required.

However, this only works if $A$ and $B$ are disjoint.

So, in the case that $A$ and $B$ are not disjoint:

$A\cup B = A \cup (B-A)$ and $A \cap (B-A) = \emptyset$

I'm not sure what to do next.

If I can show that $B-A$ is denumerable, then I can use the above workings to conclude that $A\cup B$ is denumerable too. So I will attempt to construct a bijection $\mathbb{Z}^+\rightarrow B-A$:

I know that $(B-A)\subset B$, so, there exists an inclusion function $i:(B-A)\rightarrow B$, which is an injection. I thought of $g^{-1}\circ i$ but that's just an injection $(B-A) \rightarrow \mathbb{Z^+}$.


How can I build the required bijection ? I also would like to seek feedback if there are other parts of my work that could be written better.

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  • $\begingroup$ When getting acquainted with the subject, one may want to show very soon that the existence of a surjection is enough when the set is infinite--then your function $h$ is enough to prove the claim. $\endgroup$ – Did Sep 1 '14 at 19:32
  • $\begingroup$ I prefer to go from $A\cup B$ to $\mathbb{Z}^+$. Use the inverse of your procedure to deal with any $a\in A$. Also to deal with any $b\in B\setminus A$. This gives a bijection to an infinite subset of $\mathbb{Z}^+$. Then map this infinite subset bijectively to $\mathbb{Z}^+$ by sending the $n$-th element to $n$. $\endgroup$ – André Nicolas Sep 1 '14 at 19:49
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Suppose that both $A$ and $B$ are not the empty set. Let $f:\mathbb{N}\rightarrow A$ and $g:\mathbb{N}\rightarrow B$ be ennumerations of $A$ and $B$; that is, $f$ and $g$ are surjective functions from $\mathbb{N}$ onto $A$ and $B$, respectively. The function $h:\mathbb{N}\rightarrow A\cup B$ defined by: for each $n\in\mathbb{N}$

$$h(n)=\begin{cases} f(k)\quad\text{if }n=2k\\ g(k)\quad\text{if }n=2k+1 \end{cases}$$

This proof is valid even when $A\cap B\not=\varnothing$; a point in their intersection might have several preimages, but as well as any other element in $A$ or $B$. There is no problem with the definition of $h$, and I think it is clear that $h$ is surjective.

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