Show that the union of two countable sets is countable?
See here:
Biola University Academics
Show that the union of two countable sets is countable.
Proof. Let A and B be the two countable sets. Then B\A is countable due to Prob. 34. From Thm. 1.7-A, the elements of A and B\A can be listed as A = {a0, a1, a2, . . . } and B\A = {c0, c1, c2, . . . } Then A È B = A È (B\A) = {a0, c0, a1, c1, a2, c2, . . . } is a list whose elements are all distinct and also includes all elements of A È B. Hence by Thm. 1.7-A, A È B is countable.
whats does B\A means here? can someone explain to me in a bit more detail
As the link in the previous posting shows (BTW: that is an awful proof!) any proof of this really dependents upon the definitions and theorems in your particular text. There many ways to show this some are easier than other. But they depend in particular definitions.
Here is one definition of countable: a set A is countable if there is a injection, one-to-one, function $\displaystyle f:A \mapsto N = \{ 0,1,2, \cdots \}$.
Now suppose that each of A & B is countable set, now there is also an injection such that $\displaystyle g:B \mapsto N$. We now define an injection $\displaystyle
h(x) = \left\{ {\begin{array}{lr}
{2f(x)} & {x \in A\backslash B} \\
{2g(x) + 1} & {x \in B\backslash A} \\
\end{array}} \right.
$.
BTW: $\displaystyle A\backslash B$ is set of elements in A but not in B.
Now you must show that h is an injection.
I am getting more confused by your explanation.. If both A and B are countable because they are finite then it doesn't matter because then the union of both of them have a cardinality which then can be proofed to be a injection.. I am confused by the example of Plato here giving such function g(x), say that A\B is {1,2,3,4,5} and B\A is {7,8,9,10,11} I am confused to show this injection
If both $\displaystyle A$ and $\displaystyle B-A$ are countably infinite, let: $\displaystyle \{ a_i, i \in \mathbb{N} \} $ $\displaystyle \{b_i, i \in \mathbb{N} \}$ be enumerations of $\displaystyle A$ and $\displaystyle B-A$ respectivly.
Then:
$\displaystyle
\{c_i, i \in \mathbb{N} \}
$
where $\displaystyle c_{2j} = a_j, \ j \in \mathbb{N},\ c_{2k+1}=b_k, \ k \in \mathbb{N}$ is an enumeration of $\displaystyle A \cup B$ and so $\displaystyle A \cup B$ is countable.
The cases where one or both of $\displaystyle A$ and $\displaystyle B$ are finite is easily disposed of by an enumeration where the elements of $\displaystyle A$ or $\displaystyle B-A$ (whichever is finite or either if both are) are taken first follwed by the elements of the other.
RonL