If A is finite, B is countable, and A and B are disjoint, prove that A union B is countable. I am having trouble with this question that appeared in my textbook. My first instinct is to try to use the theorem that states: if A, B are countable then A union B is countable.  
The proof would then depend on showing that A is countable.  I want to construct a bijection from the Natural numbers to A but am not sure how. Are my thoughts so far correct? Thanks!
 A: An easier proof is to use the injection $n \mapsto n+1$ To show that a countable set 'has room' for an extra element. Then use induction to see that countable sets 'have room' for finite sets.
This is a good proof in my opinion because it avoids the full strength of the theorem you cited.
A: You can use that theorem. Another way is like the following.
Let $\varphi:{\bf{N}}\rightarrow B$ be a bijection and $\psi:{\bf{N}}\rightarrow A\cup B$, $\psi(i)=a_{i}$, $i=1,...,n$, $\psi(k)=\varphi(k-n)$ for $k\geq n+1$, here $A=\{a_{1},...,a_{n}\}$ where all $a_{i}$ are distinct.
A: $ A$ is finite so $$A =\{ a_1,a_2,...,a_n\}$$
$B$ is countable, so $$ B =\{ b_1,b_2,.......\}$$
$$ A\cup B =\{a_1,a_2,...,a_n, b_1,b_2,.......\}$$
which is countable. 
A: Suppose $A$ has $k$ elements, denote the elements of $A$ as $\{a_1,\ldots, a_k\}$. Now given $B$ is countable, there is a bijection $f :\mathbb N \to B$.
We need to find a bijection  from $\mathbb N$ to $A\cup B$. We can define a map $ g: \mathbb N \to A\cup B$ by $g(i) = a_i$ if $1\le i\le k$ and $g(i) = f(i-k)$ for $i\ge k+1$. 
It can be easily checked that $g$ is a bijection. Hence the proof.
A: $A$ is finite.  So there exist $\mathbb N_{k} = \{1,.....,k\}$ and a bijection $h: \mathbb N_{k} \to A$.
$B$ is countable.  So there exist a bijection $i: \mathbb N \to B$.
Let $\phi: \mathbb N \to A\cup B$ via:  if $n \le k; \phi (n) = h(k)$.  If $n > k$ then $\phi(n) = i(n-k)$.
We just need to show $\phi$ is a bijection.
1) $\phi$ clearly maps to $A \cup B$ as for any $n$ either $n \le k$ and $\phi(n) \in A$ or $n > k$ and $\phi(n) \in B$.  
2) $\phi$ is surjective as for any $a \in A$ $\phi(h^{-1}( a)) = a$ and for any $b \in B$, $\phi(i^{-1}(b) + k) = b$.
3)  And $\phi$ is injective: if $n \ne m$ then case 1: $n,m \le k$ then $\phi(n)=h(n)\ne h(m)= \phi(m)$.  Case 2; $n,m > k$ then $\phi(n) = i(n-k) \ne i(m-k)=\phi(m)$. and if $m < k$ and $n > k$ then $\phi(m) = h(m) \in A$ while $\phi(n) = i(n-k) \in B$ but $A \cap B = \emptyset$ so $h(m) \ne i(n-k)$.  And Case 4; $n < k; m > k$ is exactly the same.
SO $\phi$ is a bijection and $A \cup B$ is countable.
