proof) if set A,B has no common elements, and n(A)=m, n(B)=n $A \cup B$ has m+n elements. Related Question
Proof that cardinality of the union of disjoints sets is the sum of cardinality of each set
If $A$ and $B$ have $n$ and $m$ elements respectively with $A\cap B=\emptyset$. Prove that $A\cup B$ has $m+n$ elements.
theorem 1.3.4 Introduce to Real Analysis, Bartle 4th
I'm stuck on prove $h$ is bijective
my proof is down below:
Let $f$ is a bijection $\mathbb{N}_m$ onto A, $g$ is bijection $\mathbb{N}_n onto B
For $A=\{a_1,a_2, \cdots ,a_m \}$, Let $\mathbb{N}_m=\{1,2,\cdots,m\}$
$f(a_1)=1, f(a_2)=2, \cdots, f(a_m)=m$. and
For $B=\{b_1,b_2, \cdots ,b_n \}$, Let $\mathbb{N}_n=\{m+1,m+2,\cdots,m+n\}$
$g(b_1)=m+1, g(b_2)=m+2, \cdots ,g(b_n)=m+n$
then defined function $h$ is bijective $\mathbb{N}_{m+n}$ onto $A\cup B$

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*How to proof Surjectivity, Injectivity with section

*What did I forgot this question...?

P.S) I can't understand after seeing solutions
PP.S) I think the hint is Definintion 1.3.1 (b): if $ n \in \mathbb{N}$ a set S is said to have $n$ elements
 A: Let me write some stuff to establish some common terms that I think will clarify things a little bit. Given a set $A$, the symbols $`` n(A)=m"$ mean that $m$ is natural number (here, the natural numbers set is $\mathbb{N}=\{0,1,2,3,4,...\})$ and there is a biyective function $f:A\to \{k\in\mathbb{N} : 0<k\leq m\}$. Let me give you some examples.
(a) Suppose $n(A)=0$. Then, by definition, we know there is a biyection between $A$ and the set $\{k\in\mathbb{N} : 0<k\leq 0\}$. This means that there is a biyection between $A$ and the empty set (there is no natural number $k$ that satisfies the relations $0<k\leq 0$). In colloquial terms, we say that $A$ has no elements. Since the empty set is the only set that satisfies this property, we deduce that $A=\emptyset$.
(b) Now let's assume that $n(A)=m$, with $m>0$. What does this mean?
If we take the biyection $f:A\to \{k\in\mathbb{N} : 0<k\leq m\}$ that exists by definition, then, if we look a little closer, we can see that the set $\{k\in\mathbb{N} : 0<k\leq m\}$ is simply the set $\{1,...,m\}$, that you denote by $\mathbb{N}_{m}$. In this case, we say that $A$ has $m$ elements.
(c) Take $a,b$ and $c$, three distinct real numbers. I claim that, if we let $A=\{a,b,c\}$, then $n(A)=3$. In order to do this, the only thing we have to show is that there is a biyective function between the sets $A$ and $\{1,2,3\}$. For example, if we send $a\to 1$, $b\to 2$, and $c\to 3$, then this produces a biyective map between $A$ and  $\{1,2,3\}$; hence, we can conclude that $n(A)=3$. Here, we say that $A$ has $3$ elements.
Now, we can try to proof the result you mention. Let $A$ and $B$ be a pair of disjoint sets (this is, $A\cap B=\emptyset$), such that $n(A)=m$ and $n(B)=n$. Fix a pair of biyective functions $f:A\to \{k\in\mathbb{N} : 0<k\leq m\}$ and $g:B\to \{k\in\mathbb{N} : 0<k\leq n\}$. Our goal is to show that $n(A\cup B)=m+n$; in other words, we must find a biyection between $A\cup B$ and $\{k\in\mathbb{N} : 0<k\leq m+n\}$ (in colloquial terms, this means that $A \cup B$ has $m+n$ elements).
First, if $n(A)=m=0$, then by the example explained in (a), we know that $A=\emptyset$. So, $A\cup B=B$, $m+n=n$, and $\{k\in\mathbb{N} : 0<k\leq m+n\}=\{k\in\mathbb{N} : 0<k\leq n\}$ . In this case, if we let $h=g$, then $h$ is the desired biyection. Similarly, if we have $n(B)=n=0$, then we can argue that $h=f$ is the biyection we are looking for.
Now, asumme that $n(A)=m>0$ and $n(B)=n>0$. Then, from (b), we know that $f$ is a biyection between $A$ and $\{1,...,m\}$, $g$ is a biyection between $B$ and $\{1,...,n\}$, and we are looking for a biyection $h$ between $A\cup B$ and $\{1,...,m,m+1,...,m+n\}$. Our aim is to construct $h$ in some way related to $f$ and $g$.
Consider $h: A\cup B \to \{1,...,m,m+1,...,m+n\}$ given by
$$ h(x) = \begin{cases} f(x), & \text{if} \hspace{2mm} x \in A, \\
m+g(x), & \text{if} \hspace{2mm} x\in B.
  \end{cases} $$
Since $A\cap B=\emptyset$, we know that $h$ is a well defined function. I claim that $h$ is biyective.
To see that $h$ is surjective, take $k\in \{1,...,m,m+1,...,m+n\}$. Then, we have two cases. If $1\leq k \leq m$, then use that $f$ is surjective to find $x\in A$ such that $f(x)=k$. It is clear that $h(x)=k$. Now, if $k=m+\ell$ form some $1\leq \ell \leq n$, employ that $g$ is surjective to find $x\in B$ such that $g(x)=\ell$. Then, again, we have that $h(x)=k$. This shows that $h$ is surjective.
In order to prove that $h$ is injective, first observe that $h[A]\subseteq \{1,...,m\}$ and $h[B]\subseteq \{m+1,...,,m+n\}$. Now, suppose that $x,y\in A\cup B$ satisfy the equality $h(x)=h(y)$. From the observation we get two cases. If $h(x),h(y)\in \{1,...,m\}$ then, from the definition of $h$, the equality $h(x)=h(y)$ becomes $f(x)=f(y)$ and, since $f$ is injective, we get that $x=y$. On the other hand, if $h(x),h(y)\in \{m+1,...,m+n\}$ then again, from the definition of $h$, the equality $h(x)=h(y)$ becomes $m+g(x)=m+g(y)$. So, $g(x)=g(y)$ and, since $g$ is injective, we get that $x=y$. We conclude that $h$ is injective.
From the two previous arguments, we get that $h$ a biyection between $A\cup B$ and the set $ \{1,...,m,m+1,...,m+n\}$, as desired.
Hope this helps.
