Suppose that $A,B,C$ are finite sets, and that $B \cup C=A$ and $B \cap C=\varnothing$. Then $|A|=|B|+|C|$

Question

Please check if my proof is fine! Thank you for your help!

Theorem:

Suppose that $A,B,C$ are finite sets, and that $B \cup C=A$ and $B \cap C=\varnothing$. Then $|A|=|B|+|C|$. Here $|X|=\operatorname{card}(X)$.

My proof:

Since $B,C$ are finite, then there exist $g:B \to I_n$ and $h:C \to I_m$ such that $g,h$ are bijective.

Let $k:C \to I_{m+n}$ such that $k(c)=h(c)+n$, then $k:C \to I_{m+n} \setminus I_{n}$ is bijective.

Let $h:A \to I_{m+n}$ such that $h(a)=g(a)$ for all $a \in B$ and $h(a)=k(a)$ for all $a \in C$.

Since $B \cap C=\varnothing$, then $(a \in A \implies a \in B$ or $a \in C$, but not both) then $h:A \to I_{m+n}$ is bijective.

Thus $|A|=m+n=|B|+|C|$.

Answer 1

Useful concepts:

In general, if $f: B \to D$ and $g: C \to D$ are any two functions we can't naturally define a function on $B \cup C$ by 'gluing' the two function together unless they agree on the intersection $B \cap C$.

In the special case where $B \cap C = \emptyset$ we are in business:

$\tag 1 h: B \cup C \to D \text{ where } h(x) = f(x) \text{ when } x \in B \text{ and } h(x) = g(x) \text{ when } x \in C$

Moreover, $h \text{ is injective iff } f \text{ is injective and } g \text{ is injective and the images of } f \text{ and } g \text{ are disjoint}$.

Answer 2

You are on the right track so far.

You still need to prove that the function $ h:A \to I_{m+n}$ is well-defined and bijective.

You have not used the fact that $A\cap B =\phi $.

Please add some detail to your proof and it will be fine.