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Define the sum and the product of two cardinal numbers and show that these are well-defined operations.

That's what I have tried:

Let $A,B$ sets with $A \cap B=\varnothing, card(A)=m, card(B)=n$.

We define the sum $m+n$ of the cardinal numbers $m,n$ as the cardinal number of the union of $A$ and $B$, i.e.

$$m+n=card(A \cup B)$$

We will show that the sum of two cardinal numbers is well-defined.

It suffices to show that if $A_1 \sim B_1, A_2 \sim B_2$ with $A_1 \cap A_2=\varnothing, B_1 \cap B_2=\varnothing$ then $A_1 \cup A_2 \sim B_1 \cup B_2$.

We know that there are bijective functions $f: A_1 \to A_2, g:B_1 \to B_2$.

We want to show that there is a bijective function $h: A_1 \cup A_2 \to B_1 \cup B_2$.

We set $ h(x)=f(x)$ if $x \in A_1, h(x)=g(x)$ if $x \in A_2$.

We will show that $h$ is 1-1.

Let $x_1, x_2 \in A_1 \cup A_2$ with $h(x_1)=h(x_2)$.

If $x_1, x_2 \in A_1$ then $h(x_1)=f(x_1), h(x_2)=f(x_2)$ and so $f(x_1)=f(x_2) \Rightarrow x_1=x_2$

If $x_1, x_2 \in A_2$ then $h(x_1)=g(x_1), h(x_2)=g(x_2)$ and so $g(x_1)=g(x_2) \Rightarrow x_1=x_2$ since $g$ is injective.

If $x_1 \in A_1, x_2 \in A_2$ then $h(x_1)=h(x_2) \Rightarrow f(x_1)=f(x_2)$ that cannot be true because $B_1 \cap B_2=\varnothing$.

We will show that $h$ is surjective, i.e. that $\forall y \in B_1 \cup B_2, \exists x \in A_1 \cup B_2$ such that $f(x)=y$.

If $y \in B_1$ then we know that there will be a $x \in A_1$ such that $h(x)=y \Rightarrow f(x)=y$ since $f$ is surjective.

If $y \in B_2$ then we know that there will be a $x \in A_2$ such that $h(x)=y \Rightarrow g(x)=y$ since $g$ is surjective.

We define the product $m \cdot n$ of the cardinal numbers $m,n$ as the cardinal number of the cartesian product of $A$ and $B$, i.e.

$$m \cdot n=card(A \times B)$$

We will show that the product of two cardinal numbers is well-defined.

It suffices to show that if $A_1 \sim B_1, A_2 \sim B_2$ with $A_1 \cap A_2=\varnothing, B_1 \cap B_2=\varnothing$ then $A_1 \times A_2 \sim B_1 \times B_2$.

We know that there are bijective functions $f: A_1 \to A_2, g:B_1 \to B_2$.

We want to show that there is a bijective function $h: A_1 \times A_2 \to B_1 \times B_2$.

We define $h: A_1 \times A_2 \to B_1 \times B_2$ such that $\langle n,m \rangle \mapsto \langle f(n),g(m) \rangle$

We will show that $h$ is 1-1.

Let $\langle m_1, n_1 \rangle, \langle m_2, n_2 \rangle \in A_1 \times A_2$ with $h(\langle m_1, n_1 \rangle)=h(\langle m_2, n_2 \rangle) \rightarrow \langle f(n_1),g(m_1) \rangle=\langle f(n_2),g(m_2) \rangle \rightarrow f(n_1))=f(n_2) \wedge g(m_1)=g(m_2) \overset{\text{f,g:} 1-1 }{\rightarrow} m_1=m_2 \wedge n_1=n_2 \rightarrow \langle m_1,n_1 \rangle=\langle m_2,n_2 \rangle$

From the surjectivity of $f,g$ we can conclude that $h$ is surjective.

Could you tell me if it is right?

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Well, almost. You need to justify why you can assume $A\cap B=\emptyset$ in the definition of the sum of two cardinals. Other then that, it all seems fine. It is a bit long though. It would be somewhat easier to show the functions you construct are bijective by exhibiting an inverse (constructed from the inverses of the each function you use).

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    $\begingroup$ How could we justify that we can assume that $A \cap B=\varnothing$ in the definition of the sum of two cardinals? Could we justify it maybe as follows? $$$$ Let $m=card(D), \ n=card(E)$. Then $$card(D \times \{0\})=m , \ card(E \times \{1\})=n$$ We set $A=D \times \{0\}$ and $B=E \times \{1\}$ and we see that the sets are dijoint. $\endgroup$ – evinda Jan 23 '15 at 12:48

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