Ordering of Cardinals Compatible with Cardinal Sum

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Theorem

Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be cardinals.


Then:

$\mathbf a \le \mathbf b \implies \mathbf a + \mathbf c \le \mathbf b + \mathbf c$

where $\mathbf a \mathbf c$ denotes the sum of $\mathbf a$ and $\mathbf c$.


Proof

Let $\mathbf a = \map \Card A$, $\mathbf b = \map \Card B$ and $\mathbf c = \map \Card C$ for some sets $A$, $B$ and $C$.

Let $C$ be chosen such that $A \cap C = \O = B \cap C$.


Let $\mathbf a \le \mathbf b$.

Then by definition of cardinal, there exists an injection $f: A \to B$.

Then the mapping $h: A \cup C \to B \cup C$ defined as:

$\forall x \in A \cup C: \map h x = \begin{cases} \map f x & : x \in A \\ x & : x \in C \end{cases}$


Let $a_1 \in A \cup C$ and $a_2 \in A \cup C$ such that:

$\map h {a_1} = \map h {a_2}$

Then:

\(\ds a_1\) \(\in\) \(\ds C\)
\(\ds \leadsto \ \ \) \(\ds a_1\) \(=\) \(\ds \map h {a_1}\) Definition of $h$
\(\ds \leadsto \ \ \) \(\ds a_1\) \(=\) \(\ds \map h {a_2}\) as $\map h {a_1} = \map h {a_2}$
\(\ds \leadsto \ \ \) \(\ds a_1\) \(=\) \(\ds a_2\) as $\map h {a_2}$ must also be in $C$


and:

\(\ds a_1\) \(\notin\) \(\ds C\)
\(\ds \leadsto \ \ \) \(\ds \map h {a_1}\) \(=\) \(\ds \map f {a_1}\) Definition of $h$
\(\ds \leadsto \ \ \) \(\ds \map h {a_2}\) \(=\) \(\ds \map f {a_1}\) as $\map h {a_1} = \map h {a_2}$
\(\ds \leadsto \ \ \) \(\ds \map h {a_2}\) \(=\) \(\ds \map f {a_2}\) as $\map h {a_2}$ must also not be in $C$
\(\ds \leadsto \ \ \) \(\ds \map f {a_1}\) \(=\) \(\ds \map f {a_2}\) as $\map h {a_1} = \map h {a_2}$
\(\ds \leadsto \ \ \) \(\ds a_1\) \(=\) \(\ds a_2\) as $f$ is an injection


So:

$\map h {a_1} = \map h {a_2} \implies a_1 = a_2$

demonstrating that $h$ is an injection.


So, by definition of sum of cardinals:

$\mathbf a + \mathbf c \le \mathbf b + \mathbf c$

$\blacksquare$


Sources