Proving $a + a = a$ if a statement is true

Question

That's the question :

Let $a$ be a cardinality such that this following statment is true :

For every $A, C$, if $ A \subseteq C$, $|A| = a$ and $|C| > a$, then $|C \setminus A| > |A|$.

Without using cardinality arithmethics, prove that $a + a = a$.

This is how the question is written; maybe i'm not reading it well because I can find a counter example. Let $C$ be $\{1,2,3,4,5\}$, and $A$ be $\{3,4\}$. Then $A \subseteq C$, the cardinality of $C$ is bigger than the cardinality of $A$, and $|C \setminus A| = 3 > 2$.

So the statment is indeed true, yet $a + a = 4 \neq 2$.

What am I missing here ? Thanks in advance.

Edit : Hagen von Eitzen and marini helped me see what i've missed, thanks. I am still trying to solve this regardless of my mistake reading the question right so a hint / help from here will be much appreciated.

I think that by saying show that $a + a = a$, it means show that $a$ has to be infinte, because when $a$ is finite than $a + a = 2a$, maybe i'm getting this all wrong...

Answer 1

You are missing the "for every". With $A=\{42,666\}$, $C=\{13,42,666\}$ we have $A\subseteq C$, $|A|=a$ and $|C|>a$, but not $|C\setminus A|>|A|$. Hence your $a=2$ does not have the required property.

Answer 2

You gave one $A$, $C$, but: The statement says all $A$, $C$. For $a=2$, $C = \{1,2,3\}$, $A = \{1,2\}$, is a counterexample, as $|C\setminus A| = 1$.

Answer 3

HINT: Take $C=A\times\{0,1\}$. If $|C|>|A|$, then it cannot be that $C$ can be split into two parts both of size $A$.