Let $X$ be a set and let $A \subseteq X$ be a subset of $X$. Asume that there is an injective function $f: X \rightarrow A$. Show that $A$ and $X$ have the same cardinality. Show that any two sets have the same cardinality $\iff$ there are injective functions $f: Y \rightarrow Z$ and $g: Z \rightarrow Y$.
Since $f: X \rightarrow A$ is injective, $\operatorname{card} X \leq \operatorname{card} A$. But since $A$ is a subset of $X$, we have the inclusion map $g: A \rightarrow X$, which is also injective. So $\operatorname{card} A \leq \operatorname{card} X \implies \operatorname{card} A= \operatorname{card} X$.
Similarly, we have $\operatorname{card} Y \leq \operatorname{card} Z$ and $\operatorname{card} Z \leq \operatorname{card} Y \implies \operatorname{card} Y = \operatorname{card} Z$.
I was just wondering if my answer was correct, because it is really short so I thought that I might be missing something…is it ok for me to assume that if we have an injective function $f: X \rightarrow Y$, then $\operatorname{card} X \leq \operatorname{card}$? Or is that what the question wants me to prove?
Thanks in advance
Best Answer
Actually this is where you need to prove the implication.
$$ |A|\leq|B|\text{ and }|B|\leq|A|\,\implies|A|=|B| $$
The symbol $\leq$ is just a notation for the existence of an injective function, and $=$ is a symbol for the existence of a bijection.
Indeed, this notation suggest the above implication, just as much as it suggest that $$ |A|\leq|B|\text{ and }|B|\leq|C|\,\implies|A|\leq|C| $$
but this needs proof that if there if the assumed injections exist then so is the implied injection from $A$ to $C$