Show that $\leq$ is transitive over the cardinality of sets

Question

$A, B, C$ are sets. I want to show $|A|\leq|B| \ \text{and} \ |B|\leq|C| \Longrightarrow |A|\leq|C|$.

I am confused as to how I would approach this, because the sets in this problem can be either finite or infinite.

Does anyone have a good way of approaching this problem? It seems like it isn't hard but I feel like I am overcomplicating it.

Answer 1

Hint: If $|A| \leq |B|$ there is an injection $f: A \to B$. If $|B| \leq |C|$ there is another injection $g: B \to C$. Now look at $g\circ f: A \to C$.

Answer 2

HINT: The composition of two injections is an injection.

Answer 3

$|A| \leq |B|$ if there is $f$ injective, $f: A \to B$. But composition of two injective functions is an injection: If $f: A \to B$ and $g: B \to C$ are injections then $g \circ f$ is an injection. As $f(x) = f(y) \implies x = y$, and $g(x) = g(y) \implies x = y$, so: $$g(f(x)) = g(f(y)) \implies f(x) = f(y) \implies x = y$$ Then, if $|A| \leq |B|$ and $|B| \leq |C| \implies |A| \leq |C|$.