I'm trying to show something similar to: $|A| \leq |B|$ iff there is an injective map from A to B, but for all sets.
I'm using the definition of cardinality that just states that $|A|$ is equal to some cardinal $k$, such that $k$ $\approx$ A (equinumerous).
I've already shown that if $|A| \leq |B|$, then there must be an injective function from A to B. Now I'm trying to prove that if there is an injective function, $|A| \leq |B|$.
I've tried to use the Axiom of Choice, that every set has a well-order, in order to show that A and B have order types of some ordinals $\alpha$ and $\beta$, and then study the different possibilities of comparisons between them; but I can't really think how that might be relevant to cardinality and showing that $|A| \leq |B|$.
Any help would be appreciated!
Suppose $A$ and $B$ are sets, $\kappa$ and $\lambda$ are cardinals, and $f \colon A \to \kappa$ and $g \colon B \to \lambda$ are bijections.
Claim: $\kappa \subseteq \lambda$ if and only if there exists an injection $h \colon A \to B$.
$\implies$: Let $\iota \colon \kappa \to \lambda$ be the inclusion map. Then $g^{-1} \circ \iota \circ f$ is an injection from $A$ to $B$.
$\impliedby$: Cardinals are linearly ordered by inclusion, so either $\kappa \subseteq \lambda$ or $\lambda \subseteq \kappa$. If the former occurs, then we're done. If $\lambda \subseteq \kappa$, then as in the "$\implies$" direction above there is an injection $j \colon B \to A$. Applying the Cantor-Schroeder-Bernstein Theorem, we can conclude that there is a bijection from $A$ onto $B$ (thus a bijection from $\kappa$ onto $\lambda$), and hence we in fact have $\lambda = \kappa$ (since we're talking about cardinals here and not just ordinals), so $\kappa \subseteq \lambda$ as desired.