Let A,B,C,D sets. if |A| $\le$|B| and |B| < |C|, show that |A| < |C|
Proof:
Case1:
suppose |A| < |B|
then there exists injection f: A$\to$B
and |B| < |C|
then there exists injection g: B$\to$C
let h(x) = f(g(x)) = h:A$\to$C
h(x) is an injection because it is the composition of two injections, thus, |A| < |C|
Case2:
suppose |A| = |B|
then there exists a bijection f:A$\to$B
(repeat for |B|<|C|)
let h(x) = f(g(x)) = h:A$\to$C
h(x) is an injection
thus |A| < |C|
This is my proof and I'm not sure I'm doing it correctly. I'm not sure this is the correct procedure
You have the right general ideas, but your proof is not quite right. In both cases, you are finding an injection from $A$ to $C$, and concluding $|A| < |C|$. However, such an injection only allows you to conclude that $|A| \leq |C|$. The canonical way to show $|A| < |C|$ is to find an injection from $A$ to $C$, and to show that there cannot be an injection from $C$ to $A$.
Here's how I would do it. What we know is that there is an injection $f : A \to B$ and an injection $g : B \to C$. We also know that there is no injection $C \to B$. Immediately, we have that $g \circ f : A \to C$ is an injection (as you observed), so $|A| \leq |C|$. To show that the inequality is strict, assume for contradiction that there is an injection $h : C \to A$. In that case, $f \circ h : C \to B$ is an injection (since it is a composition of injections). This is a contradiction, since we were told that no such injection exists. Thus concludes the proof.