Prove $|A| < |B| \leq |C| \implies |A| < |C|$ I was wondering if I actually have to construct functions here and compose them, or if I can simply argue based on cardinality.
Let's say we have injections $f: A \rightarrow B$, $g: B \rightarrow C$.
The composition of these injective functions is the injection $g \circ f: A \rightarrow C$ which gives me $|A| \leq |C|$.
I have to show this is only an injection; that is prove it is not bijective.
Suppose for the sake of contraction that $|A| = |C| \iff A \approx C$.
If $|B| \leq |C| \land |A| = |C|$
$\implies |B| \leq |A|$.
But our hypothesis gives $|A| < |B|$, which is a contradiction.
So $|A| \neq |C|$. 
So the function cannot be bijective although it is injective. 
 A: As suggested before, you can argue this simply by cardinality. Since the cardinality operator maps a set to a non-negative integer, you can argue that, if $G = \mathbb{Z}_{+} \cup \{0\}$, since
$$|A|,|B|,|C| \in G,$$
We can argue by simple by simple integer inequalities and the transitivity of "less than" that
$$|A| < |B| \leq |C| \implies |A| < |C|.$$
A: You've framed the proof in a way that doesn't quite work, but the heart of the matter is fine. You have to show that there is no bijection from $A$ to $C$, not that some particular injection is not a bijection. That is to say, you need to show that for any $h:A\rightarrow C$ it is not the case that $h$ is a bijection, but your proof only says it treats $g\circ f$ (though it never uses this, so the proof works out okay anyways).
You proceed to work directly with the relations $\leq$ and $<$ on the given sets in a way that is valid, so long as you know that $|A|<|B|$ and $|B|\leq |A|$ is a contradiction. Since you seem to be trying to prove things about the order, you might accidentally make a circular argument by using such properties, if not previously and separately proved - it would be good you to think of an argument on the underlying functions for why this the case; as Brian M. Scott points out in the comments, the Schröder-Bernstein theorem implies this.
A: It is clear that $|A|\leq |C|$. Now, if $|A|=|C|$ then $|A|\leq |B|$ and $|B|\leq |A|$ thus the Schröder-Bernstein theorem implies $|A|=|B|$ that is a contradiction.
