cardinality of sets Prove if $ |A| < |B| $ and $ |B| \leq |C|$ 
then $ |A| < |C|$
I know that $|A| < |B|$ means there is a one to one mapping of A onto a SUBSET of B but no one to one mapping from A to B.
I also know that $|B| \leq |C|$ means there is a one to one mapping of B onto C.
I am going to try to prove this, but I am fairly confident its going to be full of holes, please help!
since $|A| < |B|$ there is $Z \subseteq B$ such that $f(a)=z$ for all $a \in A$ and $z\in Z$
since $|B| \leq |C|$ we have $f(b)=c$ for all $b \in B$ and $c\in C$
since $Z \subseteq B$ we have $f(z)=y$ for all $z \in Z$ and $y \in Y$ where $Y$ is a subset of $C$
since $f$ is one to one, then the inverse exists.
therefore $z = f^{-1}(y)$ for $z \in Z$ and $y \in Y$
therefore $f(a) = f^{-1}(y)$ 
okay now i'm lost.  
Please help!
 A: First, let's clarify something:
The notation $|A|<|B|$ means that:


*

*There is a 1-1 function from $A$ into $B$.

*There is no bijection from $A$ onto $B$.


Whereas the notation $|A|\leq|B|$ means only the first one. It may be that the second condition holds, or that it fails. For example $|\{0\}|\leq|\{0,1\}|$, and also $|\{1,2\}|\leq|\{0,1\}|$.
To show now that $|A|<|C|$ use the fact that there are two injections, $f\colon A\to B$ and $g\colon B\to C$ to come up with an injection $h\colon A\to C$. Next show that if there was a bijection from $A$ onto $C$ then there had to be one onto $B$ as well, which is absurd.
A: Here is a slightly different approach.  The conceptually simplest definition of $|\cdot|<|\cdot|$ I know is $$|A|<|B| \;\equiv\; |A|\le|B| \land |B|\not\le|A|$$  Using this definition we can rewrite our demonstrandum as follows:
\begin{align}
& |A|<|B| \land |B|\le|C| \;\Rightarrow\; |A|<|C| \\
\equiv & \;\;\;\;\;\text{"definition of $|\cdot|<|\cdot|$, twice"} \\
& |A|\le|B| \land |B|\not\le|A| \land |B|\le|C| \;\Rightarrow\; |A|\le|C| \land |C|\not\le|A| \\
\equiv & \;\;\;\;\;\text{"split RHS of $\Rightarrow$"} \\
& (|A|\le|B| \land |B|\not\le|A| \land |B|\le|C| \;\Rightarrow\; |A|\le|C| ) \;\land \\
& (|A|\le|B| \land |B|\not\le|A| \land |B|\le|C| \;\Rightarrow\; |C|\not\le|A|) \\
\equiv & \;\;\;\;\;\text{"simplify second part using contraposition"} \\
& (|A|\le|B| \land |B|\not\le|A| \land |B|\le|C| \;\Rightarrow\; |A|\le|C| ) \;\land \\
& (|A|\le|B| \land |C|\le|A| \land |B|\le|C| \;\Rightarrow\; |B|\le|A|) \\
\Leftarrow & \;\;\;\;\;\text{"strengthen by weakening LHS of $\Rightarrow$, twice"} \\
& \;\;\;\;\;\phantom{\text{"}}\text{-- because the shape of these formulas suggest transitivity"} \\
& (|A|\le|B| \land |B|\le|C| \;\Rightarrow\; |A|\le|C| ) \;\land \\
& (|B|\le|C| \land |C|\le|A| \;\Rightarrow\; |B|\le|A|) \\
\end{align}
Therefore all that is left to do is to prove that $|\cdot|\le|\cdot|$ is transitive, which should be easy enough.
