I am almost finished with this proof, but I am having some trouble.
$\#$ denotes cardinality, and the sets are arbitrary, so finite or infinite.
First, I know that since there is a bijection between $A$ and $C$, there is an injection between those sets as well. Also, we have the identity functions $I(a)=a \in A \subseteq B$, and $I(b)=b \in B \subseteq C$ and since the identity function is bijetive, there is an injection between the sets $A$ and $B$, and the sets $B$, and $C$.
I am trying to prove that there is also an injection from $B$ to $A$ or $C$ to $A$ to invoke the Schroeder-Bernstein theorem and declare a bijection between those two sets, finally proving that the cardinality of $A$, $B$, and $C$ is equal. I am having trouble with this last part though. Could you give me some help please? Thank you! :)
Because $A \subseteq B$, we have the identity function $I(a)=a∈A⊆B∴I(a)=a∈B$. The identity function is injective. Thus, there is an injection between A and B. Similarly, we use this same argument with an identity function $I(b)=b∈B⊆C∴I(b)=b∈C$. Thus, there is an injection between B and C. By definition of an injection, the elements of the domain might be equal to those in the range if it is also a bijection, and less than the elements of the domain if it is not. Thus, because these injections exist, $\#A≤\#B≤\#C∴\#A≤\#C$, but, by definition, because there exists a bijection from A to C, $\#A=\#C$, so $\#A=\#B=\#C$.