# If $A\subseteq B\subseteq C$ and there is a bijection between $A$ and $C$, $\#A=\#B$

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! :)

UPDATE:

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 $\#A=\#B=\#C$.

## 1 Answer

Good work so far, now to finish: There is a bijection $C\to A;$ resitrict it to $B$ to get an injection $B\to A.$

• So if we have a bijection between the sets $A$ and $C$, that means that the bijection goes both ways right? (From $A$ to $C$ and from $C$ to $A$). Also, why can I restrict it to $B$? – The Bosco Mar 6 '17 at 12:45
• Yes, a bijection $f\colon A\to C$ has an inverse function $f^{-1}\colon C\to A,$ given by $f(a)\mapsto a$ for all $a\in A$ (this definition makes sense precisely because $f$ is bijective), and $f^{-1}$ is also a bijection. By "restrict it to $B$" I mean "consider the function $g\colon B\to A$ given by $b\mapsto f^{-1}(b)$" (note that $b\in B\subseteq C$ so $f^{-1}(b)$ is defined). It's up to you to prove that this $g$ is injective. – Will R Mar 6 '17 at 12:56
• So can I just take a composition of functions, such as: $h: C \to B$ and $g: B\to A$, and $f^{-1} = g(h(c)): C \to A$, and because $h$ has to be injective for the composition to be injective, then $h$ is injective. Is this what you mean? – The Bosco Mar 6 '17 at 13:03
• I don't understand what you are trying to prove there. – Will R Mar 6 '17 at 13:19
• Can I say that the function $f^{-1}=g \circ h$? Thus, because $f^{-1}$ is a bijection, it is also an injection. And, in composition of functions, only $g$ has to be injective for $g \circ h = f^{-1}$ to be injective. – The Bosco Mar 6 '17 at 13:40