cardinality with finite sets

Question

$A,B,C$ are finite sets. Suppose $A\subseteq B \subseteq C$ and $\#A=\#C$. Prove that $\#A=\#B$ and $\#B=\#C$.

Should I prove this by showing that there exist an element in $A$ that exist in $B$ and $C$? Or could I use the relation of transitivity to prove that $\#A=\#B$ and $\#B=\#C$ since we can assume $\#A=\#C$ and go with backwards proof?

Answer 1

The identity maps $i_1:A \to B$, $i_2:B\to C$ are injective. Since $\operatorname{card} A = \operatorname{card} C $, there is a bijection $\phi:C \to A$.

The key here is that if $\alpha:X \to X$ is an injection, and $X$ is finite, then $\alpha$ is a bijection (finiteness of $X$ means that we do not need to appeal to the Schroeder Bernstein theorem).

Claim $\eta:A \to A$ given by $\eta = \phi \circ i_2 \circ i_1$ is a bijection. Since $\phi, i_2, i_1$ are injections, so is $\eta$. Since $A$ is finite, $\eta$ is a bijection.

Hence $i_1:A \to B$ is a bijection (since $i_1^{-1} = \eta^{-1} \circ \phi \circ i_2$). Hence $\operatorname{card} B = \operatorname{card} A $.

Similarly, $i_2:B \to C$ is a bijection (since $i_2^{-1} = i_1 \circ \eta^{-1} \circ \phi $). Hence $\operatorname{card} B = \operatorname{card} C $.

Answer 2

HINT: Show that if $A$ is a proper subset of $B$ then $\#A\leq\#B$. In particular it follows that $A=C$. (Note that this is only true for finite sets, which is the case here.)