How to prove this truism: an infinite set cannot be included in a finite set?

Question

The question may seem naive, but how to prove that an infinite set A cannot be included in a finite set B ?

case 1

A is included in B and also equal to B. Therefore the infinite cardinal of A is equal to the finite cardinality of B, which seems absurd ( but very close to what has to be proved, hence a suspicion of circularity).

case 2

A is a proper subset of B. How to prove that a finite set cannot have an infinite proper subset? Maybe one could say : the biggest subset of B is B itself, and B is finite. Since A is a proper subset of B, it must be smaller than B's biggest proper subset. A set that is smaller than a finite set is finite. Therefore, A is finite.

Answer 1

Using your definition of "finite" as "not infinite" and the definition of "infinite" as "having at least one proper subset such that there is a bijection between this subset and the original set," this proof works:

Suppose $A \subseteq B$ where $A$ is infinite.

Then there is some proper subset $C \subsetneq A$ and a bijection $\phi : C \to A.$

Now define $\psi : C \cup (B\setminus A) \to B$ by $$\psi(x) = \begin{cases}\phi(x), & \text{ if } x \in C \\ x, & \text{ if } x \in B\setminus A\end{cases}$$ It's easy to check that this is a well-defined bijection and that $C \cup (B \setminus A)$ is a proper subset of $B,$ showing that $B$ is infinite.

Answer 2

Given a bijection $f : B \to \{ 1,2,\dots,n \}$, the restriction $f|_A$ is a bijection between $A$ and $f(A) \subset \{ 1,2,\dots,n \}$. To finish you can show that if $C \subset \{ 1,\dots,n \}$ then $C$ is finite, from which you could conclude that $f(A)$ is finite and thus so is $A$. One way to do that is to enumerate the elements of $C$ in sorted order (which can be done by the well ordering principle); this furnishes a bijection to some initial segment of $\mathbb{N}$; you can argue that this segment must be contained in $\{ 1,\dots,n \}$ (and thus equal to some $\{ 1,\dots,m \}$).

This assumes that "finite" is defined as "can be put into bijection with $\{ 1,\dots,n \}$ for some $n$". If you instead define it as "not Dedekind-infinite", i.e. not in bijection with any of its proper subsets, then you can proceed by first taking a bijection $f$ from some proper subset $C \subsetneq A$ to $A$. Then you can extend $f$ to be onto $B$ by appropriately defining it on $B \setminus A$.