Proper subset of a finite set is finite

Question

I'm having trouble proving the following result without making circular arguments. I'm assuming no prior machinery for this problem other than the definition of finite (in bijection with $\{1, \ldots, n\}$ for some positive integer $n$ or empty) and of a subset.

Let $B$ be a finite set and $A$ a proper subset. Then $|A| \leq |B|$ and $|A|$ is finite.

I'm going to assume $A$ is nonempty; I don't think this makes too big of a difference, because the result is immediately satisfied for the empty set. So there exists a bijection $f: B \to \{1, \ldots, n\}$ for some $n$. To somewhat get a map between $A$ and a set of positive integers, the most natural thing seems to be the inclusion $g: A \to B$, which is an injection. If I compose maps, I get an injection $f \circ g: A \to \{1, \ldots, n\}$. This in turn implies there is a bijection from $A$ to a proper subset of $\{1, \ldots, n\}$. I don't know what these elements are, but it seems I can assume, relabelling if necessary, that the image of $A$ is a list of elements of the form $\{1, \ldots, m\}$ for $m < n$. This sounds circular, however, but it seems "obvious" that if $A$ is a proper subset of $B$, then $g(f(A))$ has to be a proper subset of $\{1, \ldots, n\}$. I know that a map between two finite sets of the same size is injective if and only if it is surjective, so I'm wondering if I can in some way work this in there.

Depends on how you define finite, but it is easy to prove of $A$ is infinite and $A\subseteq B$ then $B$ is infinite. At least if you define "$X$ is infinite" iff there is a function $f:X\to X$ with $f$ injective but not bijective. — Thomas Andrews, Sep 21 '23 at 20:07
Does this answer your question? Subset of a finite set is finite or one of these many other posts about this FAQ? — Anne Bauval, Sep 21 '23 at 21:27

score 0 · Answer 1 · answered Sep 21 '23 at 20:41

You can assume WLOG that $B = \{ 1, \dots n \}$. This means $B$ is equipped with a natural order, namely the usual one, and hence $A$ inherits this order. Now enumerate the elements of $A$ from least to greatest: that is, take $a_1$ to be the least element, $a_2$ to be the least element not equal to $a_1$, and so forth. By induction it's clear that $a_k \ge k$, meaning this process terminates after at most $n$ steps, and so produces a bijection $k \leftrightarrow a_k$ between $\{ 1, \dots m \}$ and $A$ for some $m \le n$. (It's not necessary to assume that $A$ is a proper subset.)

Essentially the same argument works for infinite ordinals too.

Proper subset of a finite set is finite

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