Does proving that a property holds for every set with $n$ elements implies it holds positive integers

Question

If we prove by using PMI that some predicate $P(\mathbf{X})$ is true for every set $\mathbf{X}$ with $n$ elements, for each $n\in \mathbb{Z}^{+}$, can we say that $P(\mathbf{X})$ holds for $\mathbb{Z}^{+}$? Or are we limited to ensure it is true only if $\mathbf{X}$ is finite?
Thanks in advance!

Answer 1

Only if it's finite!

To see this in the quickest way possible, let $P(X)$ be the property "X is a finite set"!

For a more substantive example, let $P(X)$ be the property, "$\sum_{x\in X}\frac{1}{x}$ is convergent." Or, "$\sum_{x\in X}\frac{(-1)^x}{x}$ is a rational number."

Answer 2

Here is an interesting but important one:

Let $P(X)$ be that every injection from $X$ to $X$ is also a surjection.

Then $P(X)$ for every finite set $X$, but $P(\mathbb{N})$ is clearly false.

This property $P$ is used in many proofs, such as that the Frobenius map $( x \mapsto x^p )$ is surjective on any finite field of characteristic $p$.

Answer 3

Every finite (non-empty) partial order has a maximal element.

We prove this by induction. Suppose that $A$ is given with a partial order. If $A$ is a singleton, we're done. If not, pick some $a\in A$ and consider $A'=A\setminus\{a\}$ with the restriction of the partial order to $A'$ (note that $A'$ is non-empty). By the induction hypothesis $A'$ has a maximal element $b$. If $b\nleq a$, then $b$ is a maximal element in $A$, otherwise $a$ is a maximal element.$\qquad\square$

Clearly, however, this does not hold for $\Bbb Z^+$, as there are many partial orders on the natural numbers without a maximal element. I will leave it to you as an exercise to come up with one.