Does *finitely many* include the option *none*?

Does finitely many include the option none?

Say I have a sequence $(x_n)$ and I want to say that there can only be $0$ or $n\in \mathbb N$ non-zero terms. Can I say that the sequence has finitely many non-zero terms?

Thanks.

Solution 1:

Yes. No question. A subset of a finite set is finite. A polynomial with real coefficients has finitely many real zeros. We do not need (or want) to require saying: "A subset of a finite set is either finite or empty".

Solution 2:

Gerd Faltings showed that $a^{n}+b^{n}=c^{n}$ has finitely many solutions $(a,b,c)$ for any $n>2$. 10 years later, Andrew Wiles showed that it had none. Faltings is not known to have contradicted.

Solution 3:

Certainly. Even though, I always wonder about how sloppily some lecturers use termini such as necessarily and finitely many. So let me explain. Finiteness means there exists a bijective map from $\mathbb{N}_{p}:=\{n\in\mathbb{N}, n < p\}.$ Now choose $p=1$.