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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.

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    Yes, completely unambiguously. $0$ is a finite number.2016-05-26

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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".

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    Note that there a numerous texts (and lecturers) in the wild that imply the opposite because including empty sets creates technical issues. Since, we cannot possible correct all sloppy texts, expect that your audience may be confused.2011-12-11
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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.

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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 $p=1$.

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    @David Heider: I just couldn't believe you belonged both to the community of people who reject the truth of $0\in\mathbb{N}$ *and* to those who prefer their intervals to stop *before* the last element specified, typically using [a,b)=\{x:a\leq x. Both communities are large and respectable, but I thought their intersection was empty. One wonders why you would want $\mathbb{N}_p$ to name a $p-1$ element set.2011-12-11