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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?


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Maybe you could circumvent any possible ambiguity by saying something like "this sequence has either zero or some finite number of non-zero terms"? – Stijn Dec 11 '11 at 13:15
You can say, the sequence has at most finitely many non-zero terms. Or better: the sequence is eventually zero. – JDH Dec 11 '11 at 13:22
I think one might say all but finitely many terms are zero. – HbCwiRoJDp Dec 11 '11 at 13:23
Yes, completely unambiguously. $0$ is a finite number. – anomaly May 26 at 5:51
up vote 18 down vote accepted

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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And a finite dimensional vector space can have dimension $0$, no matter how many texts on linear algebra opt to avoid confronting this possibility. – Marc van Leeuwen Dec 11 '11 at 13:42
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. – Jan Dec 11 '11 at 14:22

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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Youn mean: choose $p=0$. – Marc van Leeuwen Dec 11 '11 at 13:37
It depends what he means by $\mathbb N$. Perhaps, like the OP, David intends that $\mathbb N$ begins with $1$. – GEdgar Dec 11 '11 at 13:50
@Marc: No, I mean 1. Then we have an empty set. So, this does imply in the case of a sequence that there are only as many members of the sequence different from zwero as there are elements in N_p. If p = 1, then none. – David Heider Dec 11 '11 at 17:33
@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 $[a,b)=\{x:a\leq x<b\}$. 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. – Marc van Leeuwen Dec 11 '11 at 18:55

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