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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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. 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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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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