You say f is a function. From the context I assume the domain is Nat or f
is a sequence. Mathematically speaking, you should form the partial sums of
f ( sum f from 1 to n), which is a monotonic sequence of nats. Now proof by
contradiction: if your conclusion doesn’t hold, for any m, you can find n >
m, such that f n = 1. Then the partial sum goes up by 1 at n. As m is
arbitrary, f is not summable.
However I don’t know what is the best way to carry out the above proof in
hol as I am not familiar with the relevant libraries yet.
Haitao
On Friday, March 1, 2019, Chun Tian (binghe) <binghe.l...@gmail.com> wrote:
> Hi,
>
> I'm blocked at the following goal:
>
> I have a function f returning either 0 or 1. Now I know the infinite
> sum of f is finite, i.e.
>
> suminf f < PosInf (or `summable f` speaking reals)
>
> How can I prove the set of {x | f x = 1} is finite, or after certain
> index m all the rest f(n) are zeros?
>
> ∃m. ∀n. m ≤ n ⇒ (f n = 0)
>
> If I use CCONTR_TAC (proof by contradiction), I can easily derive the
> following 2 assumptions using results I established in my previous
> similar questions:
>
> INFINITE N
> ∀n. n ∈ N ⇒ (f n = 1)
>
> but still I've no idea how to derive a contradiction with `suminf f <
> PosInf` by proving `suminf f = PosInf`...
>
> Thanks,
>
> Chun Tian
>
>
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