I assume you are using the theory from src/real/seqScript.sml and it looks
like that summable is defined as a convergence of partial sums. That means
you should be able to directly contradict the convergence criterion if (f
n) does not go to 0. For example, SER_CAUCHY gives you:
“!f. summable f =
!e. &0 < e ==> ?N. !m n. m >= N ==> abs(sum(m,n) f) < e”
Choose e < 1, for any N, find n>N such that f n = 1, then abs(sum(n,n+1)
f)=1 > e and contradiction. (I don't know the indexing of sum so it could
be sum(n,n)).
Hope this helps,
Haitao
On Fri, Mar 1, 2019 at 12:46 PM Chun Tian (binghe) <[email protected]>
wrote:
> there’s is a typo in my previous post. I wanted to ask “…, why f must
> *not* be summable?” (i.e. divergent)
>
> On a second thought, your approach could work for proving the partial sum
> is unbounded: for any real number e > 0, I can first use
> Archimedean property (simple version) of reals to find an integer N such
> that e < N, then I repeat N times your approach, first let the partial sum
> >= 1, then >= 2, … finally >= N, by induction, then the whole proof
> finished. However to implement this idea it’s not easy to me. Still want
> to know better proofs.
>
> —Chun
>
> > Il giorno 01 mar 2019, alle ore 20:54, Chun Tian (binghe) <
> [email protected]> ha scritto:
> >
> > Hi Haitao,
> >
> > thanks, and yes, f is a sequence of reals. I'm following a similar path
> > (proof by contradiction), but I don't understand the last step: for any
> > m, the partial sum of f goes up by 1 at n > m, why f must be summable? I
> > think for every monotonic sequence such properties hold.
> >
> > --Chun
> >
> > Il 01/03/19 17:24, Haitao Zhang ha scritto:
> >> 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) <[email protected]
> >> <mailto:[email protected]>> 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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