Dear Professors Tian and Zhang: I am very interesting your communication on summable, however, I would like to read them carefully. We got a new idea; the area of disc with radius r is of course \pi r^2, however for r = infinity; in the entire plane, its area is zero!!.
I will send some related article by separate e-mail. However, please look this: [5] *viXra:1902.0240 <http://vixra.org/abs/1902.0240>* *submitted on 2019-02-13 22:57:25*, (0 unique-IP downloads) *Zero and Infinity; Their Interrelation by Means of Division by Zero* *Authors:* *Saburou Saitoh <http://vixra.org/author/saburou_saitoh>* *Category:* General Mathematics <http://vixra.org/math> With best regards, Sincerely yours, Saburou Saitoh 2019.3.2.6:11 2019年3月2日(土) 5:47 Chun Tian (binghe) <binghe.l...@gmail.com>: > 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) < > binghe.l...@gmail.com> 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) <binghe.l...@gmail.com > >> <mailto: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 > >> > > > > _______________________________________________ > hol-info mailing list > hol-info@lists.sourceforge.net > https://lists.sourceforge.net/lists/listinfo/hol-info >
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