∀ ɛ > 0, Ǝ N(ɛ)>0 and N(ɛ) ϵ ℤ where n,m > N(ɛ) ϶: |a_n-a_m|<ɛ (assuming a regular metric space ℠of course ;-) ).
for this comical instance I will attempt to show the sequence is cauchy sequence (lazily):
1. The sequence a_n = π + (π/n) is strictly monotonically decreasing.
2. By Coldowl’s obvious and observation theorem (yea I just did it), This implies the sequence is bounded above by 2Ï€ and below by Ï€.
3. from baby rudin 3rd ed, page 55 theorem 3.14: Suppose {s_n} is monotonic. Then {s_n} converges iff it is bounded.
3a. This implies {a_n} from (1) is convergent.
4. Every Cauchy Sequence is Converget. (baby rudin 3rd ed page 53, theorem 3.11(c) ).
5. Q.E.D. (Boo-Ya!)
Note: I ran into some difficulty finding some N(É›) formula by manipulation from the definition. Could somebody please prove it by the “choosing some N(É›)” method? I’d love to see it.
@clinton: If n, m >= N: |(pi+pi/n)-(pi+pi/m)| = pi |1/n – 1/m| <= pi*(1/n + 1/m) <= 2pi/N 2pi/eps. Also, I think you have a typo in step 4, although it’s true, you need the reverse statement, right? (This will probably already have been answered, but the moderation currently hides any answers. Sorry.)
@courtney: 1<3~algebra comics, thanks for posting them!
(Under the typical topology on R…)
It’s clear there’s only one accumulation point not in the sequence. (It’s a cluster point, even.)
It’s clear this special accumulation point is at \pi.
So clearly, it’s clear.
(Zipping through the archives and “clearly” was only a few days ago.)
Alt. proof: Let N(\epsilon) = \ceiling{1/{2 \epsilon}}. Insert in standard proof form. Turn crank. Get cookie.
Oh, that is so sweet. It puts a little weepy pi in my eye.
∀ ɛ > 0, Ǝ N(ɛ)>0 and N(ɛ) ϵ ℤ where n,m > N(ɛ) ϶: |a_n-a_m|<ɛ (assuming a regular metric space ℠of course ;-) ).
for this comical instance I will attempt to show the sequence is cauchy sequence (lazily):
1. The sequence a_n = π + (π/n) is strictly monotonically decreasing.
2. By Coldowl’s obvious and observation theorem (yea I just did it), This implies the sequence is bounded above by 2Ï€ and below by Ï€.
3. from baby rudin 3rd ed, page 55 theorem 3.14: Suppose {s_n} is monotonic. Then {s_n} converges iff it is bounded.
3a. This implies {a_n} from (1) is convergent.
4. Every Cauchy Sequence is Converget. (baby rudin 3rd ed page 53, theorem 3.11(c) ).
5. Q.E.D. (Boo-Ya!)
Note: I ran into some difficulty finding some N(É›) formula by manipulation from the definition. Could somebody please prove it by the “choosing some N(É›)” method? I’d love to see it.
@clinton: If n, m >= N: |(pi+pi/n)-(pi+pi/m)| = pi |1/n – 1/m| <= pi*(1/n + 1/m) <= 2pi/N 2pi/eps. Also, I think you have a typo in step 4, although it’s true, you need the reverse statement, right? (This will probably already have been answered, but the moderation currently hides any answers. Sorry.)
@courtney: 1<3~algebra comics, thanks for posting them!
HTML parsing screwed that up… After 2 pi/N it should be “strictly less than eps for N strictly larger than 2 pi/eps”.
Preview button, please? Sorry for double-posting.
Pi looks like a wig. I always imagined Cauchy with a wig. Therefore, the sequence, which has pi, must be Cauchy.
(Under the typical topology on R…)
It’s clear there’s only one accumulation point not in the sequence. (It’s a cluster point, even.)
It’s clear this special accumulation point is at \pi.
So clearly, it’s clear.
(Zipping through the archives and “clearly” was only a few days ago.)
Alt. proof: Let N(\epsilon) = \ceiling{1/{2 \epsilon}}. Insert in standard proof form. Turn crank. Get cookie.