- #1
Bachelier
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I know about the proof using lim inf and lim sup and the proof using a convergent subsequent, however I thought about this proof. Can you tell me if it is correct, and if not why?
Thank you
let Sn be Cauchy seq in R
Let S be its range. Then S is bounded.
Since R is complete, sup S exists. Let x= sup S
then for all ε > 0, ∃ N1 in N st
let N = max (N1, N2)
then d(Sn, x) <= d(SN, Sn) + d(SN, x)< ε
Hence Sn converges to x.
and we're done.
Thank you
let Sn be Cauchy seq in R
Let S be its range. Then S is bounded.
Since R is complete, sup S exists. Let x= sup S
then for all ε > 0, ∃ N1 in N st
x - ε/2 <SN1 <= x
⇒ x - ε/2 <SN1 < x + ε/2
so d(SN1, x) < ε/2
now for all ε > 0, ∃ N2 in N st for all n=>N2 implies d(SN2, Sn)<ε/2let N = max (N1, N2)
then d(Sn, x) <= d(SN, Sn) + d(SN, x)< ε
Hence Sn converges to x.
and we're done.