Question-math-analysis-X-d-is-a-metric-space-and-p-n-n-1-is-a-sequence-in-X-p-n-n-1-is-cauchy-if-and-only-if-lim-N-diam-E-N-0-where-E-N-p-N-

Question Number 205559 by mnjuly1970 last updated on 24/Mar/24

Q u e s t i o n . (m a t h a n a l y s i s)

(X, d) i s a m e t r i c s p a c e a n d

{(p_{n})}_{n = 1}^{\infty} i s a s e q u e n c e i n X .

{(p_{n})}_{n = 1}^{\infty} i s c a u c h y i f a n d o n l y i f

\lim_{N \to \infty} d i a m (E_{N}) = 0 .

w h e r e, E_{N} = {p_{N}, p_{N + 1}, \dots}

d i a m E := s u p {d (x, y) ∣ x, y \in E}

Answered by Berbere last updated on 24/Mar/24

d i m (E) = s u p d (x_{i}, x_{j}); (x_{i}, x_{j}) \in E^{2}

L e t p_{n} a c a u c h y s e q u e n c e \Rightarrow \forall ϵ > 0 \exists N \in N ∣ \forall k, m ⩾ 0 d (P_{N + k}, P_{N + m}) ⩽ ϵ \dots \dots 1

E_{N} {P_{N}, P_{N + 1}, \dots \dots .}; l e t S_{N} = s u p (d (x, y) ∣ (x, y) \in E_{N}^{2})

S_{N} \in R s i n c e P_{N} i s c a u c h y S e q u e n c e \exists a > 0 \Rightarrow d (P_{k}, P_{l}) < a

\forall (k, l) \in N^{2}

i f N_{1} > N_{2} \Rightarrow E_{N_{2}} \subset E_{N_{1}} \Rightarrow S_{N_{2}} < S_{N_{1}} S_{N} d e c r e a s e b i u n d e d S e q u e n c e

\Rightarrow D i a m (E_{N}) C v

1 \dots \Rightarrow \forall ϵ ⩾ 0 \exists N \in N ∣ d i a m (E_{N}) < ϵ

\lim_{N \to \infty} d i a m (E_{N}) = a \in R_{+}

a < d i a m (E_{N}) (E_{N} d e c r e a s e) \Rightarrow \forall ϵ > 0 0 ⩽ a < ϵ

\forall n > 0 0 ⩽ a ⩽ \frac{1}{n} \Rightarrow a \in \overset{n \in N^{*}}{\cap} [0, \frac{1}{n}] = {0}

\Rightarrow \lim_{n \to \infty} d i a m (E_{N}) = 0

Commented by mnjuly1970 last updated on 24/Mar/24

t h a n k s a l o t s i r \dots

Commented by Berbere last updated on 24/Mar/24

W i t h e P l e a s u r

Answered by aleks041103 last updated on 04/Apr/24

L e t a_{N} = d i a m ({p_{k}}_{k = N}^{\infty}) .

W e a r e a s k e d t o p r o v e

{p_{k}}_{k = 1}^{\infty} i s C a u c h y \Leftrightarrow \underset{N \to \infty}{l i m} a_{N} = 0

1 . C a u c h y \Rightarrow a_{N} \to 0

C a u c h y \Rightarrow \forall ϵ > 0, \exists M : d (p_{m}, p_{n}) < ϵ / 2, \forall m, n > M

\Rightarrow a_{M} = d i a m ({p_{k}}_{k = M + 1}^{\infty}) = s u p {d (p_{m}, p_{n}) ∣ m, n > M} ⩽ ϵ / 2 < ϵ

i t i s e a s y t o s e e t h a t a_{n} > 0 i s d e c r e a s i n g

\Rightarrow \forall m ⩾ M, 0 < a_{m} ⩽ a_{M} < ϵ, i . e . ∣ a_{m} ∣< ϵ

\Rightarrow \forall ϵ > 0, \exists M :∣ a_{m} - 0 ∣< ϵ, \forall m > M

\Rightarrow \underset{N \to \infty}{l i m} a_{N} = 0

2 . a_{n} \to 0 \Rightarrow C a u c h y

a_{N} \to 0 (a_{N} > 0) \Rightarrow \forall ϵ > 0, \exists M : a_{N} < ϵ, \forall N ⩾ M

\Rightarrow a_{M} = \underset{m, n ⩾ M}{s u p} d (p_{m}, p_{n}) < ϵ

\Rightarrow d (p_{m}, p_{n}) < ϵ, \forall m, n > M

\Rightarrow \forall ϵ > 0, \exists M : d (p_{m}, p_{n}) < ϵ, \forall m, n > M

\Rightarrow C a u c h y