9-Mathematical-Analysis-I-X-d-is-a-metric-space-and-p-n-n-1-is-a-sequence-in-X-such-that-p-n-convergent-p-If-K-p-n-n-1-p-

Question Number 205971 by mnjuly1970 last updated on 03/Apr/24

9 M a t h e m a t i c a l A n a l y s i s (I)

(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

s u c h t h a t, p_{n} \overset{c o n v e r g e n t}{\to} p . I f, K = {p_{n}}_{n = 1}^{\infty} \cup {p} t h e n

p r o v e K, i s c o m p a c t i n X .

Commented by aleks041103 last updated on 03/Apr/24

I s u p p o s e y o u m e a n t

K = {p_{k}}_{k = 1}^{\infty} \cup {p}

Answered by aleks041103 last updated on 03/Apr/24

I f K = {p} \cup {p_{k}}_{k = 1}^{\infty} t h e n t h e p r o o f i s a s

f o l l o w s :

L e t O b e a n y o p e n c o v e r o f K .

p \in K \Rightarrow \exists V \in O : p \in V

\begin{matrix} V i s o p e n \\ p \in V \\ p_{k} \to p \end{matrix}} \Rightarrow \exists N \in N : \forall k > N, p_{k} \in V

F o r \forall k = 1, \dots, N, \exists U_{k} \in O : p_{k} \in U_{k}, s i n c e

O i s a c o v e r o f K (a n d p_{k} \in K) .

T h e r e f o r e i f O^{'} = {U_{1}, \dots, U_{N}, V} \subseteq O,

t h e n b y t h e a b o v e c o n s t r u c t i o n,

O^{'} i s a f i n i t e o p e n s u b c o v e r o f K .

\Rightarrow K i s c o m p a c t .

Commented by Frix last updated on 04/Apr/24

I am he as you are he and you are me and

we are all together [John Lennon 1967]

Commented by mnjuly1970 last updated on 03/Apr/24

t h a n k s a l o t s i r F r i x

Commented by aleks041103 last updated on 03/Apr/24

N o p r o b l e m .

B i t I^{'} m n o t M r F r i x, a l t h o u g h I f e e l f l a t t e r e d

t o b e c o m p a r e d t o h i m

;)

Commented by mnjuly1970 last updated on 04/Apr/24

t h a n k y o u s o m u c h s i r a l e k s .

I a m s o s o r r y s i r \dots \underset{―}{\underset{⏟}{⋚}}

Commented by mnjuly1970 last updated on 04/Apr/24

⋚

Answered by Berbere last updated on 04/Apr/24

l e t u (n) \in N^{K}

l e t s s h o w T h a t \exists φ N \to N i n c r e a d i n g

s u c h T h a t U_{φ (n)} c v

1 i f n T a c k f i n i t v a l u e T h e r e x i s t N

s u c h T h a t \forall n \in N U_{n} \in {U_{1}, \dots \dots, U_{N}}; U_{1}, U_{2}, \dots U_{N}

S o e x i s t U_{1 ⩽ m ⩽ N} T h a t a p p a i r i n f i n i t y t i m e s

φ N \to N

φ (n) =

e i t h e n i s d e f i n d b y {\begin{cases} φ (1) = m \\ φ (n + 1) = {i n f n \in N ∣ φ (n + 1) > φ (n) & U_{φ (n + 1)} = U_{m} \end{cases}

φ e x i s t s i n c e U_{m} r e p e t i n f i n i t y t i m e s

\forall n \in N U_{φ (n)} = U_{m} c v

i f U_{n} T a c k i n f i n i t y v a l u e

w e d e f i n d φ b y

φ (1) = U_{1} = p_{n_{1}}

u_{n} = p_{v (n)}

φ (n + 1) = i n f (m \in N ∣ m > φ (n), v_{m} > v_{φ (n)}}

φ i s w h i l e d e f i n d e s i n c e u_{n} T a c k i n f i n i t y v a l u e o f p

w e c a n s t r a n t {p_{v (1)}, p_{v (2)} \dots . P_{v (n)} \dots}

s i n c p_{n} C v U_{φ (n)} c v

s o e v r e y s e q u e n c e H a s e a s u b s e q u e n c e ? C v

K i s c o m p a c t S o r r y f o r m y e n g l i s h

Commented by mnjuly1970 last updated on 05/Apr/24

t h a n k y o u s o m u c h s i r

t h a n k y o u s o m u c h s i r