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 } then prove K , is compact in X .


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Question Number 205971 by mnjuly1970 last updated on 03/Apr/24
$9 Mathematical Analysis ( I ) (X , d ) is a metric space and { p_n }_(n=1) ^∞ is a sequence in X such that , p_n →^(convergent) p . If , K= {p_n }_(n=1) ^∞ ∪ { p } then prove K , is compact in X .$
$9 M a t h e m a t i c a l A n a l y s i s (I)$ $(X, d) i s a m e t r i c s p a c e a n d$ ${p_{n}}_{n = 1}^{\infty} i s a s e q u e n c e i n X$ $s u c h t h a t, p_{n} \overset{c o n v e r g e n t}{\to} p . I f, K = {p_{n}}_{n = 1}^{\infty} \cup {p} t h e n$ $p r o v e K, i s c o m p a c t i n X .$
Commented by aleks041103 last updated on 03/Apr/24
$I suppose you meant K={p_k }_(k=1) ^∞ ∪{p}$
$I s u p p o s e y o u m e a n t$ $K = {p_{k}}_{k = 1}^{\infty} \cup {p}$
	Answered by aleks041103 last updated on 03/Apr/24
	$If K={p}∪{p_k }_(k=1) ^∞ then the proof is as follows: Let O be any open cover of K. p∈K⇒∃V∈O:p∈V {: ((V is open)),((p∈V)),((p_k →p)) } ⇒ ∃N∈N: ∀k>N, p_k ∈V For ∀k=1,...,N, ∃U_k ∈O: p_k ∈U_k , since O is a cover of K (and p_k ∈K). Therefore if O′={U_1 ,...,U_N ,V}⊆O, then by the above construction, O′ is a finite open subcover of K. ⇒ K is compact.$
	$I f K = {p} \cup {p_{k}}_{k = 1}^{\infty} t h e n t h e p r o o f i s a s$ $f o l l o w s :$ $L e t O b e a n y o p e n c o v e r o f K .$ $p \in K \Rightarrow \exists V \in O : p \in V$ $\begin{matrix} V i s o p e n \\ p \in V \\ p_{k} \to p \end{matrix}} \Rightarrow \exists N \in N : \forall k > N, p_{k} \in V$ $F o r \forall k = 1, . . ., N, \exists U_{k} \in O : p_{k} \in U_{k}, s i n c e$ $O i s a c o v e r o f K (a n d p_{k} \in K) .$ $T h e r e f o r e i f O^{'} = {U_{1}, . . ., U_{N}, V} \subseteq O,$ $t h e n b y t h e a b o v e c o n s t r u c t i o n,$ $O^{'} i s a f i n i t e o p e n s u b c o v e r o f K .$ $\Rightarrow K i s c o m p a c t .$
	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

	$t h a n k y o u s o m u c h s i r a l e k s .$ $I a m s o s o r r y s i r . . . \underset{―}{\underset{⏟}{⋚}}$
	Commented by mnjuly1970 last updated on 04/Apr/24

	$⋚$
	Answered by Berbere last updated on 04/Apr/24
	$let u(n)∈N^K lets show That ∃ϕ N→N increading such That U_(ϕ(n)) cv 1 if n Tack finit value Ther exist N such That ∀n∈N U_n ∈{U_1 ,......,U_N };U_1 ,U_2 ,...U_N So exist U_(1≤m≤N) That appair infinity times ϕ N→N ϕ(n)= eithe n is defind by { ((ϕ(1)=m)),((ϕ(n+1)={inf n∈N ∣ϕ(n+1)>ϕ(n)&U_(ϕ(n+1)) =U_m )) :} ϕ exist since U_m repet infinity times ∀n∈N U_(ϕ(n)) =U_m cv if U_n Tack infinity value we defind ϕ by ϕ(1)=U_1 =p_n_1 u_n =p_(v(n)) ϕ(n+1)=inf(m∈N∣ m>ϕ(n),v_m >v_(ϕ(n)) } ϕ is while definde since u_n Tack infinity value of p we canstrant {p_(v(1)) ,p_(v(2)) ....P_(v(n)) ...} sinc p_n Cv U_(ϕ(n)) cv so evrey sequence Hase a sub sequence?Cv K is compact Sorry for my english$
	$l e t u (n) \in N^{K}$ $l e t s s h o w T h a t \exists φ N \to N i n c r e a d i n g$ $s u c h T h a t U_{φ (n)} c v$ $1 i f n T a c k f i n i t v a l u e T h e r e x i s t N$ $s u c h T h a t \forall n \in N U_{n} \in {U_{1}, . . . . . ., U_{N}}; U_{1}, U_{2}, . . . U_{N}$ $S o e x i s t U_{1 ⩽ m ⩽ N} T h a t a p p a i r i n f i n i t y t i m e s$ $φ N \to N$ $φ (n) =$ $e i t h e n i s d e f i n d b y {\begin{cases} φ (1) = m \\ φ (n + 1) = {i n f n \in N ∣ φ (n + 1) > φ (n) & U_{φ (n + 1)} = U_{m} \end{cases}$ $φ e x i s t s i n c e U_{m} r e p e t i n f i n i t y t i m e s$ $\forall n \in N U_{φ (n)} = U_{m} c v$ $i f U_{n} T a c k i n f i n i t y v a l u e$ $w e d e f i n d φ b y$ $φ (1) = U_{1} = p_{n_{1}}$ $u_{n} = p_{v (n)}$ $φ (n + 1) = i n f (m \in N ∣ m > φ (n), v_{m} > v_{φ (n)}}$ $φ i s w h i l e d e f i n d e s i n c e u_{n} T a c k i n f i n i t y v a l u e o f p$ $w e c a n s t r a n t {p_{v (1)}, p_{v (2)} . . . . P_{v (n)} . . .}$ $s i n c p_{n} C v U_{φ (n)} c v$ $s o e v r e y s e q u e n c e H a s e a s u b s e q u e n c e ? C v$ $K i s c o m p a c t S o r r y f o r m y e n g l i s h$
	Commented by mnjuly1970 last updated on 05/Apr/24

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