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

$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\mathrm{9}\:\:\mathscr{M}{athematical}\:\:\:\:\mathscr{A}{nalysis}\:\left(\:{I}\:\right) \\ $$$$\:\:\left({X}\:,\:{d}\:\right)\:{is}\:{a}\:{metric}\:{space}\:{and}\:\:\:\:\:\:\: \\ $$$$\:\:\:\left\{\:{p}_{{n}} \right\}_{{n}=\mathrm{1}} ^{\infty} {is}\:{a}\:{sequence}\:{in}\:{X}\: \\ $$$$\:\:\:\:\:{such}\:{that}\:,\:{p}_{{n}} \overset{{convergent}} {\rightarrow}\:{p}\:.\:{If}\:,\:{K}=\:\left\{{p}_{{n}} \right\}_{{n}=\mathrm{1}} ^{\infty} \cup\:\left\{\:{p}\:\right\}\:{then} \\ $$$$\:\:\:\:\:{prove}\:\:{K}\:,\:{is}\:{compact}\:{in}\:{X}\:.\: \\ $$$$\:\:\:\: \\ $$

Commented by aleks041103 last updated on 03/Apr/24

$${I}\:{suppose}\:{you}\:{meant} \\ $$$${K}=\left\{{p}_{{k}} \right\}_{{k}=\mathrm{1}} ^{\infty} \cup\left\{{p}\right\} \\ $$

Answered by aleks041103 last updated on 03/Apr/24

$${If}\:{K}=\left\{{p}\right\}\cup\left\{{p}_{{k}} \right\}_{{k}=\mathrm{1}} ^{\infty} \:{then}\:{the}\:{proof}\:{is}\:{as} \\ $$$${follows}: \\ $$$${Let}\:\mathcal{O}\:{be}\:{any}\:{open}\:{cover}\:{of}\:{K}. \\ $$$${p}\in{K}\Rightarrow\exists{V}\in\mathcal{O}:{p}\in{V} \\ $$$$\left.\begin{matrix}{{V}\:{is}\:{open}}\\{{p}\in{V}}\\{{p}_{{k}} \rightarrow{p}}\end{matrix}\right\}\:\Rightarrow\:\exists{N}\in\mathbb{N}:\:\forall{k}>{N},\:{p}_{{k}} \in{V} \\ $$$${For}\:\forall{k}=\mathrm{1},…,{N},\:\exists{U}_{{k}} \in\mathcal{O}:\:{p}_{{k}} \in{U}_{{k}} ,\:{since} \\ $$$$\mathcal{O}\:{is}\:{a}\:{cover}\:{of}\:{K}\:\left({and}\:{p}_{{k}} \in{K}\right). \\ $$$${Therefore}\:{if}\:\mathcal{O}'=\left\{{U}_{\mathrm{1}} ,…,{U}_{{N}} ,{V}\right\}\subseteq\mathcal{O}, \\ $$$${then}\:{by}\:{the}\:{above}\:{construction}, \\ $$$$\mathcal{O}'\:{is}\:{a}\:{finite}\:{open}\:{subcover}\:{of}\:{K}. \\ $$$$ \\ $$$$\Rightarrow\:{K}\:{is}\:{compact}. \\ $$

Commented by Frix last updated on 04/Apr/24

$$\mathrm{I}\:\mathrm{am}\:\mathrm{he}\:\mathrm{as}\:\mathrm{you}\:\mathrm{are}\:\mathrm{he}\:\mathrm{and}\:\mathrm{you}\:\mathrm{are}\:\mathrm{me}\:\mathrm{and} \\ $$$$\mathrm{we}\:\mathrm{are}\:\mathrm{all}\:\mathrm{together}\:\left[\mathrm{John}\:\mathrm{Lennon}\:\mathrm{1967}\right] \\ $$

Commented by mnjuly1970 last updated on 03/Apr/24

$$\:\:{thanks}\:{alot}\:{sir}\:{Frix} \\ $$

Commented by aleks041103 last updated on 03/Apr/24

$${No}\:{problem}. \\ $$$${Bit}\:{I}'{m}\:{not}\:{Mr}\:{Frix},\:{although}\:{I}\:{feel}\:{flattered} \\ $$$${to}\:{be}\:{compared}\:{to}\:{him} \\ $$$$\left.;\right) \\ $$

Commented by mnjuly1970 last updated on 04/Apr/24

$$\:\:\:\:{thank}\:{you}\:{so}\:{much}\:{sir}\:{aleks}. \\ $$$$\:\:\:\:\:{I}\:{am}\:{so}\:{sorry}\:{sir}…\:\underline{\underbrace{\lesseqgtr}} \\ $$

Commented by mnjuly1970 last updated on 04/Apr/24

$$\:\cancel{\lesseqgtr} \\ $$

Answered by Berbere last updated on 04/Apr/24

$${let}\:{u}\left({n}\right)\in\mathbb{N}^{{K}} \\ $$$${lets}\:{show}\:{That}\:\exists\varphi\:{N}\rightarrow{N}\:{increading} \\ $$$${such}\:{That}\:{U}_{\varphi\left({n}\right)} \:{cv} \\ $$$$\mathrm{1}\:{if}\:\:{n}\:{Tack}\:{finit}\:{value}\:\:{Ther}\:{exist}\:{N} \\ $$$${such}\:{That}\:\forall{n}\in\mathbb{N}\:\:{U}_{{n}} \in\left\{{U}_{\mathrm{1}} ,……,{U}_{{N}} \right\};{U}_{\mathrm{1}} ,{U}_{\mathrm{2}} ,…{U}_{{N}} \\ $$$${So}\:{exist}\:{U}_{\mathrm{1}\leqslant{m}\leqslant{N}} \:{That}\:{appair}\:{infinity}\:{times} \\ $$$$\varphi\:{N}\rightarrow{N} \\ $$$$\varphi\left({n}\right)= \\ $$$${eithe}\:{n}\:{is}\:{defind}\:{by}\:\begin{cases}{\varphi\left(\mathrm{1}\right)={m}}\\{\varphi\left({n}+\mathrm{1}\right)=\left\{{inf}\:{n}\in\mathbb{N}\:\mid\varphi\left({n}+\mathrm{1}\right)>\varphi\left({n}\right)\&{U}_{\varphi\left({n}+\mathrm{1}\right)} ={U}_{{m}} \right.}\end{cases} \\ $$$$\varphi\:{exist}\:{since}\:\:{U}_{{m}} \:{repet}\:{infinity}\:{times} \\ $$$$\forall{n}\in\mathbb{N}\:{U}_{\varphi\left({n}\right)} ={U}_{{m}} \:{cv} \\ $$$${if}\:{U}_{{n}} \:{Tack}\:{infinity}\:{value} \\ $$$${we}\:{defind}\:\varphi\:{by} \\ $$$$\varphi\left(\mathrm{1}\right)={U}_{\mathrm{1}} ={p}_{{n}_{\mathrm{1}} } \\ $$$${u}_{{n}} ={p}_{{v}\left({n}\right)} \\ $$$$\varphi\left({n}+\mathrm{1}\right)={inf}\left({m}\in\mathbb{N}\mid\:{m}>\varphi\left({n}\right),{v}_{{m}} >{v}_{\varphi\left({n}\right)} \right\} \\ $$$$\varphi\:{is}\:{while}\:{definde}\:{since}\:{u}_{{n}} \:{Tack}\:{infinity}\:{value}\:{of}\:{p} \\ $$$${we}\:{canstrant}\:\left\{{p}_{{v}\left(\mathrm{1}\right)} ,{p}_{{v}\left(\mathrm{2}\right)} ….{P}_{{v}\left({n}\right)} …\right\} \\ $$$${sinc}\:{p}_{{n}} \:{Cv}\:{U}_{\varphi\left({n}\right)} \:{cv} \\ $$$${so}\:{evrey}\:{sequence}\:{Hase}\:{a}\:{sub}\:{sequence}?{Cv} \\ $$$${K}\:{is}\:{compact}\:{Sorry}\:{for}\:{my}\:{english} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 05/Apr/24

$${thank}\:{you}\:{so}\:{much}\:{sir} \\ $$

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