show that $K=[x_n, n \in N] \cup [l]$ is compact.Using only subsequences.

Question

let $(x_n) \in E^N$ a sequence such that :

$E$ is a normed vectoriel space

$(x_n) \rightarrow l$

I wanna show that $K=[x_n, n \in N] \cup [l]$ is compact.

For every sequence $(y_n)_{n \in N} \in K$ I have to find a subsequence $(z_n)_{n \in N}$ of $(y_n)_{n \in N}$ which converge. the limit of $(z_n)_{n \in N}$ can be a $u_i$, $i \in N$ or $l$ because of what terms of $(x_n)_{n \in N}$ can $(y_n)_{n \in N}$ take.

if $(y_n)_{n \in N}$ takes a finite number terms of $(x_n)_{n \in N}$ there is certainly a subsequence of $(y_n)_{n \in N}$ which is equal to a $u_i$.

To demontrate it.If $\forall n \in N \ y_n \in [x_1,...,x_m]$ then we have for $j \in [1,m]$ . $N=\bigcup_{j=1}^m A_j$ with $A_j=[n \in N , y_n=y_j]$ so if all $J_i$ are finite the $N$ is finite. so we will use $y_{\varphi(n)}=u_i$ as a constant subsequence.

for the other case I want to take a subsequence $(y_{\varphi(n)})_{n \in N}$ such that : if $y_{\varphi(n)}=x_k$ then $y_{\varphi(n+1)}=x_l$ with $l \geq m$.

The following application $\varphi \ :N \longrightarrow N$ such that :

$B=[y_n, n \in N] $

$\varphi(0)=min[k \in N ,\ x_k \in B]$

$\varphi(n+1)=min[k \in N ,\ x_k \in B-[y_{i_{n-1}}]]$

may satisfy the condition I want ! by construction I can't formulate in an explicite level why it converges.

Answer 1

Let $\Bbb N_0=\Bbb N\cup \{0\}.$ Let $l=x_0.$

Suppose $(y_n)_{n\in \Bbb N}$ has no constant sub-sequences. For $n\in \Bbb N$ let $A(n)=\{m\in \Bbb N_0: y_n=x_m\}.$

(I). For any $m\in \Bbb N_0$ the set $\{n\in \Bbb N:\min A(n)\le m\}$ is finite.

Otherwise, for some $j\le m$ there would be infinitely many $n\in \Bbb N$ such that $j=\min A(n)\in A(n),$ implying $y_n=x_j$ for infinitely many $n\in \Bbb N,$ implying $(y_n)_{n\in \Bbb N}$ has a constant sub-sequence.

(II).Let $f(1)=1$ and take $g(1)\in A(1),$ that is, $y_{f(1)}=y_1=x_{g(1)}.$

For $n\in \Bbb N$ let $f(n+1)\in \Bbb N$ be large enough that $f(n+1)>f(n)$ and also large enough that $\min A(f(n+1))>g(n)$ (...possible by (I) with $m=g(n)$...). And take $g(n+1)\in A(f(n+1)).$ That is, $y_{f(n+1)}=x_{g(n+1)}. $

Note that $g(n+1)\ge \min A(f(n+1))>g(n).$

(III).Let $S=(y_{f(n)})_{n\in \Bbb N}.$ Then $S$ is a sub-sequence of $(y_n)_{n\in \Bbb N}$ because $f(n+1)>f(n).$

And $y_{f(n)}=x_{g(n)}$ with $g(n+1)>g(n).$ So $S=(x_{g(n)})_{n\in \Bbb N}$ is a sub-sequence of the $convergent$ sequence $(x_n)_{n\in \Bbb N_0}$. So $S$ is convergent.

Answer 2

It's much more natural to show compactness directly:

let $\mathcal{U}$ be an open cover of $K=\{x_n\mid n \in \mathbb{N}\} \cup \{l\}$.

There must be a $U_0 \in \mathcal{U}$ such that $l \in U_0$.

By convergence of $x_n$ to $l$ there must be some $N \in \mathbb{N}$ such that $\forall n \ge N: l \in U_0$.

For all (finitely many!) $n < N$ we can also find $U_n \in \mathcal{U}$ such that $x_n \in U_n$ as well (it's a cover of $K$ after all).

Then $\{U_0, U_1, \ldots, U_{N_1}\} \subseteq \mathcal{U}$ is the required finite subcover of $\mathcal{U}$ and we're done.

The fact that compactness (in the standard open cover meaning) is equivalent in metric spaces (so in particular in normed vector spaces) to sequential compactness can then be invoked if you like.