# Thread: Show there exists a SUBSEQUENCE converging to L

1. ## Show there exists a SUBSEQUENCE converging to L

[note: also under discussion in Math Links forum]

2. Originally Posted by kingwinner

[note: also under discussion in Math Links forum]

** There is a subs. $\displaystyle \{x_{n_k}^{(1)}\}$ converging to $\displaystyle L_1\Longrightarrow\,\,\,for\,\,\,\epsilon=1\,\,\ex ists\,K_1\in\mathbb{N}\,\,\,s.t.\,\,\,|x_{n_k}^{(1 )}-L_1|<1\,\,\,\forall\,\,n_k^{(1)}>K_1$ . Choose now $\displaystyle n_1:=n^{(1)}_{K_1+1}$ .

** There is a subs. $\displaystyle \{x_{n_k}^{(2)}\}$ converging to $\displaystyle L_2\Longrightarrow\,\,\,for\,\,\,\epsilon=1\slash 2\,\,\exists\,K_2\in\mathbb{N}\,\,\,s.t.\,\,\,|x_{ n_k}^{(2)}-L_2|<1\slash 2\,\,\,\forall\,\,n_k^{(2)}>K_2$ . Choose now $\displaystyle n_2:=n^{(2)}_{K_2+h_2}$ in such a way that $\displaystyle n_2>n_1$ (we can do this because we've an infinite subsequence here and thus there exists $\displaystyle h_2 \,\,s.t.\,\,\,K_2+h_2>K_1+1$ !)

** There is a subs. $\displaystyle \{x_{n_k}^{(3)}\}$ converging to $\displaystyle L_3\Longrightarrow\,\,\,for\,\,\,\epsilon=1\slash 3\,\,\exists\,K_3\in\mathbb{N}\,\,\,s.t.\,\,\,|x_{ n_k}^{(3)}-L_3|<1\slash 3\,\,\,\forall\,\,n_k^{(3)}>K_3$ . Choose now $\displaystyle n_3:=n^{(3)}_{K_3+h_3}$ in such a way that $\displaystyle n_3>n_2$ (we can do this because we've an infinite subsequence here and thus there exists $\displaystyle h_3 \,\,s.t.\,\,\,K_3+h_3>K_2+h_2$ !)

Continue inductively in the above way, and in the m-th step:

** There is a subs. $\displaystyle \{x_{n_k}^{(m)}\}$ converging to $\displaystyle L_m\Longrightarrow\,\,\,for\,\,\,\epsilon=1\slash m\,\,\exists\,K_m\in\mathbb{N}\,\,\,s.t.\,\,\,|x_{ n_k}^{(m)}-L_m|<1\slash m\,\,\,\forall\,\,n_k^{(m)}>K_m$ . Choose now $\displaystyle n_m:=n^{(m)}_{K_m+h_m}$ in such a way that $\displaystyle n_m>n_{m-1}$ (we can do this because we've an infinite subsequence here and thus there exists $\displaystyle h_m \,\,s.t.\,\,\,K_m+h_m>K_{m-1}+h_{m-1}$ !)

Take it from here...

Tonio

3. Originally Posted by tonio
** There is a subs. $\displaystyle \{x_{n_k}^{(1)}\}$ converging to $\displaystyle L_1\Longrightarrow\,\,\,for\,\,\,\epsilon=1\,\,\ex ists\,K_1\in\mathbb{N}\,\,\,s.t.\,\,\,|x_{n_k}^{(1 )}-L_1|<1\,\,\,\forall\,\,n_k^{(1)}>K_1$ . Choose now $\displaystyle n_1:=n^{(1)}_{K_1+1}$ .

** There is a subs. $\displaystyle \{x_{n_k}^{(2)}\}$ converging to $\displaystyle L_2\Longrightarrow\,\,\,for\,\,\,\epsilon=1\slash 2\,\,\exists\,K_2\in\mathbb{N}\,\,\,s.t.\,\,\,|x_{ n_k}^{(2)}-L_2|<1\slash 2\,\,\,\forall\,\,n_k^{(2)}>K_2$ . Choose now $\displaystyle n_2:=n^{(2)}_{K_2+h_2}$ in such a way that $\displaystyle n_2>n_1$ (we can do this because we've an infinite subsequence here and thus there exists $\displaystyle h_2 \,\,s.t.\,\,\,K_2+h_2>K_1+1$ !)

** There is a subs. $\displaystyle \{x_{n_k}^{(3)}\}$ converging to $\displaystyle L_3\Longrightarrow\,\,\,for\,\,\,\epsilon=1\slash 3\,\,\exists\,K_3\in\mathbb{N}\,\,\,s.t.\,\,\,|x_{ n_k}^{(3)}-L_3|<1\slash 3\,\,\,\forall\,\,n_k^{(3)}>K_3$ . Choose now $\displaystyle n_3:=n^{(3)}_{K_3+h_3}$ in such a way that $\displaystyle n_3>n_2$ (we can do this because we've an infinite subsequence here and thus there exists $\displaystyle h_3 \,\,s.t.\,\,\,K_3+h_3>K_2+h_2$ !)

Continue inductively in the above way, and in the m-th step:

** There is a subs. $\displaystyle \{x_{n_k}^{(m)}\}$ converging to $\displaystyle L_m\Longrightarrow\,\,\,for\,\,\,\epsilon=1\slash m\,\,\exists\,K_m\in\mathbb{N}\,\,\,s.t.\,\,\,|x_{ n_k}^{(m)}-L_m|<1\slash m\,\,\,\forall\,\,n_k^{(m)}>K_m$ . Choose now $\displaystyle n_m:=n^{(m)}_{K_m+h_m}$ in such a way that $\displaystyle n_m>n_{m-1}$ (we can do this because we've an infinite subsequence here and thus there exists $\displaystyle h_m \,\,s.t.\,\,\,K_m+h_m>K_{m-1}+h_{m-1}$ !)

Take it from here...

Tonio
Thanks for your great help! But I still have some concerns about extracting the information given in the problem and the notations for different subsequences.

As stated in the problem, we are using the notation (x_n_k) for the target subsequence (the one that we need to construct at the end), so I'm going to stick with that.

"For each k≥1, there is a subsequence of (x_n) converging to L_k...".
So if you're using (x_n_k (1) ), (x_n_k (2) ), ..., (x_n_k (m) ),..., wouldn't this notation mean that it is a (further) subsequence of our target subsequence (x_n_k)???
Also the question says "For each k≥1...", so I think k should occur in the bracket (x_n_k (k) ), right? But then, we'll have two k's in it. Is that OK?

So now I am confused with all those subscripts, indices and the relation between them. k seems to have a special meaning in this question that is tied to the subscript of our target subsequence. So we are going to fix k=1,k=2,k=3,... one at a time to construct our target subsequence.

What is the usual (and correct) way to denote the DIFFERENT subsequences of (x_n)?

Thanks for clarifying this! It has been a nightmare for me...