I have to prove this exercise for my math study. It's form Terence Tao's Analysis I:

Assume $(x_{n})_{n = 0}^\infty$ is a sequence of real numbers, and $K \in \mathbb{R}$.
I have to prove the following:

$K$ is a limit point of $(x_{n})_{n = 0}^\infty \Rightarrow$ There is a subsequence of $(x_{n})_{n = 0}^\infty$ that converges to K.

This is my proof:

$K$ is a limit point of $(x_{n})_{n = 0}^\infty$
$\Rightarrow$ $\forall$ $N \in \mathbb{N}$, $\forall$ $\epsilon > 0 \in \mathbb{R}$ there is a $n \geqslant N $ such that $|x_{n} - K| < \epsilon$
$\Rightarrow$ Now take the subsequence of all $x_{n}$ with $n \geqslant N$ for wich $|x_{n} - K| < \epsilon$ and call this sequence $(a_{n})_{n = 0}^\infty$
$\Rightarrow$ $(a_{n})_{n = 0}^\infty$ is a subsequence of $(x_{n})_{n = 0}^\infty$ and it converges to K
$\Rightarrow$ There is a subsequence of $(x_{n})_{n = 0}^\infty$ that converges to K

Is this correct? and if not, how should I do it?

Thanks in advance!


Let $\varepsilon > 0.$ For every $N \in \mathbb{N},$ let $n_{N}$ be an integer such that $$|x_{n_{N}} - K| < \varepsilon.$$ Then the sequence $(x_{n_{N}})$ is a subsequence of $(x_{n})$ and $\to K.$

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  • $\begingroup$ you mean $x_{n}$? and K instead of L? and it's not give that $x_{n}$ converges to K, only that it's limit point is K $\endgroup$ – Peter Mar 15 '15 at 11:39
  • $\begingroup$ @Peter: YES, sorry, wrote this after a glance of your title. I just saw "limit" without "point" $\endgroup$ – Megadeth Mar 15 '15 at 11:52
  • $\begingroup$ @Peter: Now you have got a new one addressed your problem. $\endgroup$ – Megadeth Mar 15 '15 at 12:10
  • $\begingroup$ @ YLChouIs it really so simple? Then I was thinking to difficult. Is my proof also correct or did I do something wrong? $\endgroup$ – Peter Mar 15 '15 at 12:31
  • $\begingroup$ Sorry but, in its present form, this (accepted and upvoted) answer literally makes no sense. Nothing in the suggested construction guarantees that the sequence $(x_{n_N})_{N\geqslant1}$ converges. One only knows that every term $x_{n_N}$ belongs to the interval $(K-\varepsilon,K+\varepsilon)$, for some given positive $\varepsilon$. $\endgroup$ – Did Mar 15 '15 at 13:41

Here there are two cases.

Case 1: Suppose $\{x_n|n\in \mathbb{N}\}$ is finite. Then there exists a sequence $(n_k)$ of positive integers such that $n_1<n_2\dots<n_k<\dots$ and $x_{n_k}=x_{n_1}=K$ for each $k\in \mathbb{N}$. Then $(x_{n_k})$ is convergent and $\lim\limits_{k\to \infty}x_{n_k}=K$.

Case 2: Now suppose $\{x_n|n\in \mathbb{N}\}$ is infinite. Since $K$ is a limit point of $(x_n)$, for each $ \epsilon >0$, $\left( B_{\epsilon}(K)\setminus \{K\}\right) \cap \{x_n|n\in \mathbb{N}\}\ne \varnothing $ where $B_{\epsilon}(K)$ is the open ball with center $K$ and radius $\epsilon$.

Let $n_1=\operatorname{least element of }\{n\in \mathbb{N}|x_n\in B_{\epsilon}(K)\setminus \{K\}\}$ and $ r_1=\min \{1/2,|x_{n_1}-K|\} $.

Let $n_2=\operatorname{least element of }\{n>n_1|x_n\in B_{r_1}(K)\setminus \{K\}\}$ and $ r_2=\min \{1/3,|x_{n_2}-K|\} $.

Let $n_3=\operatorname{least element of }\{n>n_2|x_n\in B_{r_2}(K)\setminus \{K\}\}$ and $ r_3=\min \{1/4,|x_{n_3}-K|\} $.


Proceeding in this manner obtain a sequence of integers $(n_k)$ such that

$n_k=\operatorname{least element of}\{n>n_{k-1}|x_n\in B_{r_{k-1}}(K)\setminus \{K\}\}$ where $ r_{k-1}=\min \{1/k,|x_{n_{k-1}}-K|\} $.

Thus $(x_{n_k})$ is a sub sequence of $(x_n)$ such that $|x_{n_k}-K|<\frac{1}{k}$ for each $k\in \mathbb{N}$.

So $\lim\limits_{k\to \infty}x_{n_k}=K$. That is $(x_{n_k})$ converges to $K$.

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