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Suppose we have two sequences $(u_n)$ and $(w_k)$ such that $(w_k)$ converges to $w$. If $\forall k \in \mathbb{N}$, $w_k$ is a limit point of $u_n$ show that $w$ is then a limit point of $(u_n)$.

So here's my attempt. Let's define a sub-sequence $\phi _{k,n}$ that for a $k \in \mathbb{N}$, $$\underset{n \infty}{\lim \phi _{k,n}} = w_k $$ $$\underset{k \infty }{\lim} \underset{n \infty}{\lim }\phi _{k,n} = w$$ and $\forall \epsilon > 0$ $|u_{\phi _{k,n}} - w_k| < \epsilon$

Then we have $\forall \epsilon$, $$|u_{\phi_{k,n}} - w| = |u_{\phi _{k,n}} - w_k + w_k - w| \leq |u_{\phi _{k,n}} - w_k| + |w_k - w|$$ As $n$ tends to $\infty$ this gives us $0 + |w_k - w|$ which equals to $0$ when $k$ tends to $\infty$.

Thus $w$ is a limit point of $u_n$. Now I know that I should write with a more precise use of $\epsilon$, but is my (non - rigorous)proof correct?

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  • $\begingroup$ The basic idea is right. But you need to choose the subsequences very carefully. $\endgroup$ – Ice sea Sep 30 '17 at 14:37
  • $\begingroup$ @Icesea What is wrong with my subsequences? $\endgroup$ – John Mayne Sep 30 '17 at 14:38
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    $\begingroup$ You will figure it yourself once you apply $\epsilon-\delta$ language $\endgroup$ – Ice sea Sep 30 '17 at 14:39
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The basic idea is right. But you need to choose the subsequences very carefully. First we will choose a subsequence of $(w_k)$, which will still be denoted by $(w_k)$, such that $|w_k-w|\leq \frac{1}{2^k}$. Then I will choose a subsequence of $(u_{k})$, which is again denoted by $(u_k)$ (because I'm lazy), such that $|w_k-u_k|\leq \frac{1}{2^k}$. Then we have $|u_k-w|\leq |u_k-w_k|+|w_k-w|\leq \frac{1}{2^{k-1}}$, which shows the desired result.

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