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I have these two definitions for an adherent value of a sequence

the first is : $a$ is a an adherent value for $(x_n)$ iff $$\displaystyle \forall \varepsilon>0,\forall n\in \mathbb{N},\exists n_0\geq n, d(x_{n_0},a)\leq \varepsilon$$ the second is $a$ is an adherent value for $(x_n)$ if there exists a sub sequence witch converge to $a$

But i don't know how to prove the equivalence between the two definitions

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If there exists a subsequence which converges to $a$, it's clear the first definition is satisfied. Conversely, suppose the first definition is satisfied. We can define a subsequence that converges to $a$ in the following way: start with $\varepsilon=1$, and let $i_0$ be the first subscript $i$ such that $\lvert x_i-a\rvert < \varepsilon$.Then set $ \varepsilon:=\dfrac\varepsilon{10}$, and let $i_1$ be the first subscript $i>i_0$ such that $\lvert x_i-a\rvert < \varepsilon$. This defines recursively a subsequence $(v_k)=(u_{i_k})$ of $(u_i)$ which converges to $a$ by construction.

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  • $\begingroup$ why $\varepsilon/10$ ? $\endgroup$ – Vrouvrou Jan 6 '15 at 20:45
  • $\begingroup$ It's just for the successive values of $\varepsilon$ to be $10^{-n}$. You might as well take $\varepsilon/2$, or any sequence that tends to $0$. $\endgroup$ – Bernard Jan 6 '15 at 20:48

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