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Show that sequence converges to accumulation point

Sol : If $\{a_{n}\}$ has an accumulation point, say, $a$, and $(a_{n})$ is convergent. Then choose some $n_{1}$ such that $a_{n_{1}}\in B_{1}(a)-\{a\}$. Then choose some $n_{2}$ such that $B_{1/2}(a)-\{a,a_{1},...,a_{n_{1}}\}$, proceed in this way we have $a_{n_{k}}\rightarrow a$. Since $(a_{n})$ is convergent, one has $a_{n}\rightarrow a$.

From

I don't understand the answer from the "choose some $n_2$ such that $B_{\frac12} (a) - \{ a, a_1...,a_{n_1} \}$". What does $n_2$ have to do with the ball with points of sequence removed..?

And, how does proceeding in this way mean that $a_{n_k} \to a$ happens..?

tryst with freedom
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2 Answers2

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I think its supposed to be $a_{n_2} \in B_{1/2}(a) - \{a_1, \ldots, a_{n_1}\}$. Then in general $a_{n_{k+1}} \in B_{(1/2^{k})}(a) - \{a_1,\ldots, a_{n_{k}}\}$

Rishi Sonthalia
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$a$ is an accumulation point of $A=\{x_n\}$ implies any open set $U$ containing $a$ intersect $A\setminus\{a\}$.

$\forall k\in \Bbb{N}$ choose a sequence $(x_{n_k})\in B(a, \frac{1}{k}) \cap A\setminus \{a\}$

$|x_{n_k}-a|<\frac{1}{k}\to \infty \text{ as } n\to \infty$

Then $(x_{n_k}) $ converges to $a$ and since $(x_{n_k})$ is a subsequence of the convergent sequence $(x_n) $ , hence $(x_n) \to a$.

Sourav Ghosh
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