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I am not sure how to deal with this Question:

Show, that the sequence $(a_{n})_{n \in \mathbb{N}}$ converges towards the limit $a \in \mathbb{R}$, exactly when every subsequence $(a_{n_{k}})_{k \in \mathbb{N}}$ of $(a_{n})_{n \in \mathbb{N}}$ owns itself a converging subsequence $(a_{n_{k_{j}}})_{j \in \mathbb{N}}$ towards $a$.

To show:

$PS(a_{n}):=$ the set of all subsequences of a given sequence.

$$a_{n}\rightarrow a \Leftrightarrow \forall a_{n_{k}} \in PS(a_{n})\; \exists \; a_{n_{k_{j}}} \in PS(a_{n_{k}}): a_{n_{k_{j}}}\rightarrow a$$

It is easy to show the "$\Rightarrow$" direction, as every subsequence of a given converging sequence, converges towards the same limit.

But what about "$\Leftarrow$"?

Jan
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1 Answers1

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Let $L$ be the set of limit points of the sequence $a_n.$ Since $a_n$ is one of the partial subsequences of $a_n$ it has a partial subsequence converging to $a$ by assumption. So $a \in L.$ Now if $L$ was not the singleton $\{a\}$ we could select a partial subsequence of $a_n$ converging to some $b \in L$ where $b \neq a$. But then this partial subsequence would have all of its partial subsequences converging to $b$ and none converging to $a.$

Since $L$ is a single point $\{a\}$ it follows that $a_n \to a.$

coffeemath
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