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## Homework Statement

## The Attempt at a Solution

If [itex](x_n)_{n\in\mathbb{N}}[/itex] is a sequence in [itex](X,d)[/itex] and [itex](n_k)_{ k \in \mathbb{N}}[/itex] is a strictly increasing sequence in [itex]\mathbb{N}[/itex] then [itex](x_ {n_k} )_{k \in \mathbb{N}}[/itex] is a

*subsequence*of [itex](x_n)_{n\in\mathbb{N}}[/itex].

How would you prove that if [itex]x_n[/itex] converges to [itex]x[/itex] in [itex]X[/itex] then any subsequence of [itex]x_n[/itex] converges to [itex]x[/itex]?

In (a) how would you justify that the sequence [tex]x_n = n\;\;(n\in\mathbb{N})[/tex] has no convergent subsequence?

For (b) the alternating sequence [tex]0, 1, 0, 1, 0, 1, 0, 1, ...[/tex] diverges but if [itex]n_k = 2k[/itex] then [itex]x_{n_k} = 1[/itex] which trivially converges to 1.

[itex](X,d)[/itex] is

*sequentially compact*if every sequence has a convergent subsequence.

For (a) the open interval [itex](0,1) \subset \mathbb{R}[/itex] is not sequentially compact as, letting [itex]x_n = \frac{1}{n}[/itex] we have that [itex]x_n \to 0[/itex] in [itex]\mathbb{R}[/itex], hence every subsequence converges to 0, but [itex]0\notin (0,1)[/itex] so by uniqueness of limits, no subsequence of [itex]\frac{1}{n}[/itex] converges in [itex](0,1)[/itex], so [itex](0,1)[/itex] is not sequentially compact.

What counterexample could I use for (b)?

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