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Let $\big<X,d\big> $

$\{x_n\} \rightarrow x \quad$ iff every subsequence of $\{x_n\}$ has a subsequence converging to $x\in X$.

Note also the proposition: $\{x_n\} \rightarrow x\quad$ iff every subsequence of $\{x_n\} $ converges to $x\in X$

When proving "$\implies $" in theorem above, should we use that proposition? Or is it trivial since my instructor skipped that step and proved the other direction.

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Leyla Alkan
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2 Answers2

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Just use the definition of the limit : Let $\{x_{\sigma (n)}\}$ be a subsequence of $\{x_n\}$. Let $\epsilon >0$. $\exists N \in \Bbb N$ such that $\forall n\ge N, d(x_n,x)<\epsilon$. And since $\sigma (n)\ge n \forall n\in \Bbb N$ then $d(x_{\sigma (n)},x)<\epsilon \forall n\ge N$ and so $x_{\sigma (n)}\rightarrow x$ as $n$ goes to infinity.

Tengen
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For $\implies$, yes that proposition easily implies the answer. If you need to ask a question about it, then it means it isn't trivial to you, which is a good indicator that you should write out the proof.

Theo Bendit
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  • Can I think of this as if x<y and y<z then x<z ? Like if sub-subsequence converges to x then subsequence converges to x also , and if subsequence converges to x then the main sequence is also convergent to x ? – Leyla Alkan Nov 05 '17 at 14:17
  • Not really. It's possible to have a non-convergent sequence with a convergent subsequence, so it's not valid to conclude, from the fact that the subsequence (or sub-subsequence) is convergent, that the entire sequence is convergent. – Theo Bendit Nov 05 '17 at 14:21
  • For proving $\impliedby$, my advice would be start with a non-convergent $(x_n){n=1}^\infty$ and construct a subsequence $(x{m_n}){n=1}^\infty$ and $\varepsilon > 0$ such that $d(x{m_n}, x) > \varepsilon$ for all $n$. – Theo Bendit Nov 05 '17 at 14:23
  • I am OK with the other direction actually – Leyla Alkan Nov 05 '17 at 14:24
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    Oh, I thought that's what you were asking about. In that case, you can conclude from the sequence converging to $x$, that every subsequence converges, and that every sub-subsequence converges similarly, to $x$. – Theo Bendit Nov 05 '17 at 14:27