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If every subsequence of a given sequence of real numbers is convergent, prove that the sequence is convergent.

Help me please. I could not understand how to solve this question.

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    $\begingroup$ The sequence is a subsequence of itself. $\endgroup$ Mar 6 '13 at 6:17
  • $\begingroup$ @user65268 Yeah, there must be something wrong in your question. Do you want to prove: If every subsequence has a further convergent subsequence and with the same limit, then the original sequence converges? $\endgroup$
    – Coiacy
    Mar 6 '13 at 7:37
  • $\begingroup$ Your question is trivial as it stands (see Andre's comment and DonAntonio's answer). You should probably change your "If" to "if and only if". $\endgroup$ Oct 23 '13 at 23:26
  • $\begingroup$ @Coiacy : I changed the "if" to "if and only if". Your version is less obvious but more interesting. $\endgroup$ Oct 23 '13 at 23:38
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Since the sequence $\,\{x_2,x_3,...,x_n,...\}\,$ converges, then also the whole sequence converges and, of course, to the very same limit.

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  • $\begingroup$ Is it true to say that one subsequence is enough to be convergent in order that the main sequence to be convergent? If so, why? (it seems to be true since every subsequence of a sequence has the same limit). Thank you. $\endgroup$
    – L.G.
    Jan 30 '15 at 8:32
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    $\begingroup$ @Ali.E. It is true that if this very particular subsequence converges then the whole one also does and to the same limit. In general it is false, though. $\endgroup$
    – Timbuc
    Feb 6 '15 at 20:13
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Your question is trivial unless you change the "If" to "If and only if". I'll prove the "if and any if" version.

(Trivial direction) Any sequence is a subsequence of itself, so if all subsequences of a given sequence converge, so does the original sequence.

(Nontrivial direction) Suppose $(x_n)$ converges to $L$. Let $(x_{n_i})_{i \geq 1}$ be a subsequence of $(x_n)$. Let $\epsilon > 0$. Since $(x_n)$ converges to $L$, there exists $N \equiv N(\epsilon) \in \mathbb{N}^+$ with the property that if $i \in \mathbb{N}^+$ with $i \geq N$, then $|x_i-L|<\epsilon$. This is from the definition of convergence of a sequence. Now $n_i \geq i$, so if $j\geq i$, then $n_j \geq n_i \geq i$, and $|x_{n_j}-L| < \epsilon$. So the subsequence $(x_{n_i})_{i \geq 1}$ converges to $L$.

There is nothing special about $\mathbb{R}$ in the proof. It would work the same in any metric space (I'm rusty when it comes to non-metric spaces).

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Hint: Recall that a sequence $\{a_n\}$ of real numbers converges to a limit $L$ when the following statement is true: for any $\epsilon>0$, there is an integer $N$ such that, for all $n>N$, we have $|L-a_n|<\epsilon$.

Now use a proof by contradiction. If the original sequence didn't converge, then there is some $\epsilon>0$ such that, no matter what integer $N$ you pick, there is some element $a_n$ of the sequence, where $n>N$, such that $|L-a_n|\geq \epsilon$.

What did we just construct ... ?

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  • $\begingroup$ Maybe to be clearer one can say take $N=1$, obtain some $a_{n_1}$ with the said property. Now take $N$ the next natural after $n_1$ and obtain an $a_{n_2}$, and so on... $\endgroup$
    – Pedro Tamaroff
    Mar 6 '13 at 5:16

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