I want to prove this and intuitively it makes sense. But I'm having a hard time coming up with a proof. So if a sequence converges, then we have a natural number for which the distance between all terms after it and the limit point get arbitrarily small. So how can I show that this also holds for every subsequence (which is like a subset of a sequence)?

(are subsequences always infinite?)

Could I suppose that there is a subsequence that doesn't converge to that limit, and find a contradiction? (and do the same for the other direction?)

  • $\begingroup$ It seems like no effort has been made by the OP to try to answer the question. $\endgroup$
    – Lost1
    Dec 10, 2014 at 0:18
  • 3
    $\begingroup$ What is the OP? $\endgroup$
    – user42
    Dec 10, 2014 at 0:19
  • $\begingroup$ OP = original poster; the one who asked the original question. BTW, subsequences are always infinite, yes. $\endgroup$ Dec 10, 2014 at 8:04

2 Answers 2


Let $x_n \to x$. Then given $\varepsilon> 0$, there exists an $N \in \mathbb N$ such that $|x_n - x| < \varepsilon$ for all $n \geq N$. In words, it means that if we go out far enough, $N$ terms, we can talk about the rest of the terms of the sequence as being close enough, within $\varepsilon$, to the limit, $x$.

If you take any subsequence, say $(x_{n_k})_{k\in\mathbb N}$, then we can say that the $N^{th}$ term of the subsequence is at least, or beyond, the $N^{th}$ term of the actual sequence. Thus, it shares the same property that the terms of the sequence are within a desired distance from the limit of the main sequence.

  • 1
    $\begingroup$ thank you! It's more simple than I thought $\endgroup$
    – user42
    Dec 10, 2014 at 0:23
  • 4
    $\begingroup$ For the other direction, note that the sequence is a subsequene of itself. $\endgroup$ Dec 10, 2014 at 8:09
  • $\begingroup$ This answer proves one direction. Any hints as to how to prove the other direction? $\endgroup$
    – user5826
    Mar 12, 2016 at 4:09
  • $\begingroup$ A sequence is a subsequence of itself - sort of in the same way that a set us a subset of itself. $\endgroup$
    – dannum
    Mar 12, 2016 at 5:49
  • $\begingroup$ Is it true that if any increasing sequence conveges then any sequence converges? $\endgroup$
    – Canjioh
    Nov 18, 2019 at 14:34

Let $\epsilon > 0$ be arbitrary. Since $x_n \rightarrow x \text{ as } n \rightarrow \infty$, $\exists \; N \in \mathbb{N}$ such that $n \geq N \Rightarrow d(x_n, x) < \epsilon$.

Now, let $\{x_{n_k}\}_{k \in \mathbb{N}}$ be a sub-sequence of $\{x_n\}$. The part I'm writing next will give you intuition as to why the sub-sequence will also thin out to $x$ like the parent sequence.

$\exists \; k \in \mathbb{N}$ such that $n_k \geq N$. If not, then $N$ would be an upper bound for the $\textbf{strictly increasing}$ sequence of indices $\{n_1, n_2, \ldots\}$ and the sub-sequence won't have infinite terms. Try to understand this. This is the definition of a subsequence: that the indices are strictly increasing, i.e, $n_1 < n_2 <\ldots$.

Since $n_k \geq N$, we have that $d(x_{n_k}, x) < \epsilon$


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