# A sequence converges if and only if every subsequence converges?

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?)

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

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.

• thank you! It's more simple than I thought Dec 10, 2014 at 0:23
• For the other direction, note that the sequence is a subsequene of itself. Dec 10, 2014 at 8:09
• This answer proves one direction. Any hints as to how to prove the other direction? Mar 12, 2016 at 4:09
• A sequence is a subsequence of itself - sort of in the same way that a set us a subset of itself. Mar 12, 2016 at 5:49
• Is it true that if any increasing sequence conveges then any sequence converges? 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 $$d(x_{n_k}, x) < \epsilon$$