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. – Lost1 Dec 10 '14 at 0:18
• What is the OP? – user42 Dec 10 '14 at 0:19
• OP = original poster; the one who asked the original question. BTW, subsequences are always infinite, yes. – Henno Brandsma Dec 10 '14 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.