Real Analysis – Sequence Convergence and Subsequence Convergence

convergence-divergencegeneral-topologyreal-analysissequences-and-series

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?)

Best Answer

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.