Question: Suppose that every subsequences of $X = (x_n)$ has a subsequence that converges to $0$. Show that $\lim(X) = 0$

My attempt:

Suppose that $\lim(X) =L \neq 0$

Let $\epsilon > 0$, and let $K(\epsilon)$ be s.t. $n \ge K(\epsilon)$

then $|x_n-L| < \epsilon$

let $n_k \ge k$, and then $k \ge K(\epsilon)$ and $n_k \ge k \ge K(\epsilon)$ s.t. $|x_{n_{k}}-L| < \epsilon$, which is a contradiction. therefore $\lim(x) =0$

  • $\begingroup$ Your sequence can fail to converge to 0 without converging to something else. It may not converge to anything at all. $\endgroup$ – siegehalver Mar 13 '16 at 6:05

Right when you say let $\lim X = L$ you've already assumed that $X$ has a limit. A priori, you can't just assume this. It may very well be that $X$ doesn't converge. The way to do this is by contradiction, assume that we don't have $\lim X = 0$. Then for some $\epsilon > 0$, we have the following: for any $k \in \mathbb N$, there is an $n_k \ge k$ such that $\|x_{n_k} \| \ge \epsilon$. Without loss of generality, $n_k$ is an increasing sequence (indeed, after finding $n_1$, there must be $n_2 > n_1$ with $\| x_{n_2} \| \ge \epsilon$ and then there must be $n_3 > n_2$, etc). Now we have assumed that every subsequence of $X$ converges to 0. However, we have constructed a subsequence which is always larger than $\epsilon$ in absolute value. This subsequence cannot converge to 0, a contradiction. Hence our original assumption was incorrect, and so we must have $\lim X = 0$.


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