$\exists\delta>0$ and subsequence $\{\vec{x}_{n_i}\}$ of $\{\vec{x}_{n}\}$ such that $\{\vec{x}_{n_i}\}\cap B_\delta(0)=\emptyset$

Question

Consider a normed vector space $(V,\lVert\rVert)$, and $\{\vec{x}_n\}\in V$, which does not converge to $\vec{0}$. Prove that $\exists\delta>0$ and a subsequence $\{\vec{x}_{n_i}\}$ of $\{\vec{x}_{n}\}$, such that $\{\vec{x}_{n_i}\}\cap B_\delta(\vec{0})=\emptyset$.

My proof:

Choose $m_1:=\min\limits_n\{\lVert \vec{x}_n\rVert\}$, $m_2:=\min\limits_{n\ne m}\{\lVert \vec{x}_n\rVert\}$, and $m=\max\{m_1,m_2\}$. Then choose any $\delta$ such that $0<\delta<m$. Now define $\{\vec{x}_{n_i}\}:=\{\vec{x_n}:n\in\mathbb{N}\text{ and } \lVert \vec{x}_n\rVert\ne 0\}$. Then $\{\vec{x}_{n_i}\}\cap B_\delta(\vec{0})=\emptyset$.

Please let me know if my proof is correct.

Answer 1

The main issue with this proof is that it's not always possible to take the minimum of an infinite number of things. For example, what's the minimum of $\{1+1/n: n\in\mathbb{N}\}$?

Recall what the definition of $x_n\to 0$ is: for all $\delta>0$ there is some $N$ such that $n\geq N$ implies $\|x_n\|<\delta$. Carefully form the negation of this statement, and you may find the correct proof is much easier than you expected!