Any sequence whose elements are non-zero, and which converges to non-zero, is bounded away from zero

Question

The question is to prove the above proposition.

I can see that this statement is obviously true, but don't know how to translate the condition "[all] elements are non-zero" into something that can help me do the proof.

Answer 1

Let $a_n$ be the sequence in question with $a_n\rightarrow a$. Now write down convergence for $a_n$: For all $\varepsilon>0$ you can find a natural number $n_0=n_0(\varepsilon)$ such that for every $n\geq n_0$ you have $$|a_n-a|<\varepsilon.$$ If you now choose $\varepsilon=\frac{|a|}{2}$, then $|a_n|\geq \frac{|a|}{2}$ for every $n\geq n_0(\frac{|a|}{2})$, hence these are bounded away from zero. Now you still have the elements $a_1,\ldots a_{n_0-1}$. Since they are all nonzero and there are only finitely many, they can be bounded by a constant $C$. Hence $$|a_n|\geq\min(C,\frac{|a|}{2}).$$

Answer 2

I think you want to show that $\exists m>0: |a_n|>m \forall n\in\mathbb{N} $

Suppose this fails to hold: Then $\forall m>0: \exists n:|a_n|\leq m $. This means that taking $m=\frac{1}{n}, n=1,2,\dots$ results in a subsequence converging to $0$. (Note how we used that the entries are non-zero - if there was a $0$ entry, then we wouldn't get a sequence). But as you initial sequence is convergent and converges to $l \neq 0$ then you get a contradiction.