# Prove there exists a positive number such that $|x_n| \geq B$

Suppose $$\{x_n\}$$ converges to $$x_0$$ and that all $$x_n$$ and $$x_0$$ are non zero. Prove that there exists a positive number B such that $$|x_n| \geq B$$ $$\forall n$$

So $$\forall \epsilon$$ $$\exists N$$ such that $$\forall n > N$$ we have $$|x_n| - |x_0| <|x_n - x_0| < \epsilon$$ which means we can make $$B = x_0 - \epsilon$$ but then $$|x_n| \geq B$$ when $$n > N$$. I'm kinda stuck on choosing B to make the inequality hold for all n.

## 4 Answers

It looks like you've proven that $$|x_n| \geq B$$ whenever $$n>N$$. Since $$x_1,\dots, x_N$$ are all nonzero, you can just let $$B'=\min(|x_1|,\dots, |x_n|, B)$$. Then $$|x_n| \geq B'$$ for all $$n$$.

$$x_n \rightarrow x_0 \implies |x_n| \rightarrow |x_0|$$

As $$|x_0|$$ is the limit and it is positive, for positive $$\varepsilon = {|x_0| \over 2}$$ there is such N, that for $$\forall n \gt N:$$

$$|x_n| \in \left(|x_0| - {|x_0| \over 2}, |x_0| + {|x_0| \over 2}\right).$$

It means that these element $$|x_n|$$ are inside the ribbon on the following image and subsequently all of them are greater than $$A = {|x_0| / 2}$$.

Now take $$B = \min(|x_1|, \dots , |x_N|, A)$$

We know that $$|x_n| \ge |x_0| - \epsilon ,\forall n > N$$

Also, clearly we have $$|x_i| \ge \min_{1\le j\le N} |x_j|>0.$$

Hence we can pick $$B= \min(\min_{1\le j\le N} |x_j|,|x_0| - \epsilon )$$ where we pick $$\epsilon=\frac{|x_0|}2$$.

Since the sequence is convergent, then it is bounded and so an upper bound exists. Let B be some upper bound. Since the sequence consists only of positive numbers, then no number can be less than -B. Since B is the upper bound and the sequence is positive, the absolute value of every number is less than B.