# $\bigcap_{n=1}^{\infty}(-\frac{1}{n}, \frac{1}{n})$ Using open ball definition

Show that $$\{0\} = \bigcap_{n=1}^{\infty}(-\frac{1}{n}, \frac{1}{n})$$ is not an open set.

This question was solved in this site, using open set definition. But I tried to solve it using open ball. Since we know, open ball is a open set we can use the open ball definition. We will use proof by contradiction.

Let $$\mathrm{S} = \bigcap_{n=1}^{\infty}(-\frac{1}{n}, \frac{1}{n})$$ is an open ball. We want to show that:

$$\forall x\in \mathrm{S},\quad \exists \delta>0, \quad B(\delta,x) \subseteq \mathrm{S}$$

Let $$x\in \mathrm{S} \rightarrow \left(-\dfrac{1}{n}, \dfrac{1}{n} \right) \in \mathrm{S}, \quad \exists \delta>0$$ then we can say there is $$\delta < x -\delta$$.

Then,

Let $$y\in B(\delta, x) \rightarrow \left|y-x\right| < \delta, x-\delta < y < x + \delta$$. And,

$$0 < x-\delta < y, \quad 0< y$$

But, $$y \notin B(\delta, x)$$. That means $$B(\delta, x) = \emptyset$$.

So $$\mathrm{S}$$ is not an open ball, rigorously, not an open set.

Is it true? Thanks in advance!

• $S$ is not an open ball doesn't mean that $S=\{0\}$.
– Surb
Oct 22, 2021 at 9:14
• Did not understand what you mean Oct 22, 2021 at 9:16
• The fact that $S$ is not an open ball doesn't imply that it's not an open set. The set $(0,\infty)$ is open, but it's not an open ball. Oct 22, 2021 at 9:19
• Oh I see! Okay I understand ! Thanks a lot! Oct 22, 2021 at 9:21

Your proof is a bit confusing. I would write it the following way. Suppose $$S$$ was an open set. Then there would exist an open ball centered at $$0$$ of a certain radius $$\delta$$ contained in $$S$$.

Suppose there exist $$\delta$$ such that $$B(\delta, 0) \subset S$$. In particular, that would mean $$0 < \delta < \frac{1}{n}$$ for every $$n \in \mathbb{N}$$. But this clearly isn't possible, since $$\lim_n \frac{1}{n} = 0$$. Therefore, no such $$\delta$$ can exist.

Clarification (following comments): you don't have to prove that your set is not an open ball. You have to show there is a point such that your set does not contain an open ball around that point. In this case, the election of the point is obvious.