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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!

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  • $\begingroup$ $S$ is not an open ball doesn't mean that $S=\{0\}$. $\endgroup$
    – Surb
    Oct 22, 2021 at 9:14
  • $\begingroup$ Did not understand what you mean $\endgroup$ Oct 22, 2021 at 9:16
  • $\begingroup$ 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. $\endgroup$ Oct 22, 2021 at 9:19
  • $\begingroup$ Oh I see! Okay I understand ! Thanks a lot! $\endgroup$ Oct 22, 2021 at 9:21

1 Answer 1

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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.

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