# Thread: prove it is closed without refering to limit points

1. ## prove it is closed without refering to limit points

I'm working on topology again and I wanted to know if this proof for a question that I came on is correct.

The question was if the set {$\displaystyle \left{\frac{1}{i}\right}$}$\displaystyle ^n_{i=1}$, let's call it $\displaystyle S$, was closed in the Euclidean topology. From prior reading I knew it wasnt since it was missing it's limit point, but I couldn't use that here so I went with this.

Take $\displaystyle S'$ and suppose that it were open. There are smaller and smaller open intervals beside and not containing smaller and smaller rational numbers approaching zero.

Let $\displaystyle (a,b)$ be the open interval that surrounds zero. To the right of it there will be an open interval but to the left of that interval will be a point of S no matter how much you shrink $\displaystyle (a,b)$.

Therefore the set is not closed.

2. Originally Posted by magus
The question was if the set {$\displaystyle \left{\frac{1}{i}\right}$}$\displaystyle ^n_{i=1}$, let's call it $\displaystyle S$, was closed in the Euclidean topology. From prior reading I knew it wasnt since it was missing it's limit point, but I couldn't use that here so I went with this.
You have given us this set $\displaystyle S = \left\{ {\dfrac{1}{i}} \right\}_{i = 1}^n$.
That is a finite set.
Therefore, it is closed in the Euclidean topology.

3. My mistake, it should have been $\displaystyle S = \left\{ {\dfrac{1}{i}} \right\}_{i = 1}^\infty$

4. Then the crucial point is: $\displaystyle \displaytype \lim_{n\to\infty}\frac{1}{n}= 0$ which is not of the form $\displaystyle \frac{1}{n}$ for any integer n.
That is essentially what you showed.

5. Originally Posted by magus
My mistake, it should have been $\displaystyle S = \left\{ {\dfrac{1}{i}} \right\}_{i = 1}^\infty$

6. I'm sorry. What I was trying to say was I needed an argument that did not include the fact that closed sets contain their limit points. Sorry.

7. Originally Posted by magus
I'm sorry. What I was trying to say was I needed an argument that did not include the fact that closed sets contain their limit points. Sorry.
I am also sorry to tell you that I have no idea what that response means.

8. Originally Posted by Plato
I am also sorry to tell you that I have no idea what that response means.

I understood what he meant, apparently, from the beginning: he needs to show that that set isn't closed

without using at all limits points.

For example, if $\displaystyle \left\{\frac{1}{n}\right\}_{n=1}^\infty$ is closed, then its complement is open, but

this isn't so since $\displaystyle 0\in Comp.\left(\left\{\frac{1}{n}\right\}_{n=1}^\infty \right)$ , but there is no

open ball around zero which is completely contained in that complement. Q.E.D.

Tonio