prove it is closed without refering to limit points

**magus** · April 6th 2011, 01:52 AM

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 { $\left{\frac{1}{i}\right}$ } $^n_{i=1}$ , let's call it $S$ , 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 $S'$ 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 $(a,b)$ 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 $(a,b)$ .

Therefore the set is not closed.

**Plato** · April 6th 2011, 02:43 AM

Originally Posted by magus

The question was if the set { $\left{\frac{1}{i}\right}$ } $^n_{i=1}$ , let's call it $S$ , 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 $S = \left\{ {\dfrac{1}{i}} \right\}_{i = 1}^n$ .
That is a finite set.
Therefore, it is closed in the Euclidean topology.

**magus** · April 6th 2011, 01:53 PM

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

**HallsofIvy** · April 6th 2011, 02:09 PM

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

**Plato** · April 6th 2011, 02:12 PM

Originally Posted by magus

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

Then your title of the OP is totally misleading.

**magus** · April 6th 2011, 02:59 PM

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.

**Plato** · April 6th 2011, 03:49 PM

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.

**tonio** · April 6th 2011, 07:16 PM

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

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

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

Tonio

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