# determine the closures of the set k={1/n| n is a positive integer}

Consider the following topologies on $\mathbb{R}$:

1. the standard topology
2. the finite complement topology
3. the lower limit topology.

The question is to determine the closures of the set $k =$ {$\displaystyle \; \frac 1n|\; n \; \text{ is a positive integer }$} under each of these topologies.

Please I would like somebody to help me.

Thank you.

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We're not here to do your homework for you. What have you tried? Where are you stuck? –  Yuval Filmus Jun 12 '11 at 6:38
@Yuval; i hv got no idea, am stil thinking –  aloyce temba Jun 12 '11 at 6:48

Let $k$ be the set of your question in your notation:

Hint for (1) [standard topology]: Prove that the point $0$ is in the closure of $k$. If $x<0$, then $x$ cannot be in the closure of $k$ since $(x-1,0)$ is a neighborhood of $x$ that does not intersect $k$. Similarly, prove that if $x>1$, then $x$ is not in the closure of $k$. Of course, every point of $k$ is in the closure of $k$. However, if $0<x<1$ is a point not in $k$, prove that $x$ is strictly between two points of $k$; that is, there is an integer $n\geq 1$ such that $\frac{1}{n+1}<x<\frac{1}{n}$. Deduce that $x$ is not in the closure of $k$. Can you now tell what the closure of $k$ is?

Hint for (2) [finite complement topology]: Note that $k$ is infinite and that the complement of any non-empty open set in this topology is finite. Can you now tell what the closure of $k$ is?

Hint for (3) [lower limit topology]: The closure of $k$ in this topology is exactly the same as in (1). Why?

The following exercises (with hints) will further your understanding of this problem:

Exercise 1:

Let $T$ be the upper limit topology on $\mathbb{R}$; that is, $T$ is the topology generated by the base of all sets of the form $(a,b]$ where $a<b$. What is the closure of $k$ in the topology $T$. (Hint: The real question is whether or not $0$ is in the closure of $k$.)

Exercise 2:

What is the closure of $S=\{1-\frac{1}{n}:n\geq 1\}$ in the lower limit topology? (Hint: Prove that $1$ is not in the closure of $S$.) What is the closure of $S$ in the upper limit topology?

Exercise 3:

The cocountable topology on $\mathbb{R}$ is the topology consisting of $\emptyset$ and all subsets of $\mathbb{R}$ whose complement is countable. What is the closure of $k$ (the set of your question) in the cocountable topology on $\mathbb{R}$? What is the closure of $(0,1)=\{x:0<x<1\}$ in the cocountable topology on $\mathbb{R}$? (Hint: $k$ is countable and $(0,1)$ is uncountable. Therefore, the complement of $k$ is open in the cocountable topology. However, the complement of $(0,1)$ is not open in the cocountable topology.)

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thanks , i have to follow through it cleary –  aloyce temba Jun 12 '11 at 6:54
For (1), if you know compactness, then it's easy to show that the set together with $\{0\}$ is a compact set and thus closed (as the usual topology is Hausdorff), and so must be its closure. –  Henno Brandsma Jun 12 '11 at 17:36
Just to nitpick, in the cofinite topology, the complement of the empty set is the entire space, and it is also closed. –  Asaf Karagila Jun 12 '11 at 17:36
Dear Henno, of course, you are correct. However, the elementary nature of the question suggests that the OP has not been introduced to compactness. If he has seen compactness, then your solution is cleanest. Dear Asaf, you are correct, of course. I was imprecise because I was thinking only about "non-empty" in any case as the open sets under consideration when one studies the closure of $k$ are non-empty. –  Amitesh Datta Jun 12 '11 at 23:13
Dear Asaf, I added "non-empty" to the relevant sentence. –  Amitesh Datta Jun 12 '11 at 23:18