Q:

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

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

the standard topology
the finite complement topology
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.

A:

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.) 

