# Closure of $(0,1) \subset \mathbb{R}$ with co-finite topology and generalisation

I'm currently working through lectures notes on a second year Topology course, and am not yet at ease with the idea of closure in different topological spaces. For example, the closure of $(0,1)$ in the Euclidean space $\mathbb{R}$ is $[0,1]$, does the closure change if $\mathbb{R}$ has co-finite topology? Can we generalise this statement for subsets of $\mathbb{R}$? Any help would be appreciated, regards.

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In the cofinite topology, a nonempty set is open if and only if its complement is finite; hence a proper subset is closed if and only if it is finite. Thus, the closure of $(0,1)$ in $\mathbb{R}$ cannot be a proper subset (since the closure contains $(0,1)$ and hence is infinite), so the closure is all of $\mathbb{R}$.

More generally, any infinite subset of $\mathbb{R}$ is dense in $\mathbb{R}$ in the cofinite topology. So the closure of a subset of $\mathbb{R}$ is the set itself, if the set is finite, or all of $\mathbb{R}$ if the set is infinite.

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I thought that may be the case but it seemed rather unconventional. Many thanks for this explanation. –  Mathmo Feb 13 '12 at 21:50
@MiamiMath: It is only "unconventional" in the sense that if no topology for $\mathbb{R}$ is specified, the convention is to take the "standard" (Euclidean) topology. But there are lots of topological spaces in which it is very easy for sets to be dense. –  Arturo Magidin Feb 13 '12 at 21:51
Thanks, I'll be sure to keep reading! –  Mathmo Feb 13 '12 at 22:00
Regardless of the ambient set $X$, in the co-finite topology on $X$ by definition we have that a subset $U \subseteq X$ is open iff either $U = \emptyset$ or $X \setminus U$ is finite. From this it follows that $F \subseteq X$ is closed iff either $F = X$ or $F$ is finite.
From this we get a simple observation. For any $A \subseteq X$ we have that $$\overline{A} = \begin{cases} A, &\text{if A is finite} \newline X, &\text{if A is infinite.} \end{cases}$$