# find closure, interior and boundary of S

For the following example:

Let the topological space $X$ be the real line $\mathbb{R}$. An open set is any set whose complement is finite. Let $S=[0,1]$. Find the closure, the interior, and the boundary of $S$.

What is meant by let the topological space $X$ be the real line $\mathbb{R}$?

• isnt it just then interval [0,1] – cele Oct 21 '13 at 23:42
• The set $X$ is the set of real numbers but the topology is not the usual one. You have other types of open sets. For example,using the standard topology, the subset $(0,1)$ is open but it is not with your topology since it does not have finite complement – Sigur Oct 21 '13 at 23:43
• The smallest closed set that contains $[0,1]$ is $X$. – André Nicolas Oct 21 '13 at 23:46

Hint. Let's $\tau$ is the topology of your question. You can think of a more concrete way of explaining the face of this open topology.If $O\in \tau$ and $O\neq \emptyset$ then $O^c$ is finite. That is, thare is $n$ numbers $x_1<x_2<\ldots, x_{n-1}<x_n$ such that $O^c=\{x_1, \ldots, x_n\}$. Then, $$O=(-\infty,x_1)\cup( x_1,x_2)\cup\ldots\cup( x_{n-1},x_n)\cup(x_n,\infty)$$ This means that each nonempty open $O\in\tau$ contains at least two intervals of infinite type $$(-\infty,a) \mbox{ and } (b,+\infty) \mbox{ with } a\leq b.$$ Therefore the only open that can be contained in the set S is empty. Since, by definition, the interior of a set $S$ is the union of all open contained in $S$ then the interior of $S$ is the empty set ($S$ contains only empty set).
What it should say is let $X$ be a topological space on $\Bbb{R}$ whose open sets consist of all subsets of $\Bbb{R}$ that have a finite complement in $\Bbb{R}$.