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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 $\mathbb{R}$?

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  • $\begingroup$ isnt it just then interval [0,1] $\endgroup$ – cele Oct 21 '13 at 23:42
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    $\begingroup$ 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 $\endgroup$ – Sigur Oct 21 '13 at 23:43
  • $\begingroup$ The smallest closed set that contains $[0,1]$ is $X$. $\endgroup$ – André Nicolas Oct 21 '13 at 23:46
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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).

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  • $\begingroup$ Im not sure if I understand. My first thinking process was that the interior was (0,1) but my definition of interior refers to x is an interior point of A if there exists an open ball B(x,r) in A so since it referes to an open ball, im assuming this topology is not in the usual R2 space and that is why it is empty? $\endgroup$ – cele Oct 22 '13 at 19:49
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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 $\Bbb{R}$.

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  • $\begingroup$ ok, so my coclusion thus far is that the closure is R, the interior is empty, and every x in [0,1] is the boundary of S. Or would it only be the pt 0 and 1?? $\endgroup$ – cele Oct 22 '13 at 20:39

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