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Consider the topological space of real numbers with the usual topology.

Consider the subset: $$Y = \{0\}\cup \left\{\frac{1}{n} \mid n\in \mathbb N\right\} \cup ((3,4)\cap \mathbb Q) \cup (6,7]$$ I have to find exterior of $Y$. I know that $\mathrm{Ext}(Y)=\mathrm{Int}(R-Y)$, but I can't find the complement of $Y$.

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The complement of $Y$ in $R$ is the set of everything in $R$ that isn't in $Y.$ Which is $R\backslash Y=A\cup B\cup C\cup D\cup E\cup F$ where $A=(-\infty,0),\; B=\cup_{n\in N} (1/(n+1),1/n), \; C=(1,3],\; D=(3,4)\backslash Q,\; E=[4,6],\;$ and $ F=(7,\infty).$

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  • $\begingroup$ $B=(0,1]\backslash Y=(0,1]\backslash \{1/n :n\in N\}.$ There are no reciprocals of positive integers between $1/n$ and $1/(n+1)$, because $(1/1,1/2,1/3,1/4,...)$ is a strictly decreasing sequence. $\endgroup$ – DanielWainfleet Jan 14 '16 at 23:24

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