# Let $D \subset \mathbb{R}$ and let $D^\circ$ be the interior of $D$. So if $x \notin D^\circ$ then $x$ must be in the closure $\mathbb{R}\setminus D$.

Let $D \subset \mathbb{R}$ and let $D^\circ$ be the interior of $D$. So if $x \notin D^\circ$ then $x$ must be in the closure $\mathbb{R}\setminus D$.

My attempt at this: Suppose $x \notin D^\circ$. Then $x$ is not an interior point of $D$, i.e there does not exist $\delta > 0$ such that $(x-\delta,x+\delta) \subset D$. So either $x \in D$ and $x$ is not an interior point or $x \in \mathbb{R}\setminus D$ and is an interior point of $\mathbb{R}\setminus D$ or $x \in \mathbb{R}\setminus D$ and is not an interior point of $\mathbb{R}\setminus D$.

From here it seems like I have to go case by case, but it seems unclear how this will even get closure for this from here. Also do I even need the first case, as it tells me nothing about $\mathbb{R}\setminus D$. Thanks in advance.

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I corrected some of your notation: the superscript symbol for interior is \circ, and set difference needs the backslash, \setminus, not the slash. – Brian M. Scott Oct 25 '12 at 13:24

$\newcommand{\int}{\operatorname{int}}\newcommand{\cl}{\operatorname{cl}}$Suppose that $x\notin\int D$. Then for every $\epsilon>0$, $(x-\epsilon,x+\epsilon)\nsubseteq D$, and therefore $$(x-\epsilon,x+\epsilon)\cap(\Bbb R\setminus D)\ne\varnothing\;.$$ This means that every open neighborhood of $x$ contains a point of $\Bbb R\setminus D$, so $x\in\cl(\Bbb R\setminus D)$.
Added: Sometimes it’s necessary to split a problem into cases, but beginners often do so when it’s not actually necessary: when the ideas are still unfamiliar, it can be easy to lose sight of the forest as a whole and see only a rather overwhelming bunch of individual trees. It’s not a bad idea to remind yourself to ask whether a division into cases is really necessary, or whether in fact there’s one observation that takes care of all of the apparent cases at once. In this problem there is: for any point $x$, $x$ is in the closure of a set $A$ if and only if every open neighborhood of $x$ has non-empty intersection with $A$. Thus, all points can be dealt with at once.