# Closure definition

I know two ways of describing the closure of a set $Y\subseteq X$:

1. The smallest closed set containing $Y$ (i.e., the intersection of all closed sets containing $Y$)

2. $x\in\overline{Y}$ iff for all open sets $U$ containing $x$, we have $U\cap Y$ is not empty

I came across another definition: $\overline{Y}=X-(X-Y)^\circ$. I can't see how to get this from the other two definitions or vice versa. Could someone please explain?

## 2 Answers

First note that $(X \setminus Y)^\circ$ is open and is a subset of $X \setminus Y$. It follows that $X \setminus ( X \setminus Y )^\circ$ is closed, and $Y = X \setminus ( X \setminus Y ) \subseteq X \setminus ( X \setminus Y )^\circ$, and from definition 1 it follows that $\overline{Y} \subseteq X \setminus ( X \setminus Y )^\circ$.

Note, now, that if $x \in X \setminus ( X \setminus Y )^\circ$, then $x \notin ( X \setminus Y )^\circ$. As $( X \setminus Y )^\circ$ is the largest open set included in $X \setminus Y$ it follows that given any open neighbourhood $U$ of $x$, $U \not\subseteq X \setminus Y$, which means that $U \cap Y \neq \varnothing$. By definition 2 it follows that $x \in \overline{Y}$. Thus $X \setminus ( X \setminus Y )^\circ \subseteq \overline{Y}$.

A good way to think about these different definitions is to realize that any set $Y\subseteq X$ in a topological space $X$ partitions $X$ into the interior of $Y$, the boundary of $Y$ and the exterior of $Y$, while those 3 parts could be defined like

1. $x\in {\rm int}(Y)$ iff there is an open set $U$ with $x\in U \subseteq Y$,
2. $x\in {\rm ext}(Y)$ iff there is an open set $U$ with $x\in U \subseteq X\setminus Y$,
3. $x\in \partial Y$ iff every open set $U$ with $x\in U$ intersects both $Y$ and $X\setminus Y$ nonempty.

From these definitions it's obvious that $X={\rm int}(Y) \cup{\rm ext}(Y)\cup\partial Y$ is in fact a partition and easy to check that your definition 2 is equivalent to $\overline Y := {\rm int}(Y) \cup \partial Y$. Further we observe that ${\rm ext}(Y) = {\rm int}(X\setminus Y)$, so we have $$X \setminus {\rm int}(X\setminus Y) = X \setminus {\rm ext}(Y) = {\rm int}(Y) \cup \partial Y = \overline Y.$$