Interiors and closures

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 21 May 2011

Interiors and closures

Let $X$ be a topological space and let $E\subseteq X$.
The interior of $E$ is the subset $E^o$ of $X$ such that

(a)   $E^o$ is open and $E^o\subseteq E$, and

(b)   if $U$ is open and $U\subseteq E$ then $U\subseteq E^o$.

The closure of $E$ is the subset $\bar{E}$ of $X$ such that

(a)   $\bar{E}$ is closed and $\bar{E}\supseteq E$, and

(b)   if $V$ is closed and $V\supseteq E$ then $V\supseteq \bar{E}$.

Let $X$ be a topological space and let $E\subseteq X$.
An interior point of $E$ is a point $x\in X$ such that there exists a neighbourhood $N$ of $x$ such that $N\subseteq E$.
A close point of $E$ is a point $x\in X$ such that if $N$ is a neighbourhood of $x$ then $N\cap E\ne \varnothing$.

Let $X$ be a topological space. Let $E\subseteq X$.

(a)   The interior of $E$ is the set of interior points of $E$.

(b)   The closure of $E$ is the set of close points of $E$.

		Proof (of part a).
	Let $I=\{x\in E|x\text{is an interior point of}E\}.$ To show that $E^o=I$, we show that (aa) $I\subseteq E^o$ and then that (ab) $E^o\subseteq I$. Let $x\in I$. Then there exists a neighbourhood $N$ of $x$ with $N\subseteq E$. So there exists an open set $U$ with $x\in U\subseteq N\subseteq E$. Since $U\subseteq E$ and $U$ is open $U\subseteq E^o$. So $x\subseteq E^o$. So $I\subseteq E^o$. We want to show that if $x\in E^o$ then $x\in I$. Assume $x\in E^o$. Then $E^o$ is open and $x\in E^o\subseteq E$. So $x$ is an interior point of $E$. So $x\in E^o$. So $I\subseteq E^o$.

Notes and References

These notes follow Bourbaki [Bou, Ch. 1 § 1.6].

References

[Bou] N. Bourbaki, General Topology, Springer-Verlag, 1989. MR1726779.

[Ru] W. Rudin, Real and complex analysis, Third edition, McGraw-Hill, 1987. MR0924157.

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