Metric and Hilbert Spaces

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

Last updated: 4 November 2014

Lecture 5: Topological spaces, interiors and closures

A topological space is a set $X$ with a collection $𝒯$ of subsets of $X$ such that

(a)	$\varnothing \in 𝒯$ and $X\in 𝒯,$
(b)	If $𝒮\subseteq 𝒯$ then $\left(\bigcup _{U\in 𝒮}U\right)\in 𝒯,$
(c)	If $n\in {\mathbb{Z}}_{>0}$ and $U_1,U_2,\dots ,U_n\in 𝒯$ then $$U_1\cap U_2\cap \cdots \cap U_n\in 𝒯\text{.}$$

Let $(X,𝒯)$ be a topological space.

An open set of $X$ is $U\in 𝒯\text{.}$

A closed set of $X$ is a subset $E\subseteq X$ such that $E^c$ is open.

Let $X$ be a set. The discrete topology on $X$ is $$𝒯=\left\{\text{subsets of}\hspace{0.17em}X\right\}$$ (the "power set of $X\text{").}$

Let $(X,d)$ be a metric space. Let $x\in X$ and let $\epsilon \in {ℝ}_{>0}\text{.}$ The ball of radius $\epsilon$ at $x$ is the set $$B_{\epsilon }\left(x\right)=\{p\in X\hspace{0.17em}|\hspace{0.17em}d(x,p)<\epsilon \}\text{.}$$ The metric space topology on $X$ is $$𝒯=\{U\subseteq X\hspace{0.17em}|\hspace{0.17em}U\hspace{0.17em}\text{is a union of open balls}\}\text{.}$$

Let $(X,𝒯)$ be a topological space. Let $Y\subseteq X\text{.}$ The subspace topology on $Y$ is $${𝒯}_Y=\{U\cap Y\hspace{0.17em}|\hspace{0.17em}U\in 𝒯\}\text{.}$$

Let $(X,{𝒯}_X)$ and $(Y,{𝒯}_Y)$ be topological spaces. The product of the sets $X$ and $Y$ is the set $$X\times Y=\left\{(x,y)\hspace{0.17em}\right|\hspace{0.17em}x\in X,y\in Y\}\text{.}$$ The product topology on $X\times Y$ is $${𝒯}_{X\times Y}=\{U\subseteq X\times Y\hspace{0.17em}|\hspace{0.17em}U\hspace{0.17em}\text{is a union of}\hspace{0.17em}A\times B\hspace{0.17em}\text{with}\hspace{0.17em}A\in {𝒯}_X\hspace{0.17em}\text{and}\hspace{0.17em}B\in {𝒯}_Y\}\text{.}$$

Examples of open and closed sets

Let $X=ℝ$ with the metric given by $$d(x,y)=\mid x-y\mid$$ and the metric space topology. Then $$\begin{array}{rcl}{\displaystyle (a,b)}& =& {\displaystyle \{x\in ℝ\hspace{0.17em}|\hspace{0.17em}a<x<b\}=B_{\frac{b-a}{2}}\left(\frac{a+b}{2}\right)\hspace{0.17em}\text{is open,}}\\ {\displaystyle [a,b]}& =& {\displaystyle \{x\in ℝ\hspace{0.17em}|\hspace{0.17em}a\le x\le b\}\hspace{0.17em}\text{is closed,}}\\ {\displaystyle [a,b)}& =& {\displaystyle \{x\in ℝ\hspace{0.17em}|\hspace{0.17em}a\le x<b\}\hspace{0.17em}\text{is not open and not closed}}\end{array}$$ (think of a door that is not open and not closed, i.e. ajar) and $$\varphi \hspace{0.17em}\text{and}\hspace{0.17em}ℝ\hspace{0.17em}\text{are both open}\hspace{0.17em}\text{and}\hspace{0.17em}\text{closed.}$$

Let $X$ be a topological space and let $x\in X\text{.}$

A neighbourhood of $X$ is a subset $N\subseteq X$ such that there exists an open set $U$ of $X$ with $x\in U$ and $U\subseteq N\text{.}$

The neighbourhood filter of $x$ is $$𝒩\left(x\right)=\left\{\text{neighbourhoods of}\hspace{0.17em}x\right\}\text{.}$$

Let $X$ be a topological space and let $E\subseteq X\text{.}$

The interior of $E$ is the subset $E^{\circ }$ of $X$ such that

(a)	$E^{\circ }$ is open and $E^{\circ }\subseteq E,$ and
(b)	if $U$ is open and $U\subseteq E$ then $U\subseteq E^{\circ }\text{.}$

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 $V\supseteq \bar{E}\text{.}$

In English: $E^{\circ }$ is the largest open set contained in $E$ and
$\bar{E}$ is the smallest closed set containing $E\text{.}$

An interior point of $E$ is a point $x\in X$ such that there exists a neighbourhood $N$ of $x$ such that $N\subseteq E\text{.}$

A close point of $E$ is a point $x\in X$ such that if $N$ is a neighbourhood of $x$ then $N\cap E\ne \varnothing \text{.}$

Let $X$ be a topological space. Let $E\subseteq X\text{.}$

(a)	The interior of $E$ is the set of interior points of $E\text{.}$
(b)	The closure of $E$ is the set of close points of $E\text{.}$

		Proof of (a).
	Let $I=\{x\in E\hspace{0.17em}|\hspace{0.17em}x\hspace{0.17em}\text{is an interior point of}\hspace{0.17em}E\}\text{.}$ To show: $I=E^{\circ }\text{.}$ To show: (aa) $I\subseteq E^{\circ }\text{.}$ (ab) $E^{\circ }\subseteq I\text{.}$ (aa) Let $x\in I\text{.}$ Then there exists a neighbourhood $N$ of $x$ with $N\subseteq E\text{.}$ So there exists an open set $U$ with $$x\in U\subseteq N\subseteq E\text{.}$$ Since $U\subseteq E$ and $U$ is open $U\subseteq E^{\circ }\text{.}$ So $x\in E^{\circ }\text{.}$ So $I\subseteq E^{\circ }\text{.}$ (ab) To show: If $x\in E^{\circ }$ then $x\in I\text{.}$ Assume $x\in E^{\circ }\text{.}$ Then $E^{\circ }$ is open and $E^{\circ }\subseteq E\text{.}$ So $x$ is an interior point of $E\text{.}$ So $x\in I\text{.}$ So $E^{\circ }\subseteq I\text{.}$ So $I=E^{\circ }\text{.}$ $\square$

Notes and References

These are a typed copy of Lecture 5 from a series of handwritten lecture notes for the class Metric and Hilbert Spaces given on August 5, 2014.

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