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) 𝒯 and X𝒯,
(b) If 𝒮𝒯 then (U𝒮U)𝒯,
(c) If n>0 and U1,U2,,Un𝒯 then U1U2 Un𝒯.

Let (X,𝒯) be a topological space.

An open set of X is U𝒯.

A closed set of X is a subset EX such that Ec is open.

Let X be a set. The discrete topology on X is 𝒯={subsets ofX} (the "power set of X").

Let (X,d) be a metric space. Let xX and let ε>0. The ball of radius ε at x is the set Bε(x)= { pX| d(x,p)<ε } . The metric space topology on X is 𝒯= { UX|U is a union of open balls } .

Let (X,𝒯) be a topological space. Let YX. The subspace topology on Y is 𝒯Y={UY|U𝒯}.

Let (X,𝒯X) and (Y,𝒯Y) be topological spaces. The product of the sets X and Y is the set X×Y= { (x,y)| xX,yY } . The product topology on X×Y is 𝒯X×Y= { UX×Y| Uis a union ofA×B withA𝒯X andB𝒯Y } .

Examples of open and closed sets

Let X= with the metric given by d(x,y)=x-y and the metric space topology. Then (a,b) = {x|a<x<b}= Bb-a2(a+b2) is open, [a,b] = {x|axb} is closed, [a,b) = {x|ax<b} is not open and not closed (think of a door that is not open and not closed, i.e. ajar) and ϕandare both openand closed.

Let X be a topological space and let xX.

A neighbourhood of X is a subset NX such that there exists an open set U of X with xU and UN.

The neighbourhood filter of x is 𝒩(x)={neighbourhoods ofx}.

Let X be a topological space and let EX.

The interior of E is the subset E of X such that

(a) E is open and EE, and
(b) if U is open and UE then UE.

The closure of E is the subset E of X such that

(a) E is closed and EE, and
(b) if V is closed and VE then VE.

In English: E is the largest open set contained in E and
E is the smallest closed set containing E.

An interior point of E is a point xX such that there exists a neighbourhood N of x such that NE.

A close point of E is a point xX such that if N is a neighbourhood of x then NE.

Let X be a topological space. Let EX.

(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 (a).

Let I={xE|xis an interior point ofE}.
To show: I=E.
To show:
(aa) IE.
(ab) EI.
(aa) Let xI.
Then there exists a neighbourhood N of x with NE.
So there exists an open set U with xUNE. Since UE and U is open UE.
So xE.
So IE.
(ab) To show: If xE then xI.
Assume xE.
Then E is open and EE.
So x is an interior point of E.
So xI.
So EI.
So I=E.

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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