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 9: Convergences, equivalent metrics, closure

Convergence

Let (X,d) be a metric space and let xX.

First definition: A function x: >0 X n xn converges to x if x satisfies: if ε>0 then there exists N>0 such that
if n>0 and n>N then d(xn,x)<ε.
Write limnxn=x if x converges to x.

Second definition: A function x: >0 X n xn converges to x if limn(xn,x)=0.

HW: Let (X,d) be a metric space and let xX. Let x:>0X be a function. Show that x satisfies (*) if and only if limnd(xn,x)=0.

Uniqueness of limits

HW: Let x:>0X and let x,yX. Show that if limnxn=x and limnxn=y then x=y.

Equivalent metrics

Let X be a set and let d1:X×X>0 and d2:X×X>0 be metrics on X. The metrics d1 and d2 are equivalent if d1 and d2 satisfy: If x: >0 X n xn and xX then limnd1(xn,x)=0 if and only if limnd2(xn,x)=0. (*)

HW: Show that if d1 and d2 satisfy if x,yX then there exist c1>0 and c2>0 such that d1(x,y)c1d2(x,y) and d2(x,y)c2d1(x,y) then d1 and d2 satisfy (*).

HW:

(a) Is the "if x,yX" in the right place of should it be after "such that".
(b) Why isn't this statement if and only if?

Convergence and closure

Let (X,d) be a metric space.

Then X is a metric space with the metric space topology.

Let AX and let A be the closure of A.

HW: Show that A= { xX|there exists a:>0 A,nansuch that limnan=x } .

Proof.

Let R= { xX|there exists a:>0 A,nansuch that limnan=x } .
To show:
(a) RA.
(b) AR.
(a) To show: If xR then xA.
Assume xR.
To show: xA.
We know: There exists a:>0A,nan with limnan=x.
To show: x is a close point of A.
To show: If V is a neighbourhood of x then there exists aA such that aV.
Assume V is a neighbourhood of x.
Then there exists ε>0 such that B(x,ε)V.
To show: There exists aA such that aV.
Let N>0 such that if n>0 and nN then d(an,x)<ε.
Let a=aN.
Then d(a,X)=d(aN,x)<ε.
So aB(x,ε)V.
So x is a close point of A.
So RA.
(b) Let xA.
To show: xR.
To show: There exists a:>0A,nan with limnan=x.
We know: x is a close point of A.
Let n>0 and let anA such that anB(x,1n).
Let a:>0A be given by a(n)=an.
To show: limnan=x.
To show: If ε>0 then there exists N>0 such that if n>0 and n>N then d(an,x)<ε.
Assume ε>0.
Let N>0 be minimal such that N>1ε.
To show: If n>0 and n>N then d(an,x)<ε.
Assume n>0 and n>N.
To show: d(an,x)<ε.
Since anB(x,1n), d(an,x)<1n<1N<ε. So limnan=x.
So xR.
So AR.

Some definitions

Let X be a topological space. Let AX.

The boundary of A is A=A(Ac).
The set A is dense in X if A=X.
The set A is nowhere dense in X if (A)=.

is dense in .

(0,1] is dense in [0,1].

The boundary of in is .

The boundary of (0,1] in is {0,1} since (0,1] ([0,1]c) = [0,1] (-,0](1,) = [0,1] ((-,0][1,)) = {0,1}.

>0 and are nowhere dense in .

is nowhere dense in 2.

The Cantor set is nowhere dense in [0,1]. The Cantor set is closed.

Notes and References

These are a typed copy of Lecture 9 from a series of handwritten lecture notes for the class Metric and Hilbert Spaces given on August 12, 2014.

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