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 x∈X.

First definition: A function x⇀: ℤ>0 ⟶ X n ⟼ xn converges to x if x⇀ satisfies: if ε∈ℝ>0 then there exists N∈ℤ>0 such that
if n∈ℤ>0 and n>N then d(xn,x)<ε.
Write limn→∞xn=x if x⇀ converges to x.

Second definition: A function x⇀: ℤ>0 ⟶ X n ⟼ xn converges to x if limn→∞(xn,x)=0.

HW: Let (X,d) be a metric space and let x∈X. Let x⇀:ℤ>0→X be a function. Show that x⇀ satisfies (*) if and only if limn→∞d(xn,x)=0.

Uniqueness of limits

HW: Let x⇀:ℤ>0→X and let x,y∈X. Show that if limn→∞xn=x and limn→∞xn=y then x=y.

Equivalent metrics

Let X be a set and let d1:X×X→ℝ>0 and d2:X×X→ℝ>0 be metrics on X. The metrics d1 and d2 are equivalent if d1 and d2 satisfy: If x⇀: ℤ>0 ⟶ X n ⟼ xn and x∈X then limn→∞d1(xn,x)=0 if and only if limn→∞d2(xn,x)=0. (*)

HW: Show that if d1 and d2 satisfy if x,y∈X then there exist c1∈ℝ>0 and c2∈ℝ>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,y∈X" 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 A⊆X and let A‾ be the closure of A.

HW: Show that A‾= { x∈X | there exists  a⇀:ℤ>0→ A,n↦an such that  limn→∞an=x } .

Proof.

Let R= { x∈X | there exists  a⇀:ℤ>0→ A,n↦an such that  limn→∞an=x } .
To show:
(a) R⊆A‾.
(b) A‾⊆R.
(a) To show: If x∈R then x∈A‾.
Assume x∈R.
To show: x∈A‾.
We know: There exists a⇀:ℤ>0→A,n↦an with limn→∞an=x.
To show: x is a close point of A.
To show: If V is a neighbourhood of x then there exists a∈A such that a∈V.
Assume V is a neighbourhood of x.
Then there exists ε∈ℝ>0 such that B(x,ε)⊆V.
To show: There exists a∈A such that a∈V.
Let N∈ℤ>0 such that if n∈ℤ>0 and n≥N then d(an,x)<ε.
Let a=aN.
Then d(a,X)=d(aN,x)<ε.
So a∈B(x,ε)⊆V.
So x is a close point of A.
So R⊆A‾.
(b) Let x∈A‾.
To show: x∈R.
To show: There exists a⇀:ℤ>0→A,n↦an with limn→∞an=x.
We know: x is a close point of A.
Let n∈ℤ>0 and let an∈A such that an∈B(x,1n).
Let a⇀:ℤ>0→A be given by a⇀(n)=an.
To show: limn→∞an=x.
To show: If ε∈ℝ>0 then there exists N∈ℤ>0 such that if n∈ℤ>0 and n>N then d(an,x)<ε.
Assume ε∈ℝ>0.
Let N∈ℤ>0 be minimal such that N>1ε.
To show: If n∈ℤ>0 and n>N then d(an,x)<ε.
Assume n∈ℤ>0 and n>N.
To show: d(an,x)<ε.
Since an∈B(x,1n), d(an,x)<1n<1N<ε. So limn→∞an=x.
So x∈R.
So A‾⊆R.

□

Some definitions

Let X be a topological space. Let A⊆X.

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‾)∘=∅.

ℚ is dense in ℝ.

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

The boundary of ℚ in ℝ is ℝ.

The boundary of (0,1] in ℝ is {0,1} since (0,1]‾∩ ([0,1]c)‾ = [0,1]∩ (-∞,0]∪(1,∞)‾ = [0,1]∩ ((-∞,0]∪[1,∞)) = {0,1}.

ℤ>0 and ℤ are nowhere dense in ℝ.

ℝ is nowhere dense in ℝ2.

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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