MAS331: Metric spaces

Lecturer: John GreenleesG16 Hicks Building

j.greenlees@sheffield.ac.uk

http://greenlees.staff.shef.ac.uk/mas331/

September 30, 2011

Introduction.

This course is entitled “Metric spaces”. You are probably wondering what such a thingshould be. We’ll see a formal definition a little later in the course, but here’s a roughdefinition: it is a space in which we can discuss convergence and do analysis, just as wasdone for the real numbers R in MAS207 Continuity and Integration. Sequences can beof many types: we can consider sequences of numbers, sequences of points in a plane,sequences of functions and so on. Whenever we talk about a sequence approaching alimit, then implicitly we are assuming that we have some notion of distance, so that wecan talk about two things being close. Roughly speaking, a metric space is a set (ofnumbers, points, functions etc.) on which we have a way to measure distance.

We’ll start the course by thinking of lots of examples of iteration and convergence.One aim of the course is to find conditions under which we can guarantee that iterativeprocesses will converge.

1 Examples of iterative processes.

Iteration is the process where an initial guess at a problem can be refined by repeatedlysubstituting it into the problem and getting better and better guesses. The process is atthe heart of many numerical, and practical, techniques.

Example 1.1. Consider the function

f(x) =1

2

✓x +

2

x

◆.

Let’s iterate f starting with the initial value x = 1. So we calculate f(1), f(f(1)),f(f(f(1))), and so on. If you are happier with sequences here is what we have done: Wehave simply constructed a sequence (x

n

) where x0 = 1 and

xn+1 =

1

2

✓x

n

+2

xn

◆.

Does it converge? What is the limit? Call the limit x, assuming it exists. Then we oughtto get a relation

x =1

2

✓x +

2

x

◆

1

which, if you solve the equation, suggests x =p

2. Here are the first few terms:

x1 = 1.50000000000000000000000

x2 = 1.41666666666666666666666

x3 = 1.41421568627450980392156

x4 = 1.41421356237468991062629

x5 = 1.41421356237309504880168

If you haven’t noticed, x5 is a correct numerical value forp

2 to 23 decimal places. Wehave used iteration to find a numerical solution of f(x) = x.

Note one last thing. All of our xn

’s are rational numbers, but their limitp

2 is nota rational number. This sort of behaviour will be important in later sections.

Example 1.2 (The Newton-Raphson Method). In the last example we used iteration tosolve the equation x = f(x). Now we will use iteration to solve f(x) = 0, i.e. to find theroots of f .

Start with a number x0 as an approximate root, and try to improve it by thefollowing idea: draw the tangent line to y = f(x) at x0, then see where the tangentcrosses the x-axis and call that point x1. If you illustrate the situation in a sketch you cansee that x1 ‘should’ be closer to the root than x0. Repeating the process in order to getcloser to the root, and doing the necessary algebra, we are led to consider the sequencegiven by

xn+1 = x

n

� f(xn

)

f 0(xn

).

Warning : We do not know when this works, if at all!

Here’s an example. Let’s try to solve the cubic equation x3 � 2x + 1 = 0. We haveto choose a value of x0 and then apply the formula

xn+1 = x

n

� x3n

� 2xn

+ 1

3x2n

� 2.

Starting with x0 = 0 we get, to three decimal places, the sequence

x0 = 0.000, x1 = 0.500, x2 = 0.600, x3 = 0.617, x4 = 0.618, x5 = 0.618,

after which the sequence is unchanging to three decimal places. By choosing di↵erentstarting values x0 we can find good approximations of the other two roots of x3�2x+1 = 0.

Example 1.3. Suppose that we want to solve the (very simple) simultaneous equations

x2 � y2 = 1,

2y � x = 0.

We can rearrange the equations as x =p

1 + y2, y = x/2, and so just as in Example 1.1we set up an iteration with

xn+1 =

p1 + y2

n

,

yn+1 = x

n

/2.

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Starting with (x1, y1) = (1, 0.5) we find that the first few terms in the iteration are, to 3decimal places,

(x1, y1) = (1.000, 0.500),

(x2, y2) = (1.118, 0.500),

(x3, y3) = (1.118, 0.559),

(x4, y4) = (1.146, 0.559),

and that by the term (x10, y10) these values are unchanging at (1.155, 0.577).

The original equations can be solved explicitly, and x = 1.115, y = 0.577 agreewith one of the solutions to three decimal places. We have solved a pair of simultaneousequations by iterating pairs of numbers.

Example 1.4. Let’s look for a continuous function x(t) defined on the open interval(�1, 1) = {t | � 1 < t < 1}, satisfying the equation

x(t) = 1�tZ

0

x(u)2 du.

First we translate this into a search for the fixed point of an iteration. Thus we define anintergral operator F which takes a function x to a new function F (x) defined by

F (x)(t) = 1�tZ

0

x(u)2 du.

As before, this lets us define a sequence (now of functions) x0, x1, ...., where xn+1 = F (x

n

)

We can start by making a guess x0 equal to the constant function 0 say (so thatx0(t) = 0 for all values of t), and defining x1 to be the result. Then we can iterate thisprocedure. So

x1(t) = 1�tZ

0

02 du = 1,

and x1(t) is the constant function 1. Next, set

x2(t) = 1�tZ

0

x1(u)2 du = 1�tZ

0

1 du = 1� t.

Repeating,

x3(t) = 1�tZ

0

x2(u)2 du = 1�tZ

0

(1� u)2 du = 1�tZ

0

(1� 2u + u2)du = 1� t + t2 � t3/3.

If you continue in this way, you should see that x4(t) begins 1 � t + t2 � t3 + · · · ,and x5(t) has even more terms of this form. This suggests (correctly!) that the limit ofthe sequence (x

n

) is

1� t + t2 � t3 + t4 � · · · =1

1 + t.

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One easily verifies that x(t) = 11+t

is a solution to the original integral equation. We havesolved an integral equation by iterating sequences of functions!

Again, a last note: all of the functions xn

were polynomials, but their limit 11+t

isnot a polynomial. This is similar to the earlier situation where a limit of rational numberswas not rational, and is an issue that we’ll come back to in later sections.

Remark 1.5. What have we done so far? We have shown how to iterate sequences ofnumbers, pairs of numbers, and functions, in the hope that the resulting number, orpair of numbers, or function will solve our original problem. You know from MAS207Continuity and Integration what it means for a sequence of numbers to converge, but notwhat it means for a sequence of pairs of numbers or of functions to converge. A metricspace is a collection of things (e.g. numbers, pairs of numbers, or functions) where wecan talk about convergence in a concrete and rigorous way. The aim of the course is todefine metric spaces and explore their properties; we will prove results that tell us wheniteration can be used to solve problems such as the ones we have seen above.

Suprema and Infima in R.

We end the first chapter of the course with a reminder on suprema and infima. Theresults we will discuss were covered in MAS207 Continuity and Integration, and so wearen’t going to prove anything here. The object is to make sure that you can work withsuprema and infima.

Definition 1.6. Let E ⇢ R be a set of real numbers.

1. A number B 2 R is an upper bound for E if x 6 B for all x 2 E.

2. E is bounded above if it has an upper bound.

3. A number S 2 R is a supremum, or least upper bound, if

• S is an upper bound for E;

• any other upper bound for E is greater than or equal to S.

Example 1.7. Let’s look at an example. Take E to be the open interval (0, 1). When isB an upper bound for E? At first glance it looks like B is an upper bound for E if andonly if B > 1. Let’s check this carefully.

• Let B > 1. Then t < 1 6 B for any t 2 (0, 1), so that t 6 B as required. Thus theB 6 1 are upper bounds for E.

• Let B be an upper bound for E. Then we want to say that 1 6 B. Suppose insteadthat B < 1, so that B < 1

2(1 + B) < 1. This means that 12(1 + B) is in E but is

greater than B, contradicting the fact that B is an upper bound for E. Thus 1 6 Bas required.

We have shown that the upper bounds for E are the numbers B > 1. The supremum ofE is the least of these upper bounds, which means that the supremum of E is 1.

Definition 1.8. Let E ⇢ R be a set of real numbers.

1. A number B 2 R is a lower bound for E if x > B for all x 2 E.

MAS331 4 Autumn 2011-12

2. E is bounded below if it has a lower bound.

3. A number I 2 R is an infimum, or greatest lower bound, if

• I is a lower bound for E;

• any other lower bound for E is less than or equal to I.

The following facts should be familiar from MAS207. If you don’t remember themfrom that course then make sure you understand them now.

1. A non-empty set E ⇢ R which is bounded above has a supremum.

2. A non-empty set E ⇢ R which is bounded below has an infimum.

3. A supremum of E, if it exists, is unique.

4. An infimum of E, if it exists, is unique.

Definition 1.9.

• The supremum of a set E that is bounded above is denoted by sup E.

• The infimum of a set E that is bounded below is denoted by inf E.

Here are some facts about suprema. These are vital for the rest of the course. Makesure you understand them!

1. If A is finite, then it is bounded, and sup A = max A is just the largest element ofA.

2. Let {a1, . . . , an

} and {b1, . . . , bn

} be sets of numbers. Then sup{ai

}, sup{bi

}, sup{ai

+bi

} all exist andsup{a

i

+ bi

} 6 sup{ai

} + sup{bi

}.

3. If f, g : R ! R are functions, bounded above, then

sup{f(x) + g(x) | x 2 R} 6 sup{f(x) | x 2 R} + sup{g(x) | x 2 R}.

Example 1.10. Let a1 = 1, a2 = 2, . . . an

= n, and let b1 = n, b2 = n � 1, . . . bn

= 1.Then a

i

+ bi

= n + 1 for all i, and so

sup{ai

+ bi

} = n + 1 < n + n = sup{ai

} + sup{bi

}.

Thus the inequality above can be strict.

2 Metric spaces

To define notions such as convergence, we need some idea of what it means for points tobe ‘close’ to one another. It is therefore necessary to consider spaces with some sort of‘distance’ on them. This leads quickly to the definition of a metric space.

We will isolate three fundamental properties of ‘distance’, and base all our deduc-tions on these three properties alone. With so few properties, this makes our proofs rather

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