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  • 8/12/2019 MathAnalysis1 13 Gs

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    ARE211, Fall2013

    ANALYSIS1: THU, AUG 29, 2013 PRINTED: AUGUST 29, 2013 (LEC# 1)

    Contents

    1. Analysis 1

    1.1. References 2

    1.2. Countable vs Uncountable infinity 2

    1.3. Sequences 4

    1.4. Distance/Metrics 5

    1. Analysis

    Heavy emphasis on proofs in this section. Many students think that the proofs are the hardest

    part of Econ 201. Only way to master the art of proofs is to do a lot of them. The best topic

    in which to learn how to do proofs is analysis. A secondary goal in this topic is to get you to

    be comfortable jumping between different notations. When you read more formal journal articles,

    every author has his/her own notation system; need to learn how to jump back and forth between

    different notational systems.

    Youve just spent a few weeks in Math Camp on analysis. Why do more of it? Consensus is that

    a few weeks is too short a time to master the topic adequately: people talk about the fire-hose

    approach to teaching math; too much too quick; one can get a mechanical understanding of what

    the material means, but its very hard in this short a time to develop intuitions for the concepts.

    1

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    2 ANALYSIS1: THU, AUG 29, 2013 PRINTED: AUGUST 29, 2013 (LEC# 1)

    The goal of the few lectures Ill spend on analysis is to consolidate what youve learned, help you

    understand the material more intuitively.

    Having said that, students in different years have different responses to math camp. Some classes

    find it excruciating and totally inadequate preparation for graduate school; others find that it gives

    them all that they need. I dont know a priori which category this classes fall into, so Ill want

    feedback on this after a couple of lectures.

    1.1. References

    Chapter 12 in Simon-Blume

    Chapter 1-2 in De La Fuente

    Appendix F in MasCollel-Whinston-Green

    Chapter 1 and 2: Elementary Classical Analysis, by J. Marsden

    1.2. Countable vs Uncountable infinity

    This is a distinction thats fundamental in math but tricky to grasp at first. Examples are easy to

    understand, but its a very difficult distinction formally. Its not much more than 100 years since it

    was formally proved that there really is a difference between them. Three part distinction between

    sets:

    (1) finite sets:

    Example:{1, 2,...N}.

    (2) countably infinite sets.

    Example:the natural numbers, denoted N, are 1,2,3,4 ..., going on for ever.

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    (3) uncountably infinite sets.

    Example:the closed unit interval, denoted [0, 1].

    Distinction between finite and infinite sets:

    Definition:A set is infiniteif it can be placed in 1-1 correspondence with a strict subset of itself.

    Example:

    {1, 2} is a strict subset of {1, 2, 3}. You cant map each element of the first set to each

    element of the second.

    The evennatural numbers, {2, 4, 6,...} are a strict subset of the natural numbers N. You

    canmap each element of the first set to each element of the second.

    Distinction between countably and uncountably infinite sets:

    Informal Definition: A set is countably infinite if you can count its elements, i.e., you can identify

    the first element, the second element, etc

    Example: Its easy to count the natural numbers, but you cant count the unit interval; for the

    closedunit interval, theres a firstelement, but there isnt a second one.

    Mapping terminology for finite, countably and uncountably infinite sets: Its useful to compare the

    following kinds of mappings. The only distinction between them is the domain of the mapping.

    The language Im going to use here is conventional but by no means universal.

    (1) v : {1,...,N} {1}, i.e., (1,..., 1). This is more commonly refered to as an N-vectorof

    ones, Distinguishing feature of a vector is that the domain of the mapping is finite.

    (2) x: N {1}. This is a sequenceof ones, i.e., xn = 1, for all n. Distinguishing feature of a

    sequence is that the domain of the mapping is countably infinite.

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    (3) f : R+ {1}. This is a continuous function, mapping the non-negative real numbers to

    1, i.e., f() = 1. Distinguishing feature of a function is that the domain of the mapping is

    uncountably infinite.

    Until youre taught to think otherwise, youd probably think of only the latter as a real function.

    Actually, all three mappings satisfy the true definition of a function, i.e., each of them assigns a

    uniquepoint in the codomain to each point in its respective domain.

    1.3. Sequences

    A sequence is a mapping from the natural numbers to a set S, i.e., f : N S; f(n) is the nth

    element of the sequence. Typically, we suppress the functional notation: instead of writing the

    image ofn under f as f(n) we denote it by xn and write the sequence as {x1, x2,...,xn,...}, i.e.,

    f(n) = xn. Ill emphasize repeatedly that what distinguishes a sequencefrom any other kind of

    mapping is the nature of its domain: it is a countably infinite set.

    A collection {y1, y2,...,yn,...}is a subsequenceof another sequence {x1, x2,...,xn,...}if there exists

    a strictly increasingmapping : N Nsuch that for all n N,yn = x(n). Note that maps the

    domainof the subsequence into the domainof the original sequence. For example, consider the two

    sequences{xn} = {3, 6,..., 3n, ...} and {yn} = {6, 12,..., 6n, ...}. Obviously{yn} is a subsequence

    of{xn}. To prove this formally, we need to come up with an appropriate function. The one we

    need is (n) = 2n, i.e., for all n,yn = x2n: e.g., y1 = x2= 6, y2 = x4= 12.

    In general, you construct a subsequence by discarding some elements of the original sequence, but

    keeping an infinitenumber of the original elements and preserving their order.

    Some examples of sequences:

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    (1) {1, 2, 3, 4...}

    (2) {1, 1/2, 4, 1/8...}

    (3) {1, 1, 1, 1...}

    (4) {1, 1/2, 1/3, 1/4...}

    (5) there are no restrictions on what can be the rangeof a sequence. In particular, sequences

    arent necessarily maps from N into scalars. For example, consider we could map N into

    the set of continuous functions: {f1, f2,...fn...}, where fn =

    1 ifx 1/n

    nx if 1/n < x

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    6 ANALYSIS1: THU, AUG 29, 2013 PRINTED: AUGUST 29, 2013 (LEC# 1)

    the context offunctions, there is a vast variety of quite different notions. Mathematicians have an

    abstract notion of what is a legitimate measure of closeness.

    Definition:a metricor distance functionon a set S is a function d :S S R satisfying, for all

    x, y S:

    (1) d(x, y) =d(y, x) (symmetry)

    (2) d(x, y) 0 (nonnegativity)

    (3) d(x, y) = 0 iffx= y (two elements are a positive distance apart iff they are different from

    each other)

    (4) d(x, y) d(x, z) +d(z, y), for all z S (the triangle inequality)

    The last property of a metric is the one that has the most bite, and the one that really captures

    the spirit of distance: it states that the shortest distance between two points is a straight line.

    Examples of metrics

    (1) on R: d1

    (x, y) =|x y|.

    (2) on Rn: d2(x, y) =

    n

    i=1(xi yi)2

    (3) on Rn: d(x, y) = max{|xi yi|: i = 1,...,n}.

    (4) on Rn: ddiscrete(x, y) =

    1 ifx=y

    0 ifx= y

    (well call this the discretemetric).

    Lets check that the function ddiscrete(x, y) =

    1 ifx=y

    0 ifx= y

    is indeed a metric. It clearly satisfies

    the first three properties. What about the triangle inequality. First observe that ifx = y, then

    ddiscrete(x, y) = 0. Since ddiscrete(x, z) +ddiscrete(z, y) is necessarily nonnegative, then the triangle

    inequality holds. Now suppose thatx=y so that ddiscrete(x, y) = 1. In this case, for allz, either

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    z= x or z = y in which case either ddiscrete(x, z) or ddiscrete(z, y) is 1 so the triangle inequality is

    again satisfied.

    An example of a function that is nota metric is e(x, y) = min{|xi yi|: i = 1,...,n}. To see this,

    first note that it fails the third condition, since e((1, 1), (1, 2)) = 0, but (1, 1) = (1, 2). Moreover,

    e also fails the last condition: set x = (1, 1), y = (2, 2), z = (1, 2), e((1, 1), (2, 2)) = 1 but

    e((1, 1), (1, 2)) =e((1, 2), (2, 2)) = 0 so that e(x, y)> e(x, z) +e(z, y).

    WhenSis a space offunctions, condition (3) in the above definition of a metric is too restrictive.

    In economics, for example, we often encounter functions that arent equalto each other, but are said

    to be of distance zero from each other. In particular, it is often natural to say that the distance,

    d, between two functions is the integralof the absolute value of the difference between them, i.e.,

    d(f, g) =

    |f(x) g(x)|dx. But if f and g differ at only a finite (indeed countable) number of

    points, then in this sense, the difference between them will be zero.

    Condition (3) is inconsistent with this usage. To deal with this problem, we define a function

    to be a pseudo-metric if it satisfies all of the conditions above except condition (3). E.g., if S

    is the set of integrable functions mapping R to R, then the function : S S R defined by

    (f, g) =|

    f(x)dx

    g(x)dx|is a pseudo-metric but not a metric. (Note that this distance notion

    is quite different from the d mentioned in the preceding paragraph!)

    (1) To see that is nota metric, consider the function f1 defined above as the first element

    of the sequence {fn}on p. 5 as example (5). Because the function is so symmetric, clearly

    f1dx =

    f1dx = 0, so that (f1, f1) = 0, but these functions are not equal to each

    other.

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    (2) On the other hand, to see that is a pseudo-metric, observe that its obviously symmetric

    and non-negative. The only thing remaining to check is that it satisfies the triangle inequal-

    ity. To prove this, we could use the following Lemma, but wont go thru it in class

    Lemma: for any x, y R,|x| + |y| |x+y|.

    Proof of the Lemma: Its obvious that ifx and y both have the same sign then |x|+ |y|=|x+y|. Now suppose without loss of generality (w.l.o.g.) that x 0 > y. In this case,

    |x| + |y| = x+ (y) > x > |x (y)| = |x+y|

    We can now check that satisfies the triangle inequality. For any functions f ,g, h S,

    (f, h) +(h, g) =

    f dx

    hdx

    +

    hdx

    gdx

    which from the lemma is

    f dx

    hdx+

    hdx

    gdx

    =

    f dx

    gdx

    = (f, g)