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Andrew McLennan
January 19, 1999
Economics 5113
Introduction to Mathematical Economics
Winter 1999
Lecture 5
Convergence, Continuity, Compactness
I. Introduction
A. Two of the most fundamental notions of the dierential calculus (recall that
Leibniz and Newton are credited with being the creators of this) `limit' and
`continuity,' were not successfully described with formal precision until the
nineteenth century.
1. After a period of evolution, the mathematics profession settled on two
principle frameworks, metric spaces and topological spaces, as the arenas
in which these concepts are dened and discussed.
2. Let X be a set. A function d : X X ! IR+ is a metric if:
a. for all x; y 2 X , d(x; y) = 0 if and only if x = y;
b. for all x; y 2 X , d(x; y) = d(y; x);
c. for all x; y; z 2 X , d(x; z) d(x; y) + d(y; z). A metric space is a
pair (X; d) where X is a set and d is a metric on X .
3. The most fundamental geometric objects in X are the open balls: for
x 2 X and > 0, let
B(x) = f x0 2 X : d(x; x0 ) < g:
II. Sequences and Convergence
A. Formally, a sequence in X is a function from f1; 2; 3; : : :g, or some similar index
set , to X .
1. Informally, a sequence is akin to a set, and we denote it by fxn g (or
sometimes x1 ; x2 ; : : : ) suggesting that we are concerned less with the
particular ordering and more with the general tendency as n gets large.
B. TRoughly speaking, the sequence fxn g converges to the limit x 2 X if it is
eventually inside each B(x).
1. In more detail, for any > 0 there is an integer N such that d(xn ; x) < for all n N .
2. The notation xn ! x indicates that the sequence fxng converges to x.
3. A sequence that converges to something is said to be convergent. Sequences that are not convergent are said to be divergent.
4. There are two ways that a sequence might be divergent:
a. The sequence might just bounce around without ever settling down.
This happens when, for some > 0, it is possible, for any integer
N , to nd n; m N , such that d(xn ; xm ) .
b. A Cauchy sequence is a sequence such that for any > 0 there is an
integer N such that d(xn ; xm ) < for all n; m N , i.e., a sequence
that does not bounce around too much.
c. If a Cauchy sequence fails to converge, the usual view is that this
is not because the sequence is in any sense ill behaved. Instead,
the fault lies with the given space. A metric space is said to be
complete if every Cauchy sequence has a limit.
C. Sequences in IR.
1. The basic reason that the real numbers are more useful than the rational
numbers is as follows:
Theorem: IR is complete.
Proof: Let ftng be a Cauchy sequence in IR. Let S be the set of real numbers S such
that tn > s for all but at most nitely many n. It is straightforward to use the denition
of a Cauchy sequence to show that S is nonempty and bounded above. Let s be the least
uppoer bound of S . We claim that tn ! s. In order to prove this by contradiction we
suppose that it is not the case, which means that there is some > 0 such that jtn , sj > for innitely many n. Suppose N is large enough that jtm , tnj < =2 for all n; m N ,
and choose M N such that jtM , sj > . Then jtn , tM j < =2 for all n M , so if
tM > s + =2, then s + =2 is an element of S , while if tM < s , =2, then s , =2 is
an upper bound. In either case we have contradicted the assumption that s is the least
uppoer bound of S .
2 It turns out that any metric space X can be \completed" by adding \the
missing points." The formal procedure for doing this is to let X be the
set of equivalence classes of Cauchy sequences in X , where two Cauchy
sequences are equivalent if the distance between them goes toi zero in the
limit. The details (dening the natural metric on X and showing that X
is complete) are lengthy, so I will not discuss them, but that is not to say
that it would be bad for you to think carefully about what is involved.
3 A sequence ftk )g in IR is increasing if tk tk+1 for all k, and it is strictly
increasing if tk < tk+1 for all k. The sequence ftk g is bounded above if
there is some t such that tk t for all k.
Theorem: An increasing sequence ftk )g that is bounded above has a limit.
Proof: We leave this as an exercise, since the ideas are very similar to those used to prove
the last result.
C. Sequences in IRn .
1. Unless we explicitly say otherwise, IRn is always endowed with the Eu3
clidean metric: d(x; y) = kx , yk.
2. Basically all you need to know about convergence in IRn is:
Theorem: A sequence fxk = (x1k ; : : : ; xnk )g in IRn converges if and only if each component
sequence fxik g is convergent.
Proof: If the sequence converges, say to x, then each component sequence converges
because, for all i = 1; : : : n, jxi , xik j kx , xk k. Conversely, suppose that each component
sequence fxik g converges to xi , and dene x to be the point (x1 ; : : : ; xn ). For any y 2 IRn
we have
kyk2 = jy1j2 + : : : + jyn j2 (jy1 j+ : : : + jynj)2 :
kxk , xk jx1k , x1 j + : : : + jxnk , x1 j ! 0:
III. Open and Closed Sets
A. A set C X is closed (or closed in X if some other containing space is possible)
if it contains all its limit points. That is, whenever fxk g is a sequence in C
that converges to some point x, x is an element of C .
1. Example: IRn+ is closed.
B. A neighborhood of a point x 2 X is any set S X that contains B(x) for
some > 0. A set U X is open (or open in X ) if it is a neighborhood of
each of its points, so tha t for each x 2 U there is > 0 for which B(x) U .
1. Example: IRn++ is open.
Theorem: For any x 2 X and > 0, B(x) is open.
Proof: For any y 2 B(x), the triangle inequality implies that B(,d(x;y))(y) B(x).
3. Exercise: Prove that, for any x and , f y 2 X : d(x; y) > g is open.
Theorem: A set U X is open if and only if its complement U c = X nU is closed.
Proof: Let C = U c . The assertion consists of two implications, the `if' and the `only if.'
Suppose that C is closed, and x 2 U . If, for each natural number k, B1=k (x) 6 U ,
we can choose a point in B1=k (x) \ C , thereby constructing a sequence fxk g in C that
converges to x. Since C is closed, this w ould imply that x 2 C , contrary to our assumption
that x 2 U . Therefore B1=k (x) U for large k, and since x was an arbitrary point of U ,
we have shown that U is open.
Suppose that U is open, and that fxk g is a sequence in C that converges to x. If
x 2 U , then B(x) U for some > 0, and xk 2 B(x) for large k since xk ! x, but this
contradicts the a ssumption that xk 2 C . Thus C is closed.
C. A topological space is a pair (T; ) in which T is a set and is a collection of
subsets of T , called the open sets of T , with the properties of the open subsets
of X given by:
(a) ; and X are open sets.
(b) The intersection of nitely many open sets is open.
(c) The union of an arbitrary collection of open sets is open.
Proof: This is all pretty obvious, so we will only mention that (b) is proved by noting
that if U1; : : : ; Up are open and B1 (x) U1; : : : ; Bp (x) Up, then
Bminf1;:::pg (x) U1 \ : : : \ Up:
1. Exercise: In a topological space a closed set is by denition a set whose
complement is open. The collection of all closed subsets of T has properties that are similar to, and immediately derivable from, (a){(c). What
are they?
2. The theory of topological spaces is much more complicated than the
theory of metric spaces, essentially because any number of things can
go wrong. For starters, in a topological space there can be sets that
contain all the limits of their convergent sequences, but are nonetheless
not closed.
3. In the future, at least, I will try to be careful to give denitions that are
valid for all topological spaces, not just metric spaces, and to be careful
to indicate what properties of metric spaces might not be true more
generally. However, we are basically going to just forget about general
topological spaces.
IV. Continuity
A. Let (X; dX ) and (Y; dY ) be metric spaces, and let f : X ! Y be a function.
1. The function f is continuous if f ,1 (V ) := f x 2 X : f (x) 2 V g is open
whenever V Y is open.
2. This denition makes sense, and is correct, for general topological spaces,
but has the unfortunate aspect of seeming strange on rst sight. Note,
however, that if f is continuous, x 2 X , and > 0, then f ,1 (B(f (x))) is
open, so that there exists some > 0 such that B (x) f ,1 (B(f (x))).
3. Exercise: The reverse implication also holds:
(8x 2 X )(8 > 0)(9 > 0) B (x) f ,1 (B(f (x)))
implies that f is continuous. Prove this.
B. The following test of continuity for functions between metric spaces is very
Theorem: f is continuous if and only if f (xn ) ! f (x) whenever fxn g is a sequence in X
that converges to x.
Proof: Suppose f is continuous. Let fxn g be a sequence in X with xn ! x. For
any > 0, f ,1 (B(f (x))) is open and contains x, so there is some > 0 such that
B (x) f ,1 (B (f (x))). Since xn ! x, for all suciently large n we have xn 2 B (x) and
f (xn ) 2 B(f (x)). Since was arbitrary, we have shown that f (xn ) ! f (x).
Suppose that f (xn ) ! f (x) whenever xn ! x. Let V Y be open. If f ,1 (V ) is
not open, it must contain a point x such that for each natural number n we can choose
xn 2 B1=n (x) n f ,1 (V ). Then xn ! x. But V is open, so there is > 0 such that
B(f (x)) V . For each n, f (xn ) 2= V , so that d(f (xn ); f (x)) > and f (xn ) 6! f (x),
contrary to assumption. Therefore f ,1 (V ) is open, and since V was arbitrary, we have
shown that f is continuous.
V. Compactness
A. The concept of a compact set, developed in the rst part of this century, is
now applied in most aspects of mathematics.
B. The denition is far from intuitive. Let (X; d) be a metric space.
1. An open cover of a set C X is a collection fUg2A of open sets that
\covers" C in the sense that C 0 [2A U.
2. A subcover is a subset of fUg2A that also covers C .
3 This set C is compact if every open cover has a nite subcover.
C. It is easy to explain why such sets might be attractive from the point of view
of optimization.
Theorem: If f : X ! IR is continuous, and C X is compact, then f attains its supremum
on C (that is, arg maxx2C f (x) is nonempty).
Proof: For each x 2 C let
Ux = fy 2 X : f (y) < f (x)g = f ,1 (,1; f (x))
Since f is continuous, Ux is open. If f does not attain its maximum on C , then every
point of C is in some Ux so fUxgx2C is a cover of C . Let Ux1 ; : : : ; Uxk be a nite subcover.
Reordering if we need to, suppose that f (x1 ) f (x2 ) f (xk ). Then Ux1 Ux2 Uxk , so we have
C Ux1 [ [ Uxk = Uxk :
Since xk 2 C , this would yield f (xk ) < f (xk ). This contradiction completes the proof.
D. What kinds of sets are compact?
1. Finite sets are obviously compact. Other examples are not so obvious.
2. A set C X is bounded if, for any x 2 C , there is M > 0 such that
d(x; x0 ) M for all x0 2 C .
Lemma: If C X is compact, it is bounded.
Proof: For any x 2 C , the sets
Uk = fx 2 C : d(x; y) < kg
(k = 1; 2; : : :)
constitute an open cover of C , and must have a nite subcover.
Lemma: If C X is compact, it is closed.
Proof: Suppose fxng is a sequence in C that converges to x. If x 2= C , the sets
Uk = y 2 C : d(x; y) > k1
(k = 1; 2; : : :)
constitute an open cover of C which cannot have a nite subcover if there is a sequence in
C converging to x.
3. A subsequence of a sequence fxng is a sequence xn1 ; xn2 ; : : : where n1 <
n2 < .
4. An accumulation point of a sequence fxn g is a point x with the property
that, for any > 0, there are innitely many n such that xn 2 B(x).
a. Exercise: Prove that a sequence fxng has a convergent subsequence if and only if it has an accumulation point.
5. A set C X is sequentially compact is every sequence in C has a convergent subsequence whose limit is in C .
Theorem: A compact set C X is sequentially compact.
Proof: By the exercise, it suces to show that a given sequence fxk g in C has an accumulation point. If not, for each x it is possible to nd an open set Ux that contains xk
for at most nitely many k. Since x 2 Ux, for each x, fUx: x 2 C g is an open cover of
C . But the union of the elements of a nite subcover contains C , and contains xk for at
most nitely many k. (The sum of nitely many nite numbers is nite.) Since this is
impossible, the proof is complete.
6. The converse is also true for metric spaces (we will prove this shortly)
but not for general topological spaces.
E. Compact Subsets of Euclidean Space.
Lemma: If fxk g is a bounded sequence in IRn, it has a convergent subsequence.
Proof: We claim that it suces to prove this in the case n = 1. If it is known to be true
for n = 1, then, in the general case, we can choose a subsequence such that the sequence of
rst components is convergent, take a further subsequence of this subsequence for which
the sequence of record components converges, and so on until, in the nth sequence, all
sequences of components are convergent. From an earlier exercise we know that this is
sucient for convergence.
So, let ftk g be a bounded sequence in IR, with lower bound a0 and upper bound b0 .
We construct a sequence [a1; b1 ]; [a2 ; b2 ]; : : : of closed intervals \inductively" by letting
[ai ; bi ] = ai,1 ; ai,1 +
if the right hand side contains tk for innitely many k. Otherwise we set
[ai ; bi ] =
; bi :
Arguing by induction, we can easily see that each [ai ; bi ] contains tk for innitely many k,
since this is true for [ai,1 ; bi,1 ].
Now choose k1 such that tk1 2 [a1 ; b1 ]. We construct fki g inductively by choosing
ki > ki,1 with tk 2 [ai; bi ]. Since tki ; tki+1 ; : : : are all contained in [ai; bi ], which has length
(b0 ,a0 ) , ft g is a Cauchy sequence, hence convergent since IR is complete.
1. All that remains is to bundle our ndings in a nice neat package.
Theorem: A set C IRn is compact if and only if it is closed and bounded.
Proof: We have already shown that a compact C is closed and bounded. Suppose that C
is closed and bounded. Consider that any sequence in C has a convergent subsequence, by
the limit, and the limit must be in C since C is closed. Thus C is sequentially compact,
so it is compact since IRn is a metric space.
V. Countable and Uncountable Sets
A. Two sets X and Y have the same cardinality if there is a bijection f : X ! Y .
B. A set is said to be countable if it has the same cardinality as the natural
numbers N := f1; 2; 3; : : :g.
1. Some authors say that a set is countable if it is either nite or has the
same cardinality as N, using the phrase \countably innite" to describe
a set with exactly the same cardinality as N.
2. A set that is neither nite nor countable is said to be uncountable .
C. Properties of Countable Sets.
1. Any innite subset of a countable set is countable. (Pf: If Y X =
fx1 ; x2 ; : : :g with Y innite, we map Y bijectively to N by letting f (xn )
be the number of i n such that xi 2 Y .)
2. The cartesian product of two countable sets is countable.
3. Examples:
a. The rational numbers are countable since they can be put in one
to one correspondence with a subset of N N.
b. The real numbers are not countable. This is proved by producing a
contradiction using Cantor's diagonalization procedure: if r1 ; r2 ; : : :
is an enumeration of the reals, create a number between 0 and 1 by
choosing a rst digit (after the decimal point) that is dierent from
the rst digit of r1 , a second digit dierent from the second digit
of r2, and so forth. All chosen digits can be dierent from 0 and 9,
to avoid ambiguity about numbers such as 1 that may be written
either 1:000 : : : of 0:999 : : :.) The number constructed in this way
is dierent from any number in the list.
D. Let fUg2A be a collection of open sets. Another collection of open sets
fV g2B is a renement of fUg2A if S2B V = S2A U and, for each
2 B, there is some 2 A such that V U .
Lemma: Any collection of fUg2A of open subsets of IRn has a countable renement.
Proof: Let B be the set of pairs (x; r) 2 IRn (0; 1) such that r and all components of
x are rational numbers, and V(x;r) := Br (x) U for some . Then B is countable, by
virtue of our remarks above, and we only need to show that (x;r)2B V(x;r) = 2A U.
Choose 2 A and y 2 U arbitrarily. Then B(y) U for some > 0. Choose a point x
with rational coordinates in B=2(y), and let r be a rational number between kx , yk and
, kx , yk. Then
y 2 Br (x) B(y) U:
Lemma: If fV g2B is a renement of fUg2A, and for some set S , nitely many
elements of fV g2B cover S , then there is a nite subcover of fUg2A that covers S .
Proof: If V1 ; : : : ; VK covers S , and for each k = 1; : : : ; K , VK Uk , then U1 ; : : : ; UK
covers S .
Theorem: If C IRn is sequentially compact, then it is compact.
Proof: Let fUg2A be an open cover of C . This cover has a countable renement, and
it suces to nd a nite subcover of the renement, which means that we only need to
prove the result when A is countable, so we may assume that A = N. That is, we assume
that the open cover is of the form fU1 ; U2; : : :g. If there is no nite subcover, then for each
k = 1; 2; : : : we may choose
xk 2 C n (U1 [ : : : [ Uk,1):
By assumption fxk g1
k=1 has a convergent subsequence whose limit x is in C . There is some
K such that x 2 UK , and there is some > 0 such that B(x) UK , so kxk , xk for
all but nitely many k, which is impossible if fxk g1
k=1 has a subsequence converging to x.
This contradiction completes the proof.