Download Introductory Analysis 1 The real numbers

MSc in Mathematics and Finance
Introductory Analysis
Mark H.A.Davis
1
1.1
The real numbers
Convergence
In these notes R denotes the set of real numbers (or the ‘real line’). For a1 , a2 ∈
R we have −|ai | ≤ ai ≤ |ai |, i = 1, 2. Adding these two inequalities we have
−(|a1 | + |a2 |) ≤ a1 + a2 ≤ (|a1 | + |a2 |)
and hence
|a1 + a2 | ≤ |a1 | + |a2 |.
This is the triangle inequality, which will be used constantly in what follows.
Let A be a set of real numbers. A is bounded if there is a number n such
that −n < y < n for all y ∈ A. A number x is the supremum of A if (i)
y ∈ A ⇒ y ≤ x and (ii) if x0 is another number such that y ∈ A ⇒ y ≤ x0 then
x0 ≥ x. It is axiomatic in the construction of the real line that every bounded
set has a supremum (there are no ’gaps’).
A sequence {xn } converges to x if for every > 0 there is a number N such
that n > N ⇒ |xn − x| < . A sequence is monotone increasing [decreasing] if
m > n ⇒ xm ≥ xn [xm ≤ xn ].
Proposition 1 A bounded monotone increasing sequence converges to its supremum
The limit superior of a sequence is defined as follows
lim sup xn = lim sup xk .
n→∞ k>n
n
If we define yn = supk>n xk then yn is a monotone decreasing sequence. Hence
the lim sup always exists. The inferior limit is defined as
lim inf xn = lim inf xk .
n
n→∞ k>n
Equivalently, lim inf xn = − lim sup(−xn ).
Proposition 2 A sequence {xn } converges to x iff
x = lim sup xn = lim inf xn .
n
n
1
Proof: It is clear that x∗ := lim sup xn ≥ lim inf xn =: x∗ . If xn → x then yn
defined as above converges to x, so that x∗ = x, and similarly x∗ = x. From the
definitions, given > 0 there exists N such that xn < x∗ + and xn > x∗ − for
all n > N . Hence the sequence converges if x∗ = x∗ . If x∗ < x∗ there cannot
be a limit point x (Check the candidates x < x∗ , x ∈ [x∗ , x∗ ], x > x∗ .)
♦
A subsequence of a sequence {xn } is a sequence {xn1 , xn2 , . . .} where n1 <
n2 < . . ..
Theorem 1 Any sequence has a monotone subsequence.
Proof: Call xm a peak of a sequence {xn } if xm ≥ supk>m xk . If {xn } has
infinitely many peaks xm1 , xm2 , . . . then these form a monotone decreasing subsequence. If {xn } has finitely many or no peaks, set n1 = m + 1, where m is
the index of the last peak (or 0 if there are none). Then xn1 is not a peak, so
there exists an index n2 such that xn2 ≥ xn1 . Now xn2 is not a peak, so there
exists n3 such that ... ♦
This result leads to the famous Bolzano-Weierstrass Theorem
Theorem 2 Every bounded sequence has a convergent subsequence
Proof: By Theorem 1, a bounded sequence {xn } has a bounded monotone
subsequence, and bounded monotone sequences are convergent.
♦
A sequence {xn } is a Cauchy sequence if for every > 0 there exists M such
that n, m > M ⇒ |xn − xm | < .
Theorem 3
1. Every Cauchy sequence is bounded.
2. A sequence {xn } is convergent if and only if it is a Cauchy sequence.
Proof: (1) If {xn } is Cauchy then there exists M such that |xn − xm | < 1 for
n, m ≥ M , so that in particular |xn − xM | < 1 for all n > M . It follows that
for any k, |xk | ≤ max{|x1 |, |x2 |, . . . , |xM | + 1}.
(2) If xn → x then for any > 0 there is an N such that |xn −x| < /2 for n > N .
Thus for n, m > N we have |xn − xm | ≤ |xn − x| + |xm − x| < . Conversely if
{xn } is Cauchy we know from (1) that it is bounded and hence from Theorem
2 that there is a convergent subsequence xni → x∗ . We claim that the whole
sequence converges to x∗ . Indeed, given > 0 take M such that |xn − xm | < /2
for n, m > M and an index i such that ni > M and |xnk − x∗ | < /2 for k ≥ i.
Then for m > ni , |xm − x∗ | ≤ |xm − xni | + |xni − x∗ | ≤ /2 + /2 = . ♦
1.2
Open, Closed and Compact Sets
An open interval in R is a set (a, b) = {x : a < x < b} while a closed interval is
[a, b] = {x : a ≤ x ≤ b}. An open set is a subset A ⊂ R with the property that
x ∈ A ⇒ B (x) ⊂ A for some > 0, where B (x) = {y : |y − x| < } (the ‘-ball’
around x). A set B is closed if its complement B c = {x ∈ R : x ∈
/ B} is open.
Theorem 4 A set A is open if and only if it is a disjoint union of countably
many open intervals.
2
We do not prove this here; see Bartle & Sherbert1 , p315. A set is countable if
it is in one-to-one correspondence with the natural numbers {1, 2, 3, . . .}.
A point x is a cluster point of a set A if B (x) ∩ A 6= ∅ for any > 0. This
gives us the following characterization of closed sets. The proof is an exercise
for ther reader!
Theorem 5 A set A ⊂ R is closed if and only if it contains all its cluster points
We will also need the following result, known as the ‘nested intervals property’:
Proposition 3 Let In = [an , bn ] be a nested sequence of closed intervals (i.e.
In ⊂ Im when n > m). Then there exists a number x such that x ∈ In for all
n.
Proof: The nested property implies that the sequence an and [bn ] are increasing and decreasing respectively. Let a∗ = sup an and b∗ = inf bn . Then a∗ ≤ b∗
and any point x ∈ [a∗ , b∗ ] satisfies x ≥ an , x ≤ bn for all n, i.e. x ∈ ∩n In . ♦
An open cover of a set A ⊂ R is a collection of sets G = {Gα } such that
[
A⊂
Gα .
α
A subcover is a smaller collection of sets in G that also covers A. A set K ⊂ R is
compact if every open cover of K has a finite subcover (i.e. a subcover consisting
of a finite number of sets). We shall see several properties of compact sets in
a more general setting later, but in the case of compact sets in R we have the
following important characterization, known as the Heine-Borel theorem.
Theorem 6 A ⊂ R is compact if and only if it is closed and bounded.
Proof: Suppose K is compact. Define open sets Bm = (−m, m). Then
k
K ⊂ ∪∞
n=1 Bn , so it must be the case that K ⊂ ∪n=1 Bn for some k. Thus
K ⊂ [−k, k], i.e. K is bounded. To show K is closed we show that K c is open.
Take y ∈ K c and let Cn = {x ∈ R : |x − y| > 1/n}. Then R \ {y} = ∪n Cn and
since y ∈
/ K the sets Cn cover K. Hence K ⊂ ∪m
1 Cn for some m, which implies
that B1/m (y) ⊂ K c , so that K c is open. We have shown that compact sets are
closed and bounded.
For the converse we argue by contradiction, supposing that K is closed and
bounded and that G = {Gα } is an open cover of K with no finite sub-cover.
Suppose a is a number such that K ⊂ [−a/2, a/2] and define I10 = [−a/2, 0] and
I100 = [0, a/2]. At least one of K ∩ I10 , K ∩ I100 must have these properties:
1. it is non-empty;
2. it is not contained in any finite sub-collection of G.
Denote by I2 whichever of I10 , I100 has these properties. Now bisect I2 into sets
I20 , I200 and denote by I3 whichever of these sets has the above property. In this
way we construct a nested sequence I1 ⊃ I2 ⊃ I3 ⊃ · · · which, by Proposition
3 must have a common point z. We claim that z is a cluster point of K. Take
1 R.G.
Bartle and D.R Sherbert: Introduction to real analysis, 3rd edn, Wiley 2000
3
> 0. Now z ∈ In and the length of In is 2−n , so In ⊂ B (z) whenever
n ≥ n = − log2 . By construction K ∩ In 6= ∅, so B (z) ∩ K 6= ∅, showing
that z is a cluster point. Since K is closed, z ∈ K by Theorem 5. Now G covers
K, so z ∈ Gα0 for some α0 . Since Gα0 is open, B (x) ⊂ Gα0 for some > 0.
This implies as above that In ⊂ Gα0 . But this contradicts property 2 in the
construction of Gα0 . Thus no such collection G can exist: every cover must have
a finite sub-cover. K is compact. ♦
2
Topological and Metric Spaces
2.1
Topological spaces
A topological space is a set X together with a collection T of subsets called open
sets with the following properties:
1. X ∈ T and ∅ ∈ T ;
2. A, B ⊂ T ⇒ A ∩ B ∈ T ;
3. If {Aj } ∈ T for j ∈ J, where J is an arbitrary index set, then ∪j Aj ∈ T .
If (2) is satisfied then evidently ∩n1 Ak ∈ T if A1 , . . . An ∈ T . Thus (1) and (2)
say that T is closed under unions and finite intersections. A subset A is closed
if its complement Ac = X \ A is open.
It is possible to have more than one topology on the same set X. The discrete
topology is the topology in which all sets are open; the trivial topology is the
one where T = {X, ∅}. If T1 ⊂ T2 we say that T1 is coarser than T2 , or T2 is
finer than T1 . The topologies are equivalent if T1 = T2 .
The standard example of a topological space is the set Rn = {(x1 , x2 , . . . , xn ) :
xi ∈ R, i = 1, . . . , n} with the usual topology: a set A is open if x ∈ A ⇒ B (x) ⊂
A for some > 0, where B (x) = {y : |x − y| < } with the Euclidean distance
v
u n
uX
|x − y| = t (xi − yi )2 .
1
For x ∈ X, A ⊂ X we say that A is a neighbourhood of x if x ∈ B ⊂ A
for some open set B. A sequence {xn } in X converges to x ∈ X if for any
neighbourhood A of x there is a number N such that xn ∈ A for all n ≥ N .
(‘xn is eventually in every neighbourhood of x’.) A point z is a cluster point of
a set A if B ∩ A 6= ∅ whenever B is a neighbourhood of z. The closure of a set
A, denoted A, is the set of all cluster points of A.
If X, Y are sets, a function f is a rule that associates one element f (x) ∈ Y
with each x ∈ X. X and Y are the domain and range of f respectively. For
A ⊂ X, B ⊂ Y we denote f (A) = {y ∈ Y : y = f (x) for some x ∈ A} and
f −1 (B) = {x ∈ A : f (x) ∈ B}. The function f is continuous if f −1 (B) ∈ S for
all B ∈ T .
Theorem 7 If f : (X, S) → (Y, T ) is continuous and {xn } is a sequence in X,
then xn → x implies f (xn ) → f (x).
4
Proof: Let B be a neighbourhood of f (x) and f (x) ∈ O ⊂ B where O ∈ T .
Then f −1 (O) ∈ S is a neighbourhood of x. If {xn } converges to x then xn ∈
f −1 (O) for large n, which is equivalent to f (xn ) ∈ O. ♦
A family G of open sets is a base for a topology T if every open set in T can
be expressed as a union of sets in G
Proposition 4 G is a base for T if and only if for any A ∈ T and x ∈ A there
exists G ∈ G such that x ∈ G ⊂ A.
Proof: Exercise!
A topological space (X, T ) satifies the 2nd axiom of countability if it has a
countable base (i.e. there is a base G consisting of a countable number of sets.)
Theorem 8 If (X, T ) has a countable base then X contains a countable dense
subset, i.e. a countable set {x1 , x2 , . . .} whose closure is equal to X
Proof: Let G = {G1 , G2 , . . .} be a countable base and choose points x1 , x2 , . . .
such that xi ∈ Gi , i = 1, 2, . . .. Let M be the set of these xi . We claim that
M̄ = X. If not, A = X \ M̄ is a non-empty open set that contains no points of
M . But A is the union of some sets Gn containing points xn , a contradiction.
♦
The 1st axiom of countability states that each x ∈ X has a countable neighbourhood base, i.e. a countable system H of neighbourhoods of x such that if
A is a neighbourhood of x there exists H ∈ H such that x ∈ H ⊂ A.
Proposition 5 If (X, T ) satisfies the 1st axiom of countability, every cluster
point of a set A ∈ X is the limit of a convergent sequence {xn } ⊂ A.
Proof: Let x be a cluster point of A and {Hi } be a countable neighbourhood
base at x. The Hi0 = ∩ik=1 Hk is also a countable neighbourhood base, and for
each i there exists xi ∈ Hi0 ∩ A because x is a cluster point. Now xi → x as
i → ∞ (Show this!) ♦
A topological space (X, T ) is compact if every open cover of X has a finite
sub-cover. There is a useful general criterion for compactness, which is as follows. A system of subsets H of X is centred if every finite intersection ∩n1 Hk of
members of H is non-empty.
Proposition 6 X is compact if and only if every centred system of closed subsets of X has non-empty intersection
Proof: Let X be compact and H a centred system of closed subsets. Then
the complementary sets Hc = {H c : H ∈ H} are open. The centred property
implies that no finite union of sets in Hc covers X, but then the whole system
Hc cannot cover X, by compactness. Hence H has non-empty intersection.
Conversely, suppose every centred system of closed subsets has non-empty
intersection. Let G = {Gα } be any open cover of X. Then ∩α Gcα = ∅. Hence the
sets {Gcα } are not centred, i.e. there are sets Gcα1 , . . . , Gcαn such that ∩n1 Gαk = ∅;
but then the corresponding sets Gk cover X. X is compact. ♦
Here now are some important properties of compact spaces. A point z is an
accumulation point 2 of a set A if every neighbourhood of z contains infinitely
2 this
terminology does not seem to be completely standardized.
5
many points of A. A subset K of a topological space X is compact if K is
compact in the relative topology (see Problems).
Theorem 9 Let (X, T ) be a compact space.
1. Every infinite subset of X has an accumulation point.
2. If f : X → R is continuous then f (X) is compact in R.
3. If f : X → R is continuous then f attains its supremum.
4. If A is a closed subset of X then A is compact.
Proof: (1) Any infinite subset contains an infinite sequence {x1 , x2 , . . .} of
distinct points. Let Fn = {xk : k > n}. Then any point in ∩n Fn is an accumulation point. However, {Fn } is a centred system of closed sets, so ∩n Fn 6= ∅.
(2) Exercise.
(3) By (2), f (X) is compact in R, i.e. closed and bounded, so f (X) contains
y ∗ = sup{f (x) : x ∈ X}. Thus there is some x∗ such that y ∗ = f (x∗ ).
(4) If {Gα } is an open cover of A then {Gα } together with the open set Ac
covers X. The result follows. ♦
Combining (1) of Theorem 9 and Proposition 5 we can say that if a space
satisfies the first axiom of countability then any sequence in a compact set
has a convergent subsequence. Property (1) of Theorem 9 is called countable
compactness. We have the following result.
Theorem 10 Countable compactness and compactness are equivalent in a space
with a countable base.
Proof: The proof will not be given here; see K&F3 p96.
In a general topological space X it is not necessarily the case that compact
sets are closed. However, this is true if X is a Hausdorff space, which is a space
in which any two distinct points have disjoint neighbourhoods.
Theorem 11 A compact subset of a Hausdorff space is closed.
Proof: Let K be compact and take a point x ∈
/ K. (If there is no such
point then K is the whole space and is automatically closed.) For each point
y ∈ K there are disjoint open neighbourhoods Ay , By of x, y respectively. Now
{By : y ∈ K} covers K, so {Byi : i = 1, . . . , n} covers K for some finite subset.
Then O = ∩n1 Ayi is an open neighbourhood of x that is disjoint from K. ♦
2.2
Metric Spaces
A metric on a set X is a function d : X × X → R such that
(1) d(x, y) = d(y, x) ≥ 0 for x, y ∈ X
(2) d(x, y) = 0 if and only if x = y
(3) For x, y, z ∈ X, d(x, z) ≤ d(x, y) + d(y, z).
The -ball around x is B (x) = {y ∈ X : d(x, y) < }. The metric topology
on X is defined by saying that a set A is open if x ∈ A implies that B ⊂ A
for some > 0. The standard example is Euclidean space Rn as defined earlier.
Cauchy sequences and completeness are also defined as before: a metric space
X is complete if every Cauchy sequence has a limit in X.
3 A.N.
Kolmogorov and S.V. Fomin, Introductory Real Analysis,Dover 1970
6
Proposition 7 Every metric space satisfies the first axiom of countability. The
second is satisfied if and only if the space is separable (has a countable dense
subset).
Proof: The family {B1/n (x), n = 1, 2, . . .} is a countable neighbourhood base
at x, proving the first statement. Each space that satisfies the 2nd axiom is
separable. For the converse, let X be separable and Y a countable dense subset
and define U = {Br (y) : y ∈ Y, r rational}. Then U is countable. If A is a
neighbourhood of a point x then Br (x) ⊂ A where r = 1/n, for some n. Take
y ∈ Y such that d(x, y) < r/3. Then x ∈ B2r/3 (y) ⊂ Br (x), showing that U is
a base for the metric topology. ♦
A simple but extremely useful result is the contraction mapping theorem.
Let (X, d) be a metric space and f : X → X be a function. f is a contraction
mapping if
d(f (x), f (y)) ≤ αd(x, y), x, y ∈ X
(1)
for some α < 1. A contraction mapping is continuous because it follows from
(1) that f (xn ) → f (x) whenever xn → x. A point x ∈ X is a fixed point of f if
f (x) = x.
Theorem 12 If (X, d) is a complete metric space then every contraction mapping has a unique fixed point.
Proof: Take an arbitrary point x0 and define xn = f (xn−1 ) = f n (x0 ) for
n = 1, 2, . . .. We claim that this is a Cauchy sequence. Indeed, for n ≤ n0 we
have
d(xn , xn0 )
0
=
≤
≤
d(f n (x0 ), f n (x0 ))
αn d(x0 , xn0 −n )
αn (d(x0 , x1 ) + d(x1 , x2 ) + · · · + d(xn0 −n−1 , xn0 −n ))
≤
αn d(x0 , x1 )(1 + α + α2 + · · · + αn −n−1 )
1
αn d(x0 , x1 )
→ 0 as n → ∞.
1−α
≤
0
Since X is complete, the limit x = limn xn exists, and by continuity f (x) =
lim f (xn ) = lim xn+1 = x. To show uniqueness, note that if f (x) = x and
f (y) = y then (1) states that d(x, y) ≤ αd(x, y), which can only be satisfied if
d(x, y) = 0. But then x = y. ♦
We now discuss compactness in metric spaces. If (X, d) is a metric space
and > 0, a set A is an -net for M ⊂ X if for every x ∈ M there is a y ∈ A
such that d(x, y) < . M is totally bounded if there exists a finite -net for each
> 0.
Theorem 13 Every countably compact metric space is totally bounded.
Proof: Suppose (X, d) is not totally bounded, i.e. for some 0 there is no
finite 0 -net. Let a1 be arbitrary, and pick a2 such that d(a1 , a2 ) > 0 . This
must be possible, else a1 is an 0 -net. Now pick a3 such that d(a1 , a3 ) > 0
and d(a2 , a3 ) > 0 . Again, this must be possible, else {a1 , a2 } is an 0 -net.
Continuing, we obtain a sequence {a1 , a2 , . . .} such that d(ai , aj ) > 0 for i 6= j.
This sequence has no accumulation point, so X is not countably compact. ♦
7
Corollary 1 Every countably compact space has a countable dense subset and
a countable base.
Proof: If X is totally bounded then X has a finite 1/n-net for n = 1, 2, . . ..
The union of these is a countable dense subset D, and {B1/n (x) : x ∈ D, n =
1, 2, . . .} is a countable base. ♦
Corollary 2 Every countably compact metric space is compact.
Proof: This follows from Corollary 1 and Theorem 10 above.
Here now is the main result of this section.
Theorem 14 A metric space (X, d) is compact if and only if it is complete and
totally bounded.
Proof: If X is not complete, there is a Cauchy sequence {xn } with no limit.
The sequence {xn } then has no accumulation points, so X is not compact. In
view of Theorem 13 this shows that if X is compact then it is totally bounded
and complete.
Conversely, suppose X is totally bounded and complete, and let {xn } be
an infinite sequence of distinct points. Let N1 be a 1-net; then for some y1 ∈
N1 , A1 = B1 (y1 ) contains an infinite subsequence x11 , x12 , . . . of {xn }. Now let N2
be a 1/2-net of A1 and A2 = B1/2 (y2 ) for some y2 ∈ N2 contain a further infinite
subsequence x21 , x22 , . . . of x11 , x12 , . . .. In this way we construct sets A1 , A2 , . . .
such that the radius of An is 2−(n−1) and each An contains infinitely many points
of the original sequence. Defining A0n to be the sphere with the same centre as
An but twice the radius, we obtain a nested sequence A01 ⊃ A02 ⊃ . . . with the
same property. The sequence A0n has non-empty intersection: this follows by an
argument very similar to the nested intervals theorem, Proposition 3 above. If
x0 ∈ ∩n A0n then clearly x0 is an accumulation point of xn . Thus X is countably
compact, and hence compact by 2. ♦
2.3
The space of continuous functions
The most important space in stochastic analysis is the space C[0, T ] of all continuous functions f : [0, T ] → R. The metric is the uniform metric
d(f, g) = max |f (t) − g(t)|.
t∈[0,T ]
It is easy to check that this is a metric.
Proposition 8 C[0, T ] is complete and separable.
Proof: Let fn be a Cauchy sequence in C[0, T ]. For each t, {fn (t)} is a Cauchy
sequence of real numbers and hence converges to a limit f (t). It remains to show
that the function t → f (t) is continuous. For s, t ∈ [0, T ],
|f (t) − f (s)| ≤ |f (t) − fn (t)| + |fn (t) − fn (s)| + |fn (s) − f (s)|.
(2)
Given > 0 we can choose n such that the first and third terms on the right
are both less than /3, whatever s, t. Now fn is continuous, so for fixed t we
can choose δ such that |t − s| < δ ⇒ |fn (t) − fn (s)| < /3; then from (2),
|f (t) − f (s)| < . Thus f is continous at all t.
8
There are many countable dense subsets. We could for example take the set
of piecewise linear functions with rational breakpoints, such a function φ being
defined by rational numbers 0 = t0 < t1 < · · · < tn = T and a0 , . . . an such that
φ(t) = ak +
t − tk
(ak+1 − ak ) for t ∈ [tk , tk+1 [
tk+1 − tk
♦
There is a standard criterion for compactness in C[0, T ] known as the Arzelà
theorem. A subset A ⊂ C[0, T ] is uniformly bounded if there is a number K
such that |f (t)| ≤ K for all f ∈ A and t ∈ [0, T ]. A is equicontinuous if for any
> 0 there exists δ such that |s − t| < δ ⇒ |f (s) − f (t)| < for all s, t ∈ [0, T ]
and f ∈ A.
Theorem 15 A necessary condition for a set A ⊂ C[0, T ] to be relatively compact is that A be uniformly bounded and equicontinuous.
Relative compactness means that the closure A is compact. The proof of this
theorem will be found on p. 102 of K&F. It is somewhat lengthy but uses only
arguments similar to those deployed in these notes.
9