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02-07-2012, 09:37 PM   #1
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bagman is so hella pwnt and mad that he quit surfing the interwebz and returned to his math55A homework

hella studying and shit!

02-07-2012, 09:46 PM   #2
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Re: bagman is so hella pwnt and mad that he quit surfing the interwebz and returned to his math55A homework

here is some notez bro!

Math 55a: Honors Advanced Calculus and Linear Algebra
Metric topology IV: Sequences and convergence;
the spaces
B(X; Y ) and C(X; Y ), and uniform convergence

Sequences and convergence in metric spaces.
[See Rudin, 3.1, 47{51.]
The notion of convergence in the metric space
R is implicit in such familiar
contexts as in nite series and even nonterminating decimals, but was not made
explicit until centuries after Euler spent considerable e
ort trying to evaluate
such series as 1!
�� 2! + 3! �� 4! +    [sic]. As with many such notions, while
our initial interest is in sequences in
R or perhaps Rn, the basics are just as
easy to formulate in the context of an arbitrary metric space, and we shall have
occasion to use this notion in that generality later in the course.
Let
fpng = fp1; p2; p3; : : :g1 be a sequence in a metric space X (i.e., with each

p
n 2 X, n = 1; 2; 3; : : :). We say that fpng converges if there is a point p 2 X

such that: for every
 > 0 there is an integer N such that d(pn; p) <  for each

n > N
. Equivalently, for every  > 0 we have d(pn; p) <  for all but nitely
many
n. [Why are the two de nitions equivalent?] Such p is called the limit

of
fpng. The de nite article must be justi ed: we must show that if p; p0 are
both limits of
fpng then p = p0. This is easily shown as follows [see Rudin,
Thm. 3.2b, p.48]. For any
 > 0, there are integers N;N0 such that d(pn; p) < 

for each
n > N and d(pn; p0) <  for each n > N0. Let n be any integer that
exceeds both
N and N0. Then by the triangle inequality d(p; p0) < 2. Since  is
an arbitrary positive number, it follows that
p = p0. We also use the following
notations for \
p is the limit of fpng": \fpng converges to p", \pn approaches p",
\
p = limn!1 pn", and \pn ! p" (or \pn ! p as n ! 1").2

A sequence that does not converge is said to
diverge. Note that this notion can
depend on
X as well as fpng; e.g. f2��ng converges as a sequence in R, but not
as a sequence in (0
; 1). We can say \fpng converges in X" if the ambient space
may otherwise be ambiguous.
Like continuity of functions, the limit of a sequence may be de ned topologically:

p
n ! p if every open set containing p also contains pn for all but nitely many n.3

This lets us de ne lim
n!1 pn in an arbitrary topological space, but with the
proviso that the limit of a sequence may not be unique without some assumption
on the topology. (Challenge: formulate such an assumption that makes limits
unique but still allows a non-metric topology!)
It should be clear that
pn ! p if and only if the sequence fd(pn; p)g converges

1
We use the braces to distinguish the sequence fpng from its typical element pn. Note
set fpnjn = 1; 2; 3; : : :g, namely which element
of that set occurs as which
pn.

2
NB In such contexts 1 is no more than a suggestive symbol; e.g., there is no point \1"
in
R for fng to approach.

3
NB this is not quite the same as \contains all but nitely many pn"! Cf. the end of
footnote 1.

1
to 0 in
R. It is easy to see that a sequence f(xn; yn)g in a product metric space

X
Y converges to (x; y) if and only if xn ! x and yn ! y.
Convergence of sequences is also related to several notions already introduced,
as follows.

Theorem.
[Cf. Rudin, 3.2d on p.48] If p is a limit point of E  X then there
exists a sequence
fpng in E with pn ! p. The closure of any E  X is the set
of limits in
X of sequences in E. In particular E is closed if and only if every
sequence in
E that converges in X also converges in E.
Proof
: We saw in the second handout that if p is a limit point then there are

q
n 2 E (n = 1; 2; 3; : : :), with each qn 6= p, such that for each r > 0 we have

d
(p; qn) < r for all but nitely many n. Then qn ! p. If p 2 E but p is not a
limit point then
p 2 E, and we obtain fpng by setting each pn = p. Conversely,
if
pn ! p for some sequence fpng in E, then every neighborhood of p contains
a point of
E, whence p 2 E. The last statement follows from Rudin 2.27b (also
done in the second handout):
E = E if and only if E is closed. 2

Theorem.
[Cf. Rudin, Thm. 4.2 on p.84] Let f : X ! Y be a function between
metric spaces, and
p 2 X. Then f is continuous at p if and only if f(pn) ! f(p)

for every sequence
fpng in X such that pn ! p.
Proof
: ) is clear from the de nition. For (, suppose on the contrary that f

is not continuous at
p. Then there exists  > 0 such that for each  > 0 some

p
0 2 X satis es d(p; p0) <  but d(f(p); f(p0))  . Let pn be a p0 that works for


= 1=n. Then pn ! p but f(pn) does not converge to f(p). 2

We deduce the following for sequences of real or complex numbers:
Theorem.
[Rudin, Thm. 3.3 on p.49] Let fsng; ftng be sequences of real or
complex numbers. Assume that
sn ! s, tn ! t. Then: the sequence fsn + tng

converges to
s + t; the sequence fsn �� tng converges to s �� t; the sequence

f
sntng converges to st; and, if s 6= 0 and sn 6= 0 for each n, the sequence f1=sng

converges to
1=s.
Proof
: This follows immediately from the previous theorem together with the
continuity of the functions (
s; t) 7! s  t, (s; t) 7! st, s 7! 1=s shown in the
previous (third) handout.
2

[Note that this looks nothing like the proof in Rudin, which nevertheless should
be familiar; what's going on here?]
2
Sequences in function spaces; uniform convergence.
Let Y be a metric
space, and
X an arbitrary set. We noted that the set Y X of functions from X

to
Y becomes a metric space under the sup metric if Y is bounded, or if X

is nite, but that there is a problem if
Y is unbounded and X is in nite: the
purported distance between two functions might be in nite! The usual way
around this problem is to restrict attention to
bounded functions from X to Y ,
i.e. functions
f : X ! Y such that f(X) is a bounded set in Y . The set of
such functions is denoted
B(X; Y ), and it becomes a metric space under the sup
metric
d(f; g) := supx2X d(f(x); g(x)). Note that B(X; Y ) is simply Y X in the
known cases of bounded
Y or nite X.
Now let
ffng be a sequence in B(X; Y ). What does it mean for ffng to converge
to
f? Unwinding the de nition, we nd that the condition is:
For each
 > 0 there exists an integer N such that d(fn(x); f(x)) < 

for all
x 2 X and n > N.
If
fn; f are any functions (not necesssarily bounded) from X to Y , we say that

f
n converges uniformly to f if the above condition is satis ed. [Cf. Rudin,
Def. 7.7 on p.147, which is the special case
Y = R or C.] Note that if each fn is
bounded then so is
f (if fn(X) is contained in a neighborhood of radius M, and

d
(f; fn) < , then f(X) is contained in a neighborhood of radius M +  with
the same center). If
fn ! f uniformly, then certainly fn(x) ! f(x) for each

x
2 X, i.e. fn also converges pointwise to f. Note that pointwise convergence
of
fn to f, de ned by
For each
x 2 X and each  > 0 there exists an integer N such that

d
(fn(x); f(x)) <  for all n > N,
is a strictly weaker notion than uniform convergence, since
N is allowed to
depend on
x as well as . (Cf. our earlier distinction between continuity and
uniform continuity.) For instance, let
X = Y = [0; 1], and fn(x) = xn. Then fn

converges pointwise, but not uniformly, to the function
f
(x) :=


1
; if x = 1;
0
; otherwise.
If
X is itself a metric space4 then we are often interested in the subspace C(X; Y )
of
B(X; Y ) consisting of continuous bounded functions. The key fact here is:

Theorem.
C(X; Y ) is closed in B(X; Y ) for any metric spaces X; Y .

That is, the uniform limit of bounded continuous functions is again bounded and
continuous. We have already shown that the uniform limit of bounded functions
4
More generally X, but not Y, can be a general topological space.

3
is bounded. We next show that the uniform limit of continuous functions is again
continuous:
Theorem.
[Cf. Rudin, Thms. 7.11 and 7.12 on p.149{150] Suppose X; Y are
metric spaces, and
ffng a sequence of continuous functions from X to Y . If

f
fng converges uniformly to a function f : X ! Y , then f is continuous.
Proof
: Fix x 2 X. We show more generally that if fn is continuous at x and

f
n ! f uniformly then f is continuous at x. Indeed, let  be any positive real
number. Since
fn ! f uniformly, there exists N such that d(fn; f) < =3 for
all
n > N. Fix one such n. Since fn is continuous at x, there exists  > 0 such
that
d(fn(x); fn(x0)) < =3 provided d(x; x0) < . Then for all such x we have

d
(f(x); f(x0))  d(f(x); fn(x)) + d(fn(x); fn(x0)) + d(fn(x0); f(x0))

<

3
+


3
+


3
=
:

Thus
f is continuous at x as claimed. 2

This also completes the proof that
C(X; Y ) is closed in B(X; Y ).
Note that the example of
fn(x) = xn shows that the pointwise limit of continuous
functions may fail to be continuous. Must the pointwise limit of bounded
functions be bounded?

4

02-07-2012, 09:48 PM   #3
in need of a μse

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Re: bagman is so hella pwnt and mad that he quit surfing the interwebz and returned to his math55A homework

I don't attend Harvard.

02-07-2012, 09:53 PM   #4
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Re: bagman is so hella pwnt and mad that he quit surfing the interwebz and returned to his math55A homework

pwnt!
are you claiming that natalie portman is teh hella be smartz0rz than you?

02-07-2012, 10:03 PM   #5
Is Butthurt.

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Re: bagman is so hella pwnt and mad that he quit surfing the interwebz and returned to his math55A homework

02-07-2012, 10:04 PM   #6
in need of a μse

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Re: bagman is so hella pwnt and mad that he quit surfing the interwebz and returned to his math55A homework

Probably. She was a psychology major at Harvard.

02-07-2012, 10:05 PM   #7
Is Butthurt.

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Re: bagman is so hella pwnt and mad that he quit surfing the interwebz and returned to his math55A homework

bagman gets psycho analyzed by pilgrims behind plymoth roc.

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