Metric Spaces, Lecture Notes - Mathematics, Study notes for Calculus. University of California (CA) - UCLA

Calculus

Description: Metric spaces, Metrics a Nation of Distance, Examples, Convergence
Showing pages  1  -  4  of  5
Metric Spaces
Adrian Down 16779577
August 2, 2005
1 Metrics: a notion of distance
So far in analysis on R, every definition (convergence, Cauchy, continuity)
has included statements like “|xy|< ”. The idea is that this |xy|
represents a distance from xto y.
A distance should satisfy the properties:
i) |xy| ≥ 0,x, y R
ii) |xy|= 0 x=y
iii) |xy|=|yx|
iv) “Triangle inequality”: |xy|+|yz| ≥ |xz|,x, y, z
Definition: a metric space is a set of “objects” Stogether with a function
don the set of pairs (x, y), x, y Ssuch that
d(x, y)R,x, y S
1)d(x, y)0,x, y S
2)d(x, y) = 0 x=y
3)d(x, y) = d(y, x),x, y S
4)d(x, y) + d(y, z)d(x, z),x, y, z S
dis called the metric on S. May write (S, d) to specify the metric.
4) is called the triangle inequality; it is the most important, and it is
usually the hardest to verify.
2 Examples
1) (R, d) is a metric space where d(x, y) = |xy|
1
2) “Discrete metric space”: Sis any set, and define
d(x, y) = (1x, y S, x 6=y
0xS, x =y
3) “Euclidean k-space”:
S={(x1, x2, . . . , xk)|xiR,1ik}
d(x,y) = k
X
i=1
(xiyi)2!
1
2
Properties 1, 2, and 3 are clear. Property 4 takes work. Best way is to
consider dot products of vectors.
This is the most important metric space that we will cover. The given
metric space is called the standard metric on Rk.
4) “Manhattan metric”:
S=Rk
d(x,y) =
k
X
i=1 |xiyi|
Properties 1, 2, and 3 are easy. For 4, use the triangle inequality on R,
X|xiyi|+X|yizi| ≥ X|xizi|
5) C[0,1] :
S={f|fis a continous function on [0,1]}
d(f, g) = sup{|f(x)g(x)||x[0,1]}
Note that f, g continuous on [0,1] fgis continuous on [0,1] d(f, g)
exists.
To verify property 2,
d(f, g) = 0 sup{|f(x)g(x)|} = 0 → |f(x)g(x)|= 0,xf(x) = g(x),x
2
To verify property 4, use the triangle inequality on R,
d(f, g) + d(g, h) = sup{|f(x)g(x)|} + sup{|g(x)h(x)|}
sup{|f(x)g(x)|+|g(x)h(x)|} ≥ sup{|f(x)h(x)|} =d(f, h)
6) l1(N) :
S={(an)
n=1|
X
n=1 |an|converges}
d((an),(bn)) =
X
n=1 |anbn|
Properties 1, 2, and 3 are easy. For 4, use the triangle inequality on R,
d((an, bn)) + d((bn),(cn)) =
X
n=1 |anbn|+
X
n=1 |bncn|
=
X
n=1 |anbn|+|bncn| ≥
X
n=1 |ancn|=d((an),(cn))
Idea is to abstract the qualities of Rto many situations. The multitude
of examples is one of the most useful aspects of metric spaces.
3 Convergence
Definition: A sequence (sn)
n=1 in a metric space (S, d) is said to converge to
sSif
 > 0,NNsuch that n>Nd(sn, s)< 
Note: snconverges to sif and only if limn→∞ d(sn, s) = 0 in R.
Definition: a sequence (sn) is (S, d) is Cauchy if
 > 0,NNsuch that n, m > N d(sn, sm)< 
Definition: A metric space (S, d) is said to be complete if every Cauchy
sequence converges.
Note: the converse statement that convergence implies Cauchy is always
true. Emulate the proof from R.
3
Examples:
1) (R, d), d(x, y) = |xy|is complete by the Completeness axiom
2) (Q, d), d(x, y) = |xy|is not complete
3) C[0,1] is complete.
4) (Rk, d), dis the Euclidean metric, is complete. To prove it, we need
lemmas
Lemma 1: (xn) is a convergent sequence in Rk(xn
i)
n=1 is convergent
in R,i, 1ik.
Proof: Assume (xn) is convergent and xnconverges to x. Let  > 0 be
given. Nsuch that
n>N k
X
i=1
(xn
ixn
j)2!
1
2
<  X(xn
ixi)2< 2
(xn
ixi)2< 2,i→ |xn
ixi|< , ilim
n→∞
xn
i=xi,i
Now assume that (xn
1),(xn
2),...,(xn
k) converge in R. Let
lim
n→∞
xn
1=x1,..., lim
n→∞
xn
k=xk
Let  > 0 be given. Then N1, . . . , NkNsuch that
n>N1→ |x1xn
1|<
k, . . . , n > Nk→ |xkxn
k|<
k
Let N= max{N1, N2, . . . , Nk}. Then
Lemma 2: (xn) is Cauchy in Rkif and only if (xn
i) is Cauchy in R,i
Proof: left to the reader
Now, if (xn) is Cauchy in Rk(xn
i) is Cauchy in R,i(xn
i) is convergent
i(xn) is convergent in Rk.
So Rkis complete.
What’s useful are the lemmas: to examine the properties of something in
Rk, look at the components of the vector one at a time.
Definition: A set AS, (S, d) is a metric space is bounded if xSand
MRsuch that d(a, x)M, aA.
Definition: the open ball around xwith radius Ris BR(x) = {s
S|d(s, x)< R}.
The closed ball =¯
BR(x) = {sS|d(s, x)R}.
Ais bounded if A¯
BR(x) for some x, R.
4
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