Euclidean space is the space in which everyone is most familiar. In Euclidean k-space, the distance between any two points is
d
(
x
,
y
)
=
∑
i
=
1
k
(
x
i
−
y
i
)
2
{\displaystyle d(x,y)={\sqrt {\sum _{i=1}^{k}{(x_{i}-y_{i})^{2}}}}}
where k is the dimension of the Euclidean space. Since the Euclidean k-space as a metric on it, it is also a topological space.
Sequences
edit
Convergent sequences are Cauchy Sequences.
Proof :
Suppose that
l
i
m
s
n
=
s
{\displaystyle lim\;s_{n}=s}
.
Then,
∣
s
n
−
s
m
∣=∣
s
n
−
s
+
s
−
s
m
∣≤∣
s
n
−
s
∣
+
∣
s
−
s
m
∣
{\displaystyle \mid s_{n}-s_{m}\mid =\mid s_{n}-s+s-s_{m}\mid \leq \mid s_{n}-s\mid +\mid s-s_{m}\mid }
Let
ϵ
>
0
{\displaystyle \epsilon >0}
. Then
∃
N
{\displaystyle \exists N}
such that
n
>
N
⇒∣
s
n
−
s
∣<
ϵ
2
{\displaystyle n>N\Rightarrow \mid s_{n}-s\mid <{\frac {\epsilon }{2}}}
Also:
m
>
N
⇒∣
s
m
−
s
∣<
ϵ
2
{\displaystyle m>N\Rightarrow \mid s_{m}-s\mid <{\frac {\epsilon }{2}}}
so
m
,
n
>
N
⇒∣
s
n
−
s
m
∣≤∣
s
n
−
s
∣
+
∣
s
−
s
m
∣<
ϵ
2
+
ϵ
2
=
ϵ
{\displaystyle m,n>N\Rightarrow \mid s_{n}-s_{m}\mid \leq \mid s_{n}-s\mid +\mid s-s_{m}\mid <{\frac {\epsilon }{2}}+{\frac {\epsilon }{2}}=\epsilon }
Hence,
{
s
n
}
{\displaystyle \{s_{n}\}}
is a Cauchy sequence.
Theorem 1
edit
Convergent sequences are bounded.
Proof: Let
{
s
n
}
{\displaystyle \{s_{n}\}}
be a convergent sequence and let
l
i
m
s
n
=
s
{\displaystyle lim\;s_{n}=s}
. From the definition of convergence and letting
ϵ
=
1
{\displaystyle \epsilon =1}
, we can find N
∈
N
{\displaystyle \in \mathbb {N} }
such that
n
>
N
⇒∣
s
n
−
s
∣<
1
{\displaystyle n>N\Rightarrow \mid s_{n}-s\mid <1}
From the triangle inequality;
n
>
N
⇒∣
s
n
∣<∣
s
∣
+
1
{\displaystyle n>N\Rightarrow \mid s_{n}\mid <\mid s\mid +1}
Let
M
=
m
a
x
{
∣
s
n
∣
+
1
,
∣
s
1
∣
,
∣
s
2
∣
,
.
.
.
,
∣
s
N
∣
}
{\displaystyle M=max\{\mid s_{n}\mid +1,\mid s_{1}\mid ,\mid s_{2}\mid ,...,\mid s_{N}\mid \}}
.
Then,
∣
s
n
∣≤
M
{\displaystyle \mid s_{n}\mid \leq M}
for all
n
∈
N
{\displaystyle n\in \mathbb {N} }
. Thus
{
s
n
}
{\displaystyle \{s_{n}\}}
is a bounded sequence.
Theorem 2
edit
In a complete space, a sequence is a convergent sequence if and only if it is a Cauchy sequence.
Proof:
⇒
{\displaystyle \Rightarrow }
Convergent sequences are Cauchy sequences. See Lemma 1.
⇐
{\displaystyle \Leftarrow }
Consider a Cauchy sequence
{
s
n
}
{\displaystyle \{s_{n}\}}
. Since Cauchy sequences are bounded, the only thing to show is:
lim inf
s
n
=
lim sup
s
n
{\displaystyle \liminf s_{n}=\limsup s_{n}}
Let
ϵ
>
0
{\displaystyle \epsilon >0}
. Since
{
s
n
}
{\displaystyle \{s_{n}\}}
is a Cauchy sequence,
∃
N
{\displaystyle \exists N}
such that
m
,
n
>
N
⇒∣
s
n
−
s
m
∣<
ϵ
{\displaystyle m,n>N\Rightarrow \mid s_{n}-s_{m}\mid <\epsilon }
So,
s
n
<
s
m
+
ϵ
{\displaystyle s_{n}<s_{m}+\epsilon }
for all
m
,
n
>
N
{\displaystyle m,n>N}
. This shows that
s
m
+
ϵ
{\displaystyle s_{m}+\epsilon }
is and upper bound for
{
s
n
:
n
>
N
}
{\displaystyle \{s_{n}:n>N\}}
and hence
v
N
=
sup
{
s
n
:
n
>
N
}
≤
s
m
+
ϵ
{\displaystyle v_{N}=\sup\{s_{n}:n>N\}\leq s_{m}+\epsilon }
for all
m
>
N
{\displaystyle m>N}
. Also
v
N
−
ϵ
{\displaystyle v_{N}-\epsilon }
is a lower bound for
{
s
m
:
m
>
N
}
{\displaystyle \{s_{m}:m>N\}}
. Therefore
v
n
−
ϵ
≤
inf
{
s
m
:
m
>
N
}
=
u
N
{\displaystyle v_{n}-\epsilon \leq \inf\{s_{m}:m>N\}=u_{N}}
. Now:
lim sup
s
n
≤
v
N
≤
u
N
+
ϵ
≤
lim inf
s
n
+
ϵ
{\displaystyle \limsup s_{n}\leq v_{N}\leq u_{N}+\epsilon \leq \liminf s_{n}+\epsilon }
Since this holds for all
ϵ
>
0
{\displaystyle \epsilon >0}
,
lim sup
s
n
≤
lim inf
s
n
{\displaystyle \limsup s_{n}\leq \liminf s_{n}}
. The opposite inequality always holds and now we have established the theorem.
Note : The preceding proof assumes that the image space is
R
{\displaystyle \mathbb {R} }
. Without this assumption, we will need more machinery to prove this.
Definition
edit
It is possible to have more than one metric prescribed to an image space. It is often better to talk about metric spaces in a more general sense. A sequence
{
s
n
}
{\displaystyle \{s_{n}\}}
in a metric space
(
S
,
d
)
{\displaystyle (S,d)}
converges to s in S if
lim
n
→
∞
d
(
s
n
,
s
)
=
0
{\displaystyle \lim _{n\rightarrow \infty }d(s_{n},s)=0}
. A sequence is called Cauchy if for each
ϵ
>
0
{\displaystyle \epsilon >0}
there exists an
N
{\displaystyle N}
such that:
m
,
n
>
N
⇒
d
(
s
m
,
s
n
)
<
ϵ
{\displaystyle m,n>N\Rightarrow d(s_{m},s_{n})<\epsilon }
.
The metric space
(
S
,
d
)
{\displaystyle (S,d)}
is called complete if every Cauchy sequence in
S
{\displaystyle S}
converges to some element in
S
{\displaystyle S}
.
Theorem 3
edit
Exercises
edit
1. Let
x
,
y
∈
R
k
{\displaystyle x,y\in \mathbb {R} ^{k}}
. Let
d
1
(
x
,
y
)
=
m
a
x
∣
x
i
−
y
i
∣
{\displaystyle d_{1}(x,y)=max\mid x_{i}-y_{i}\mid }
where
{
i
=
1
,
2
,
.
.
.
,
k
}
{\displaystyle \{i=1,2,...,k\}}
and
d
2
(
x
,
y
)
=
∑
i
k
∣
x
i
−
y
i
∣
{\displaystyle d_{2}(x,y)=\sum _{i}^{k}\mid x_{i}-y_{i}\mid }
. Show:
a.)
d
1
{\displaystyle d_{1}}
and
d
2
{\displaystyle d_{2}}
are metrics for
R
k
{\displaystyle \mathbb {R} ^{k}}
.
b.)
d
1
{\displaystyle d_{1}}
and
d
2
{\displaystyle d_{2}}
form a complete metric space.
2. Show that every open set in
R
{\displaystyle \mathbb {R} }
is the disjoint union of a finite or infinite sequence of open intervals.
3. Complete the proof for theorem 3.
4.Consider: Let
X
{\displaystyle X}
and
Y
{\displaystyle Y}
be metric spaces.
a.) A mapping
T
:
X
→
Y
{\displaystyle T:X\rightarrow Y}
is said to be a Lipschitz mapping provided that there is some non-negative number c called a Lipschitz constant for the mapping such that:
d
[
T
(
x
)
,
T
(
y
)
]
≤
c
d
(
x
,
y
)
∀
x
,
y
∈
X
{\displaystyle d[T(x),T(y)]\leq cd(x,y)\;\forall x,y\in X}
b.) A Lipschitz mapping
T
:
X
→
Y
{\displaystyle T:X\rightarrow Y}
that has a Lipschitz constant less than 1 is called a contraction .
Suppose that
f
:
R
→
R
{\displaystyle f:\mathbb {R} \rightarrow \mathbb {R} }
and
g
:
R
→
R
{\displaystyle g:\mathbb {R} \rightarrow \mathbb {R} }
are both Lipschitz. Is the product of
these functions Lipschitz?