2E
l
e
m
e
n
t
a
r
yG
e
o
m
e
t
r
y
o
f
R
N
Avector
v
!
R
N
of norm one is called a
direction
.G
iv
enapo
in
t
x
0
!
R
N
and
ad
irect
ion
v
,the
line
through
x
0
in the direction
v
is given by
L
:=
!
x
!
R
N
:
x
=
x
0
+
t
v
,t
!
R
"
.
Given two points
y
,
z
!
R
N
with
y
"
=
z
line
through
y
and
z
is given by
L
:=
!
x
!
R
N
:
x
=
t
y
+(1
#
t
)
z
!
R
"
,
while the
segment
joining
y
and
z
is given by
S
:=
!
x
!
R
N
:
x
=
t
y
#
t
)
z
!
[0
,
1]
"
.
Aset
E
$
R
N
is
convex
if for every
x
,
y
!
E
,thesegmentjo
ing
x
and
y
is contained in
E
.
Remark 3
Given a point
x
0
!
R
N
and
r>
0
ball
centered at
x
0
and of
radius
r
is the set
B
(
x
0
,r
):=
!
x
!
R
N
:
%
x
#
x
0
%
<r
"
.
Let’s prove that the ball
B
(
x
0
)
is convex. Consider two points
x
,
y
!
B
(
x
0
)
and let
t
!
[0
,
1]
. We want to prove that
t
x
#
t
)
y
belongs to
B
(
x
0
)
.W
r
i
te
x
0
=
t
x
0
#
t
)
x
0
. Then we have
%
t
x
#
t
)
y
#
x
0
%
=
%
t
x
#
t
)
y
#
(
t
x
0
#
t
)
x
0
)
%
=
%
t
(
x
#
x
0
)+(1
#
t
)(
y
#
x
0
)
%
&%
t
(
x
#
x
0
)
%
+
%
(1
#
t
y
#
x
0
)
%
=
t
%
x
#
x
0
%
#
t
)
%
y
#
x
0
%
<tr
#
t
)
r
=
r,
where we have used the triangle inequality and the positive homogeneity of the
norm. This shows that the segment joining
x
and
y
is contained in
B
(
x
0
)
.
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View Full Document2bis Continuous functions in
R
N
Let
E
⊂
R
n
,let
f
:
E
→
R
and let
x
0
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 Spring '14
 Geometry, Derivative, Continuous function

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