QUESTION
Pa}t
A Afine subsets and
affne nappinAs
Tbroughout
Pafi A,
y
will
be a rea.l vector spa.e and, for
a
nor
empty subset
5
of
7 and o €
{ii+a
:j’
€ S} will be denoted
by S+a.
An affine subset of
V
is
a non-empty subset M ofy with
the
Foperty
that )o+
(1-))V
€ M s,henever !x,U
e
M
ald,\ER,
furl@
ro r*trutu
tlis corcept, shoq’ that
M
=
{a:
(a:-…,h)
e
JRa :2rr 12+a3:1ard.q+4q
2aa=3}
is an affine
subset of lRa.
A.2
Let M be an a,frne
subBet of Y.
Do,lE
-@pt*”
ihat M
+a
js
afrne for every a
€
(ii)
Deduce that
therc exist a subspace
U
of
y
y
and that, if
0 € M, then M is a, subspac€.
and a
€ V such that
y,
the set
M=u+a. (1)
lHint:
s’hai can be said about
such all a, assuming that ii exisis?]
Show futher thst the subspace
U
in (1)
tu udquely d€iermined bI, M
and describe the €xt€nt to which a is
d€terrnined by M.
(iii)
Ilustlate the description (1) in
the ca,se of the afrne subset of lRa
deffned h A.1.
4,3 Aa affine mapping
ftom
y
to W, where t{/ is a second
rcal vector spa.e, is a mapping
l:
V
-W
sucL
that
lor a.ll r,g
€ Y alld a.ll )
€
R.
(i)
Prcve that
wber€ ? :
y
f
:
f(M+(1
-))y)
:I/(r)+(1
y
+
tlz is afrne if and or
y
if it is of the form
+
ty is linear aild 6
e
W.
LHht:
))l(e)
l@):T*+4,
Q)
to
prcv€
the
‘only
if’pa,-t, corFider the mapping
/
–
l(o).1
Show turttrer thai
”
and b in
(2)
are uniquely determined
by
l.
(ii)
Prove that the image of an a.frne sub€t und€I an a,frne mappiag
b a.mrc and that the composition
ot two affne maps is again afrne.
Part
B Fractal
sets in the
plane
Ther€
is no single deffnition
ot a
Jractal
set in the
p1arc
lR,
(or,
for that matter,
in Rn
or
in
more
general
spaces). N€veriheless,
they ax€ informally
c.hara.te zed as havine
a fine
(al]d
olien irlesdax)
struciue
that
persists
11nd€r
arbitrary scaling they’look
the sameJ however much
they de magnified.
I.\dher,
a semible
mea”‘ring fot thei
dimension can be
given
ard this will, tl”icatly,
not be
an
integer.
Afrne
mappings
car sometim€s
be used to
geneBte
Facta.l
sets a.rd thi6 is
illustrated here.
Anr
stmilirzde on lR’? i6
an afrne mappinsl: R,
+
R,
such ihai
lll(s)-/fu)ll:,)),-y
foraltj’,s,
€
R,,
where
ll
.ll
denotes
rhe Euclidean
distarce in Rr. Note that
the tengrh of
a
tine
sesment , in
R,
(ihe
1 dimensional
measure
of ,) is scaled by a factor r
under an r-simiiitude,
whilsi rhe
arca of a region
S
in
Rz
(ihe
2
dinensional medure
of S) is scaled by a factor
r, under
/.
An
example Let
A,B,C be the vertices
of an equiiateral
tdangte in the
first
quadrart
of lRr, where
A
:
(0,0)
and B
:
(1,0),
ald let ,, .o, r be ihe midpoints
of ,4-B, -Bd
and C,4 respectivetv.
8.1 Find
an ro-similitude
ll
oflR, mapping th€
tiangular region A
wirh vertices
trc onto the
tdanaular
resion /rF for
an appropriate va.lue
of r0.
Simildly, find ro-simititudes
the triangulax
rcgions with, respectively,
vstices DBE
a\d FEC.
What is the va.tue
of ro?
8.2 Now
define sets Aa
for n
>
0 by settins
Ao: A a,,rd A,+r:
/2
ard
/3
mappnig A
orto
/1(4.)
u/r(A”)U/3(A”)
for n
>
0.
Sketch Ar and A2
and
prove
that 4,,+r
c
A” for all
4
>
0. Define
S
js
non-empty
(consider
the
boundary of A) and
thai
[The
inciusion
s:
i(s)ul,(s)u/3(s).
]
herc is
straightforwaxd but
the inclusion
=
nff:rA,.
g
is
tdckier. For
prove
rhat
3
q,
show rhat, if,
€ S,
then.foratleaston€t€{1,2,3},oei(44)forinfinitetymanyn.J}ddedrcethato€i(S)forsuch
P.ove
a.lso that the length
of the
pedmete.
of An tends to co
a”5 n
+
co ivhilst
the area of A-
tends
to0asn-co.
B.3
Supposing that
a sensible meaning can be
gi\,€n
to the statement
that S has linite positive
3-dimensional
measue,
show ihat
(3)
$lodd susgest
that 316
:
1. Motivated
by rhis,
the dimension
s of S is deffned
by ihis equation.
Calculat€ s and note
thar it lies stricrty
berw€en 1
and
2.
Explain
briefly
why this might
be expected tuom th€ ffnal
result in B.2.
ADother
exarnple
B.4
D€scribe briefly
(that
is,
q’ithout
givine
detaiied
proofs)
a similax construction
to thai in B.1
3
in
which
the triangular rcgion A
is replaced by a squaxe
region I that
is divided into nine
conguent
sub-squares
in a 3 x
3
grid
ald one conside$ a$ne mappings
of I onto
aU but the centlai
sub-squaxe.
Compute the dimension
of the resulting set in this
case.
Comment
The sets in 8.1-3
and 8.4 werc first considered
by the Polish ma.thematician
W. Sieryiiiski;
they
are usua.lly rcferred
to as, respectively, the
SierpiiEki
gasket
ard the
Sierpinski ca.rpet.
SOLUTION
❆✷
✐✐✮ ■❢ M = U + a
t❤❡♥ a 2 M
❛s 0 2 U ✭✐t ✐s ❛ ✈❡❝t♦r s✉❜s♣❛❝❡✮
❛♥❞ s❡tt✐♥❣ u = 0 u + a = a 2 M
❈♦♥s✐❞❡r t❤❡ s❡t M a = M + (a)
❲❡ ❝❧❛✐♠ t❤❛t M a ✐s ❛ ✈❡❝t♦r s✉❜s♣❛❝❡
❝❧❡❛r❧②✱ 0 2 M a ❛s a 2 M
▲❡t u; v 2 M a
u = u
1
a ❛♥❞ v = v
1
a ✇❤❡r❡ u
2 M
❆s M a ✐s ❛✣♥❡ ✭❜② ♣❛rt ✐✮
u+v = f (u
1
a) + (1 ) (v
1
1
; v
1
a)g+(1 ) f (v
1
a) + (1 ) (u
a)g
a
(u
1
a) + (1 ) (v
1
a) ; (v
1
a) + (1 ) (u
a) 2 M a
❍❡♥❝❡✱ u + v = u
0
+ (1 ) v
0
1
a 2 M a
❚❤❡r❡❢♦r❡ M a ✐s ❝❧♦s❡❞ ✉♥❞❡r ✈❡❝t♦r ❛❞❞✐t✐♦♥
❙✐♥❝❡✱ M a ✐s ❛✣♥❡✱ s❝❛❧✐♥❣ ❧❛✇s ❤♦❧❞✳
❆✳✸
✐✮
)
f : V ! W ✐s ❛✣♥❡
❈♦♥s✐❞❡r t❤❡ ♠❛♣
g : V ! W
✇❤❡r❡ g = f f(0)
◆♦✇✱ g (x + (1 ) y) = f (x + (1 ) y) f(0)
= f(x) + (1 ) f(y) f(0) + (1 ) f(0)
= (f(x) f(0)) + (1 ) (f(y) f(0)) = g(x) (1 ) g(y)
❍❡♥❝❡✱ g ✐s ❛✣♥❡
t❛❦❡ = 1
g(x) = f(x) f(0) = f(x 0)
❛♥❞
g( x) = f( x) f(0) = f ( x 0) = f ( (x 0)) = f (x 0) = g(x)
❍❡♥❝❡✱ g(x) s❛t✐s✜❡s ♣r♦♣❡rt✐❡s ♦❢ ❛ ❧✐♥❡❛r ♦♣❡r❛t♦r
9T ✇❤✐❝❤ ✐s ❛ ✉♥✐q✉❡ ❧✐♥❡❛r tr❛♥s❢♦r♠❛t✐♦♥ s✉❝❤ t❤❛t
g(x) = Tx
) f(x) f(0) = Tx
f(x) = Tx + b
✇❤❡r❡ b = f(0) ❛♥❞ ✐s ✉♥✐q✉❡
(
f(x) = Tx + b
❈♦♥s✐❞❡r
f (x + (1 ) y) ❢♦r s♦♠❡ x; y 2 V ❛♥❞ s❝❛❧❛r
f (x + (1 ) y) = T (x + (1 ) y) + b
s✐♥❝❡ T ✐s ❧✐♥❡❛r
f (x + (1 ) y) = T(x) + (1 ) T(y) + b + (1 ) b
= [T(x) + b] + (1 ) [T(y) + b] = f(x) + (1 ) f(y)
✶
1
❍❡♥❝❡✱ f ✐s ❛✣♥❡
✐✐✮
■♠❛❣❡ ♦❢ ❛♥ ❛✣♥❡ s✉❜s❡t ✉♥❞❡r ❛♥ ❛✣♥❡ ♠❛♣
▲❡t M ❜❡ ❛ ❛✣♥❡ s✉❜s❡t ♦❢ V
▲❡t N ❜❡ t❤❡ ✐♠❛❣❡ ♦❢ M ✉♥❞❡r f ✐♥ W
N = f(M)
▲❡t x; y 2 M ) x + (1 ) y 2 M
❙✐♥❝❡✱ f ✐s ❛✣♥❡
f (x + (1 ) y) = f(x) + (1 ) f(y) 2 N
❍❡♥❝❡✱ N ✐s ❛✣♥❡
❈♦♠♣♦s✐t✐♦♥ ♦❢ ❛✣♥❡ ♠❛♣s✳
▲❡t f; g ❜❡ t✇♦ ❛✣♥❡ ♠❛♣s
f : V ! W✱ g : U ! V
▲❡t x; y 2 U ❛♥❞ ❜❡ ❛♥ s❝❛❧❛r
❙✐♥❝❡✱ ❣ ✐s ❛✣♥❡
f g (x + (1 ) y) = f fg (x + (1 ) y)g = f fg(x) + (1 ) g(y)g
❛♥❞ s✐♥❝❡✱ ❢ ✐s ❛✣♥❡
= f fg(x)g + (1 ) f fg (y)g
❍❡♥❝❡✱ f g ✐s ❛♥ ❛✣♥❡ ❢✉♥❝t✐♦♥
❇✶✳
C ✐s
f
1
1
2
(x) =
;
p
3
1
2
2
x
♠❛♣s ABC t♦ ADF
f
2
ABC t♦ DBE
f
2
=
1
2
((1; 0) + x)
ABC t♦ FEC
f
r
3
0
=
✐s
1
2
1
2
h
x +
❤❡r❡
❇✷✳
❙❦❡t❝❤ ❢♦r 4
1
1
2
;
p
3
2
i
❣r❡② ❛r❡❛ ✐♥❞✐❝❛t❡s t❤❡ r❡❣✐♦♥ 4
1
✷
❙❦❡t❝❤ ❢♦r 4
2
❣r❡② ❛r❡❛ ✐♥❞✐❝❛t❡s r❡❣✐♦♥ 4
2
✸
◆♦t❡ t❤❛t f
1
(4) 4 ❛♥❞ f
2
(4) 4 ❛♥❞ f
(4) 4
❛♥❞ ✐♥❞✉❝t✐✈❡❧② s✐♥❝❡ f
1
; f
2
and f
3
3
❛❧✇❛②s ❧❡❛❞ t♦ ❝r❡❛t✐♦♥ ♦❢
s♠❛❧❧❡r ❛✣♥❡ s✉❜s❡ts ♦❢ ❡❛❝❤ ❛✣♥❡ s❡t✱
4
n+1
= f
1
(4
n
)
S
f
2
(4
S ✐s ♥♦♥✲❡♠♣t② ❛s (0; 0) 2 4
n
)
S
f
n
3
(4
n
) 4
n
❛♥❞ ♠♦r❡ s♣❡❝✐✜❝❛❧❧② t♦ f
❢♦r ❛❧❧ n
S = f
1
(S)
S
f
2
(S)
S
f
(S)
▲❡t x 2 f
1
(S)
S
f
2
3
(S)
S
f
3
(S) ) 91 i 3 s✉❝❤ t❤❛t x 2 f
(S)
) 9y 2 S s✉❝❤ t❤❛t f
(y) = x
❍❡♥❝❡✱ x 2 f
i
(4
n
i
) ❢♦r ✐♥✜♥✐t❡❧② ♠❛♥② n
❍❡♥❝❡✱ x 2 S
❚❤❡r❡❢♦r❡✱
f
1
(S)
S
f
2
(S)
S
(S) S
❈♦♥s✐❞❡r
x 2 S; t❤❡♥ ❢♦r ❛t❧❡❛st ♦♥ i 2 f1; 2; 3g
x
i
2 f
i
(4
n
f
3
) ❢♦r ✐♥✜♥✐t❡❧② ♠❛♥② n
❙✐♥❝❡ f
i
(4
n
) 4
✱
x 2 4
n
n
❢♦r ✐♥✜♥✐t❡❧② ♠❛♥② n
) x
i
2 f
(S)
❆r❡❛✭4
1
i
✮❂
p
3
4
■♥❞✉❝t✐✈❡❧② Area(4
p
3
4
n
1
4
) =
=
p
3
p
3
4
4
1
1
1
2
2
1
4
=
✹
3
2
4
3
p
3
16
:::
3
2
2n
1
i
lim
n!1
1
2
2
+
3
2
4
::: +
3
2
n
2n
+ ::: =
1
4
= 1
❍❡♥❝❡✱
lim
n!1
Area(4
) = 0
P❡r✐♠❡t❡r✭4
1
) = 3
3
■♥❞✉❝t✐✈❡❧② Perimeter(4
n
1
2
n
1
1
) = 3
3
4
3
2
✺
n
n
! 1 ❛s n ! 1
J010
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