Random IFS
Same Picture
Observe
(x
0
, y
0
)
=
T
1
(x
0
, y
0
)
=
T
1
(T
1
(x
0
, y
0
))
=
T
1
(T
1
(T
1
(x
0
, y
0
)))
= ...
So
(x
0
, y
0
)
belongs to the square with address 1, the square with address 11, the square with address 111, and so on.
Because the diameters of these squares are
shrinking to 0
, the address
111... = 1
∞
corresponds to the single point
(x
0
, y
0
)
.
Alternately, note that the only point left unchanged by repeated application of T
1
is the point with address
111... = 1
∞
Return to
Iteration and address shift
.
