Shape

286 II Seeing How It Works
the same parts in the topology of C, then their mappings are parts of the same parts in
the topology of C þ . The rule changes the shape C to make the shape C þ , so that the
parts of C þ and their relationships are consistent with the parts of C. There’s no division
in C þ that implies a division that’s not already in C.
The trick is to define topologies for shapes, so that whenever I apply a rule
it’s continuous. This is possible for different mappings—each in a host of different
ways—working backward after I finish calculating. (If I’m impatient, I can define topologies
after every rule application, or intermittently, but these definitions may
change as I go on.) The mapping
h 1 ðxÞ ¼x
tðAÞ
is an obvious choice. It preserves every part x of the shape C tðAÞ that isn’t erased
when a rule A fi B is used to change a shape C. It’s right there as the leading term in
the formula
ðC
tðAÞÞ þ tðBÞ
that gives the result C þ in which the supporting term tðBÞ plays its part. Better yet, I
can use the topology of the shape C þ to get a minimal topology for C that guarantees
continuity. In particular, x is a closed part of C just in case it’s empty or y is a closed
part of C þ and
x ¼ tðAÞþððC
tðAÞÞ yÞ
The divisions in the piece C tðAÞ are fixed in the topology of C þ . Still, I may want
something more profligate—perhaps a Boolean algebra for each shape. Then, for y in
the Boolean algebra of C þ , x is as before, or
x ¼ðC
tðAÞÞ y
In both cases, distinguished values of x are easy to find. The empty shape is given
explicitly, and C is given for y ¼ C þ . And notice that tðAÞ is defined when y is
empty—condition 1 is included in condition 2. Then, there are sums and products.
And in the Boolean case, the shape x ¼ tðAÞþððC tðAÞÞ yÞ has the complement
xA ¼ðC tðAÞÞ yA, where y and yA are complements. But be careful. Even though it
looks it, the topology of C needn’t be bigger than the topology of C þ —either by a
single part or twice. Values of y may be equivalent in the sense that they define the
same shapes.
If I try the mapping h 1 with identities, I get something rather different from what
I described in my second scenario. This gives equal insight in an alternative way, and
it’s fun to compare. Even so, the mapping
h 2 ðxÞ ¼x ðtðAÞ tðBÞÞ
works to include the second scenario in this one. Now
h 2 ðxÞ ¼x