Truth about 1

Here is

1 = {•}

a singleton set with one element (•).

Here’s a part

g: 1 –> 1

of 1 i.e. g(•) = (•).

This part g is the inverse image of another part

v: 1 –> W

with W = {true, false} and v (•) = true…

Now this is beginning to read a lot like Ramayana in three sentences, which is all the excuse I need to go back—way back to my very good friend function

f: A –> B

Let’s say we have a function

color: A –> B

with A = {Rayudu} and B = {Black, Brown} telling us the hair color of people in A.  For example

color(Rayudu) = Black

which says something like: the hair color of Rayudu is black.

A recurring nightmare of mine [in mathematics] is inversion—fortunately, for now and here, it’s not that scary; it simply asks: who’s that in A who got black hair?

Rayudu

Rayudu, not unlike any other answer, is boring; it’s how we got to ‘Rayudu’ from ‘Black’ that’s worth a penny or two.

But, first, who’s Rayudu?

Rayudu is a part of the domain set A i.e.

Rayudu: 1 –> A

What, while we are at it, is black?

Black is a part of the codomain set B i.e.

Black: 1 –> B

Going from the color Black to the person who’s color is black is, as far as I can tell, going from a given part

Black: 1 –> B

to a part

Rayudu: 1 –> A

along the function

color: A –> B

The question then is: given

color: A –> B, color(Rayudu) = Black

and

Black: 1 –> B

how did we end up with this

Rayudu: 1 –> A

What’s so special about Rayudu?

It’s a part of A; but so is

0: 0 –> A, 0 = {}

Rayudu is special in that any generalized element of A (Sets for Mathematics, page 16) i.e.

x: X –> A

is in

Rayudu: 1 –> A

if and only if

fx: X –> B

is in

Black: 1 –> B

which is just about all the confusion I can cook up out of

Rayudu: 1 –> A is the inverse image of Black: 1 –> B along f: A –> B

Conceptual Mathematics, page 336

Since we are least interested in Rayudu, his color, or for that matter, any other values of physical variables associated with him, let’s get off this tangent and get back to business pronto.

The part

g: 1 –> 1, g(•) = •

is the inverse image of the part

v: 1 –> W, v(•) = true

but along what?

Even more bluntly, only if I specify a function

f: 1 –> W

and say something like

g is the inverse of v along f

then somebody else can tell me if it’s true or false.

Given that

W = {true, false}

we have two functions from 1 to W

f: 1 –> W, f (•) = true

f’: 1 –> W, f’ (•) = false

Now the question we have staring into our pupils:

Is g the inverse image of v along f or along f’ or along both f and f’?

We can say

g is the inverse image of v along f

because the generalized element

x: 1 –> 1

is in (Conceptual Mathematics, page 335)

g: 1 –> 1

i.e. there exists a p such that

x = gp

and

fx: 1 –> W

is in

v: 1 –> W

i.e. there exists a q such that

fx = vq

 and because

 f’x: 1 –> W

is not in

v: 1 –> W

i.e. because there is exactly one function

f: 1 –> W

along which

g is the inverse image of v

Just in case you feeling bad about the other function

f’: 1 –> W, f’(•) = false

being left out of our story, here’s fairness-in-action: there’s another part of 1

g’: 0 –> 1, 0 = {}

that is the inverse image of v along f’ (but not along f).

That’s enough pillow-fights and here comes the hardcore definition (Lawvere element):

An object W together with a given part v: 1 –> W is called a truth value object (for the category of sets) if and only if for every part g of any set X there is exactly one function f: X –> W for which g is the inverse image of v along f (Conceptual Mathematics, page 337).

Here’s the part I like in the above definition:

for every part g… there is exactly one function f

As much as I just do things, here’s the plan: redo this exercise

1. for D (dot) in the category of graphs
2. for U (= 0 –> 1) in the category of two-stage variable sets (Sets for Mathematics, pp. 114 – 9)
3. for 0: 0 –> 1 in the category of labeled sets

The above 3 cases have something in common with the case of 1 = {•} of the present post which is the one dot (in visual depiction) of the objects

1, D, U, and 0

which means… nothin 😉