Climbing a ladder called

The Connected Components Functor

You start at a.

You climb 2 stairs (α).

You reach α²a.

I start at b.

I climb 1 stair.

I reach αb.

I see you on the same stair i.e.

α²a = αb

It starts at c.

It climbs 1 stair.

It reaches αc.

I can’t sit still on αb; I’m like a monkey whose tail has been set on fire.

I start at b.

I climb 2 stairs.

I reach α²b.

I see It on the same stair i.e.

α²b = αc

You, upon watching me run around like a headless chicken (which is what my good Mexican friend Alicia says) being with you at times and with It at other times, begin to wonder how many steps you have to climb up from your a to be with It.  After some action potentials you realize that you can also be with It if the stair you reach after climbing some n stairs from your a is same as the stair that It reaches after climbing some m stairs from its c i.e. if

αⁿa = α^mc

(Please see the internal diagram on top of page 183 of Conceptual Mathematics textbook to see how the ladder we are climbing now looks like.)

So you go looking for the above numbers (n, m) in what you have seen

α²a = αb

α²b = αc

Since I climbed 2 steps to get to be with It, you climb 1 more step with me from where we (you & I) met i.e.

α²a = αb

α (α²a) = α (αb)

α³a = α²b

α³a = αc (since α²b = αc)

So if you climb 3 stairs from a you will be with It which got there after climbing 1 stair from c.

Meeting-point of You and I

a -> αa -> (α²a = αb)

^

|

b

Meeting-point of I and It

(α²b = αc) <- c

^

|

αb

^

|

b

Meeting-point of You and It

                         (α²b = αc) <- c

                   ^

                   |

a -> αa -> (α²a = αb)

I have a vague feeling that there’s something that’s not quite right with this ‘physical interpretation’ of Exercise 1 (Conceptual Mathematics, page 358):

If

x is connected to y

and

y is connected to z

then

x is connected to z

In other words, if

α^px = α^qy

and

α^ry = α^sz

then there exist numbers t, u such that

α^tx = α^uz

Multiplying both sides of the given equation

α^px = α^qy

with α^r (from the given α^ry = α^sz)

α^r (α^px) = α^r (α^qy)

α^{(r + p)}x = α^{(r + q)}y

(given the commutativity of addition i.e. r + q = q + r)

= α^{(q + r)}y

= α^q (α^ry)

(given α^ry = α^sz)

= α^q (α^sz)

α^{(r + p)}x = α^{(q + s)}z

α^tx = α^uz (with t = r + p, u = q + s)

So we say x, z are connected.

Having said that let me go work on my (aforementioned) feelings!