Fixed Point (Special Version)

Give me a complete metric space.

No missing limits.

No unfinished points.

Give me a map T

and a number c

with

0 ≤ c < 1

such that

d(Tx, Ty) ≤ c · d(x,y)

for every x and y.

[whisper repeats, each time quieter]

c less than one

less than one

less than one

Start anywhere

Call it x-zero

No privileged beginning

No perfect place to stand

Take one step through T

x₁ = T(x₀)

Then do it again

x₂ = T(x₁)

Again

Again

Again

Define:

xₙ₊₁ = T(xₙ)

Now measure every movement

Distance from the next to last

The contraction makes it smaller

Every iteration passed

d(xₙ₊₁, xₙ)

at most

c · d(xₙ, xₙ₋₁)

Then again

and again

until induction gives:

d(xₙ₊₁, xₙ) ≤ cⁿ d(x₁,x₀)

Every step is smaller

Than the step that came before

c times

c times

c times

Closer to the floor

Closer every time

Closer every turn

Run the same transformation

Watch the distance burn

Start wherever you want

Take whatever road

If every step contracts us

There is somewhere

we must go

Closer every time

Nothing left to prove

Not because we chose it

Because the space

won't let us move

forever.

Take two distant iterates

m greater than n

The distance between them

Can be bounded by the steps between

Triangle inequality

Stack the gaps in line:

d(xₘ,xₙ)

no greater than

the sum from k equals n

to m minus one

of

d(xₖ₊₁,xₖ)

Each gap bounded geometrically:

≤ d(x₁,x₀) Σ cᵏ

So the tail is at most

cⁿ / (1-c) · d(x₁,x₀)

[background vocal begins counting downward]

And when n goes to infinity—

cⁿ → 0

The tail disappears

So for every epsilon

There is somewhere far enough

Beyond it every pair is closer

Than epsilon demands

[whisper]

Cauchy.

The sequence is Cauchy.

The space is complete.

So there exists a point

x-star

such that

xₙ → x*

Closer every time

Closer every turn

Run the same transformation

Watch the distance burn

Start wherever you want

Take whatever road

If every step contracts us

There is somewhere

we must go

But reaching a limit

Is not enough

We still have to prove

The limit stays

T is Lipschitz

So T is continuous

Therefore—

T(x*)

equals

T(lim xₙ)

equals

lim T(xₙ)

But T of x-n

is x-n-plus-one

and that limit

is x-star

[single piano note]

So:

T(x*) = x*

FIXED

POINT

FIXED

POINT

Now suppose—

THERE ARE TWO.

x-star

and y-star

Both fixed.

Then:

T(x*) = x*

T(y*) = y*

Measure the distance:

d(x*,y*)

equals

d(Tx*,Ty*)

and contraction says

d(x*,y*) ≤ c · d(x*,y*)

[guitars stop]

But

c < 1

[whisper]

There is only one way.

[huge impact]

d(x*,y*) = 0

SO—

x* = y*

ONE LIMIT

ONE FIXED POINT

NO SECOND PLACE

NO SECOND CHOICE

Closer every time

Closer every turn

Run the same transformation

Watch the distance burn

Start wherever you want

Take whatever road

If every step contracts us

There is somewhere

we must go

Closer every time

Past the fear and noise

Not because the universe

Was listening to our choice

If the map keeps pulling

Every distance into one—

There is only one fixed point

when the motion

is done.

Start anywhere.

Apply the map.

Again.

Again.

Again.

[each word moves toward the stereo center]

A contraction on a complete metric space

has exactly one fixed point.

And the iteration converges to it.

[very quiet]

Existence.

Uniqueness.

Convergence.

[pause]

Q.E.D.

[last clean vocal]

Maybe some things don't find each other

because they were meant to.

Maybe the rules of the space

left nowhere else to go.