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Today I learned about a cute new variety of algebraic structure, the generalized ultrametric space, aka "gum space". It's just like an ultrametric space, (which in turn is just like a metric space with a stronger triangle inequality that says d(x,z) ≤ max(d(x,y), d(y,z)) rather than d(x,z) ≤ d(x,y) + d(y,z)) except instead of the metric d taking values in the nonnegative real numbers, you let it take values in some ol' poset-with-least-element.

A nice example of a gum space that is not trivially an ultrametric space is a collection of functions A → B whose "distance from one another" is just the

Formally (X, d, (P, ≤, 0)) is a gum space with underlying set X if (P, ≤, 0) is a poset with least element 0, and d : X × X → P has the properties

Exercise: when P=

However, and I think this is rather cute, there's still enough machinery in them to prove a standard contracting-map fixed point theorem.

Define a contracting map from one gum space to another to be a map f of the underlying sets such that d(f(x),f(y)) < d(x,y). Say a gum space is spherically complete if the intersection of any chain of balls in it is nonempty. (This condition is a stand-in for the usual notion of Cauchy-completeness)

For existence, make the following definitions. A pair (x,p) is a

Putting these three facts together via the ultrametric inequality, d(x,f(x)) ≤ r, so f(x) ∈ B(y,r). Hence f(x) ∈ ∩C.

Now if we go just one step further iterating f on x, notice that d(f(x),f(f(x))) < d(x, f(x)) since f is contracting, so (f(x), d(f(x),f(f(x)))) is a good pair whose ball

A nice example of a gum space that is not trivially an ultrametric space is a collection of functions A → B whose "distance from one another" is just the

*subset of A on which they have different outputs*. Here the powerset of A is standing in for**R**^{≥0}as the type of "how different can two things be" --- it's actually telling you*how*they're different, not just how much.Formally (X, d, (P, ≤, 0)) is a gum space with underlying set X if (P, ≤, 0) is a poset with least element 0, and d : X × X → P has the properties

- (
**Zero**) d(x,y) = 0 iff x = y - (
**Symmetry**) d(x,y) = d(y,x) - (
**Ultrametric Inequality**) if d(x,y) ≤ p and d(y,z) ≤ p, then d(x,z) ≤ p

Exercise: when P=

**R**^{≥ 0}with the usual order, then the ultrametric inequality is the same thing as requiring d(x,z) ≤ max(d(x,y), d(y,z)).However, and I think this is rather cute, there's still enough machinery in them to prove a standard contracting-map fixed point theorem.

Define a contracting map from one gum space to another to be a map f of the underlying sets such that d(f(x),f(y)) < d(x,y). Say a gum space is spherically complete if the intersection of any chain of balls in it is nonempty. (This condition is a stand-in for the usual notion of Cauchy-completeness)

**Theorem**(Priess-Crampe and Ribenboim '93) Any contracting map from a spherically complete gum space to itself has a unique fixed point.**Proof**(I couldn't find the original paper, so I just reconstructed the following proof from some guesswork, hints in papers that cited it, and vague memories of seeing the Banach fixed point theorem in a topology class long ago) Uniqueness of the fixed point follows immediately from the contracting condition on the function, and the zero axiom of gum spaces.For existence, make the following definitions. A pair (x,p) is a

*good pair*if d(x,f(x)) ≤ p. We abuse notation and let a good pair (x,p) stand for the closed ball B(x,p). So for instance a set S of good pairs is a*chain*if the set of closed balls {B(x,p)|(x,p) ∈ S} is a chain in the usual sense. Consider the collection of all chains of good pairs in the space, ordered by inclusion. The union of any chain of chains is again a chain, so Zorn's lemma yields a maximum chain C. By spherical completeness, ∩C is nonempty. We argue by contradiction that any point in ∩C must be a fixed point. For suppose otherwise; we have x ∈∩C such that f(x) ≠ x. First of all, also f(x) ∈ ∩C. Why? Let any good pair (y,r) ∈ C be given. We know that x ∈ B(y,r) so d(x,y) ≤ r. Because f is contracting, d(f(x),f(y)) < r. Because (y,r) is a good pair, d(y,f(y)) ≤ r.Putting these three facts together via the ultrametric inequality, d(x,f(x)) ≤ r, so f(x) ∈ B(y,r). Hence f(x) ∈ ∩C.

Now if we go just one step further iterating f on x, notice that d(f(x),f(f(x))) < d(x, f(x)) since f is contracting, so (f(x), d(f(x),f(f(x)))) is a good pair whose ball

*does not include x*, yet is contained in every ball in C. (This claim tacitly uses the weird ultrametric-specific fact that every point in a ball is its center) We have refuted the maximality of C, QED.