If X and Y are metric spaces, we say that a function f from X is an embedding if f(x)f(y) = xy for all x,y in X.
Let P be an hereditary property of metric spaces, that is, if M is a P metric space, then any subspace of M is also P. Here are some examples of hereditary properties of metric spaces.
We say that an unordered triple (an unorded tuple may contain duplicates) of real numbers {x,y,z} is P-permissible if there is some P metric space M and points p,q,r in M such that pq = x, pr = y, and qr = z. Note that, regardless of P, the conditions x<=y+z, y<=x+z, and z<=x+y must be met in order for {x,y,z} to be P-permissible.
Exercise: List the set of all P-permissible triples, if P is the property Checkered and Distance bounded by 2. Answer.
Given any cardinality alpha, and given any hereditary property P of metric spaces, we say that a P metric space M is an alpha-universal P metric space if, given any P metric space Y of cardinality alpha, given any point y in Y, and given any embedding f from X=Y-y to M, f can be extended to an embedding of Y to M.
We use the phrase finitely universal to mean m universal for every finite cardinal m.
Exercise. Prove that universal is not an hereditary property of metric spaces.
Exercise. Prove that a countably universal metric space must be complete. (That is, every Cauchy sequence converges.)
Exercise. Find an example of a finitely universal metric space that is not complete.
Exercise. Prove the existence of finitely universal P metric spaces for all the hereditary properties mentioned above.
Exercise. Not every hereditary property has universal spaces. The property bounded is clearly hereditary. Prove that there is no finitely universal bounded metric space.
Exercise. Let M be the infinite hypercube, that is, the direct limit of all finite hypercubes. More formally, M is the set of all infinite sequences over the binary alphabet {0,1) where all but finitely many values are zero, with the Hamming metric. Prove, or disprove, that M is a finitely universal checkered space.