3 ms·

> because I'm not in like n dimensional non Euclidean space huh. I was sure that DF used taxicab distances

• pdpi 5y ago
  Taxicab distances are perfectly fine in Euclidean spaces
  • mananaysiempre 5y ago
    Many things go by the name of “Euclidean space” (basically the “flat” case of every kind of space, linear=vector, affine, topological, uniform, metric, topological manifold, algebraic variety, differential manifold, Riemannian, topological linear=vector, Banach manifold, countless others). I (and probably the GP) think of “R^n with taxicab distance” as a metric space, which is indeed non-isometric to the “Euclidean metric space” as in “R^n with Euclidean distance”, thus can meaningfully be called non-Euclidean in this context. You seem to be thinking of whether taxicab distance agrees with some preexisting, weaker structure on this structure’s notion of “Euclidean space”, like how the taxicab metric topology agrees with the topology of the “Euclidean topological space”, that is, “R^n with standard topology”, or how taxicab distance is translation-invariant thus an admissible norm on the “Euclidean vector space”, that is, “R^n with standard R-vector space structure”. In those senses taxicab distances are in fact fine in Euclidean spaces. (Note that this ceases to work as soon as you go into infinite dimensions, probably in any of the many, many ways to do so.)
    • pdpi 5y ago
      I think we're just working with slightly different definitions of "works fine". Isometry is a pretty strong definition of "works fine", but R^n with 1-norm is isomorphic and quasi-isometric to R^n with 2-norm which is still fairly strong.