= Metric Space = A '''metric space''' is a [[Analysis/Sets|set]] for which a distance [[Analysis/Functions|function]] can be defined. Not to be confused with a [[Analysis/MeasureSpace|measure space]]. <> ---- == Description == A metric space is the double ''(X,d)'' composed of: * a given space ''X'' * a '''distance''' or '''metric''' [[Analysis/Functions|function]] ''d: X × X -> [0,∞)'' The distance function must satisfy these conditions: for any points ''a'', ''b'', and ''c'' in ''X''... * ''d(a,b) ≥ 0'', and ''d(a,b) = 0'' only if ''a = b'' * symmetry: ''d(a,b) = d(b,a)'' * triangle inequality: ''d(a,b) + d(b,c) ≥ d(a,c)'' === Compactness === A set is '''compact''' when every cover has a finite subcover. In a metric space like ''R'', it is always possible to design an infinite series that converges to the limit point of a set without including that limit point. Crucially however, no finite subset of these can converge to that limit point. Therefore whenever a set does not include its own limit points, there is an infinite cover with no finite subcover. Thus, in a metric space... * all compact sets are closed. * given a compact set, all closed subsets are also compact. In ''R^n^'' specifically, per the '''Heine-Borel theorem''', a set being closed and bounded implies compactness, and vice versa. === Comparison to Topological Spaces === [[Analysis/TopologicalSpace|Topological spaces]] are a similar concept that use 'closeness' rather than a distance function. There are however some advantages to using a metric space. Within a metric space, a set is '''open''' if every point within the set can be perturbed in any direction and remain within the set. Clearly this is only ever ''not'' the case if a point is a [[Analysis/LimitPoint|limit point]] of a set, so an open set does not include any of its limit points while a closed set includes all of them. The topological definition of openness relies on a neighborhood topology function, i.e. for a set ''A'' the neighborhood of point ''p'' is given by ''N,,A,,(p)''. A similar function can be defined for metric spaces; a ball containing all points less than ''r'' distance away from point ''p'' is the neighborhood ''N,,r,,(p)''. This has a distinct advantage in that neighborhoods are always open. This is proven by demonstrating that for every point in a neighborhood, the neighborhood of that new point is a subset of the original neighborhood. * Consider the aforementioned neighborhood of ''p ∈ A'', i.e. ''N,,r,,(p)'' * Choose any point ''q ∈ N,,r,,(p)'' * Let ''a = d(p,q)'' * Note that ''d(p,q) = a < r'' * Let ''r' = r - a'' * Define the neighborhood of ''q'' as ''N,,r',,(q)'' * Choose any point ''s ∈ N,,r',,(q)'' * Note that ''d(q,s) < r' = r - a'' * Via the triangle inequality, ''d(p,q) + d(q,s) ≥ d(p,s)'' * Since ''d(q,s) < r - a'' and ''d(p,q) = a'', it follows that ''d(p,q) + d(q,s) < (r - a) + a = r'' * Altogether ''d(p,s) < r'', proving that ''N,,r',,(q) ⊆ N,,r,,(p)'' ---- CategoryRicottone