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| 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. Compare to the more fundamental definition of openness in a [[Analysis/TopologicalSpace|topological space]]. | This is sufficient to define a '''triangle inequality''': ''d(a,b) + d(b,c) ≥ d(a,c)'' for any points ''a'', ''b'', and ''c'' in the metric space. === 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)'' |
Metric Space
A metric space is a set for which a distance function can be defined.
Not to be confused with a measure space.
Description
A metric space is a set with a distance (or metric) function as d: M × M -> R.
For any p and q in M, the distance must feature these properties:
d(p,q) ≥ 0, and d(p,q) = 0 only if p = q
Symmetry: d(p,q) = d(q,p)
This is sufficient to define a triangle inequality: d(a,b) + d(b,c) ≥ d(a,c) for any points a, b, and c in the metric space.
Comparison to Topological Spaces
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 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 NA(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 Nr(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. Nr(p)
Choose any point q ∈ Nr(p)
Let a = d(p,q)
Note that d(p,q) = a < r
Let r' = r - a
Define the neighborhood of q as Nr'(q)
Choose any point s ∈ Nr'(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 Nr'(q) ⊆ Nr(p)
