Last modified on 4 October 2013, at 06:56

Famous Theorems of Mathematics/Analysis/Metric Spaces

A metric space is a tuple (M,d) where M is a set and d is a metric on M, that is, a function

d : M \times M \rightarrow \mathbb{R}

such that

  1. d(x, y) ≥ 0     (non-negativity)
  2. d(x, y) = 0   if and only if   x = y     (identity of indiscernibles)
  3. d(x, y) = d(y, x)     (symmetry)
  4. d(x, z) ≤ d(x, y) + d(y, z)     (triangle inequality).

The function d is also called distance function or simply distance. Often d is omitted and one just writes M for a metric space if it is clear from the context what metric is used.

Basic definitionsEdit

Let X be a metric space. All points and sets are elements and subsets of X.

  1. A neighborhood of a point p is a set N_r(p) consisting of all points q such that d(p,q) < r. The number r is called the radius of N_r(p). If the metric space is R^k (here the metric is assumed to be the Euclidean metric) then N_r(p) is known as the open ball with center p and radius r. The closed ball is defined for d(p,q) \le r.
  2. A point p is a limit point of the set E if every neighbourhood of p contains a point q\nep such that q \in E.
  3. If p \in E and p is not a limit point of E then p is called an isolated point of E.
  4. E is closed if every limit point of E is a point of E.
  5. A point p is an interior point of E if there is a neighborhood N of p such that N \subset E.
  6. E is open if every point of E is an interior point of E.
  7. E is perfect if E is closed and if every point of E is a limit point of E.
  8. E is bounded if there is a real number M and a point q \in X such that d(p,q) < M for all p \in E.
  9. E is dense in X every point of X is a limit point of E or a point of E (or both).

Basic proofsEdit

1. Every neighborhood is an open set

Proof: Consider a neighborhood N = N_r(p). Now if q \in N then as d(p,q) < r we have h = r - d(p,q) > 0. Consider s \in N_h(q). Now d(p,s) \le d(p,q) + d(q,s) < r - h + h = r, and so N_h(q) \subset N. Thus q is an interior point of N.

2. If p is a limit point of a set E, then every neighborhood of p contains infinitely many points of E

Proof: Suppose there is a neighborhood N of p which contains only a finite number of points of E. Let r be the minimum of the distances of these points from p. The minimum of a finite set of positive numbers is clearly positive so that r > 0. The neighborhood N_r(p) contains no point q of E such that q \ne p which contradicts the fact that p is a limit point of E.

3. A finite set has no limit points

Proof: This is obvious from the proof 2.

4. A set is open if and only if its complement is closed.

Proof: Suppose E is open and x is a limit point of E^c. We need to show that x \in E^c. Now every neighborhood of x contains a point of E^c so that x is not an interior point of E. Since E is open it means x \notin E and so x \in E^c. So E^c is closed.
Now suppose E^c is closed. Choose x \in E. Then x \notin E^c, and so x is not a limit point of E^c. So there must be a neighborhood of x entirely inside E. So x is an interior point of E and so E is open.

5. A set is closed if and only if its complement is open.

Proof: This is obvious from the proof 4.