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(x, y) ≥ 0 (non-negativity)
- d(x, y) = 0 if and only if x = y (identity of indiscernibles)
- d(x, y) = d(y, x) (symmetry)
- 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.
Let X be a metric space. All points and sets are elements and subsets of X.
- A neighborhood of a point p is a set consisting of all points q such that d(p,q) < r. The number r is called the radius of . If the metric space is (here the metric is assumed to be the Euclidean metric) then is known as the open ball with center p and radius r. The closed ball is defined for d(p,q) r.
- A point p is a limit point of the set E if every neighbourhood of p contains a point qp such that q E.
- If p E and p is not a limit point of E then p is called an isolated point of E.
- E is closed if every limit point of E is a point of E.
- A point p is an interior point of E if there is a neighborhood N of p such that N E.
- E is open if every point of E is an interior point of E.
- E is perfect if E is closed and if every point of E is a limit point of E.
- E is bounded if there is a real number M and a point q X such that d(p,q) < M for all p E.
- E is dense in X every point of X is a limit point of E or a point of E (or both).
1. Every neighborhood is an open set
- Proof: Consider a neighborhood N = . Now if q N then as d(p,q) < r we have h = r - d(p,q) > 0. Consider s . Now d(p,s) d(p,q) + d(q,s) < r - h + h = r, and so 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 contains no point q of E such that q 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 . We need to show that x . Now every neighborhood of x contains a point of so that x is not an interior point of E. Since E is open it means x E and so x . So is closed.
- Now suppose is closed. Choose x E. Then x , and so x is not a limit point of . 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.