Famous Theorems of Mathematics/Analysis/Metric Spaces

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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×M

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 definitions

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 Nr(p) consisting of all points q such that d(p,q) < r. The number r is called the radius of Nr(p). If the metric space is Rk (here the metric is assumed to be the Euclidean metric) then Nr(p) is known as the open ball with center p and radius r. The closed ball is defined for d(p,q) r.
  2. A point p is a limit point of the set E if every neighbourhood of p contains a point qp such that q E.
  3. If p 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 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 X such that d(p,q) < M for all p 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 proofs

1. Every neighborhood is an open set

Proof: Consider a neighborhood N = Nr(p). Now if q N then as d(p,q) < r we have h = r - d(p,q) > 0. Consider s Nh(q). Now d(p,s) d(p,q) + d(q,s) < r - h + h = r, and so Nh(q) 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 Nr(p) 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 Ec. We need to show that x Ec. Now every neighborhood of x contains a point of Ec so that x is not an interior point of E. Since E is open it means x E and so x Ec. So Ec is closed.
Now suppose Ec is closed. Choose x E. Then x Ec, and so x is not a limit point of Ec. 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.

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