# 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

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

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 ${\displaystyle N_{r}(p)}$  consisting of all points q such that d(p,q) < r. The number r is called the radius of ${\displaystyle N_{r}(p)}$ . If the metric space is ${\displaystyle R^{k}}$  (here the metric is assumed to be the Euclidean metric) then ${\displaystyle N_{r}(p)}$  is known as the open ball with center p and radius r. The closed ball is defined for d(p,q) ${\displaystyle \leq }$  r.
2. A point p is a limit point of the set E if every neighbourhood of p contains a point q${\displaystyle \neq }$ p such that q ${\displaystyle \in }$  E.
3. If p ${\displaystyle \in }$  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 ${\displaystyle \subset }$  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 ${\displaystyle \in }$  X such that d(p,q) < M for all p ${\displaystyle \in }$  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 = ${\displaystyle N_{r}(p)}$ . Now if q ${\displaystyle \in }$  N then as d(p,q) < r we have h = r - d(p,q) > 0. Consider s ${\displaystyle \in }$  ${\displaystyle N_{h}(q)}$ . Now d(p,s) ${\displaystyle \leq }$  d(p,q) + d(q,s) < r - h + h = r, and so ${\displaystyle N_{h}(q)}$  ${\displaystyle \subset }$  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 ${\displaystyle N_{r}(p)}$  contains no point q of E such that q ${\displaystyle \neq }$  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 ${\displaystyle E^{c}}$ . We need to show that x ${\displaystyle \in E^{c}}$ . Now every neighborhood of x contains a point of ${\displaystyle E^{c}}$  so that x is not an interior point of E. Since E is open it means x ${\displaystyle \notin }$  E and so x ${\displaystyle \in E^{c}}$ . So ${\displaystyle E^{c}}$  is closed.
Now suppose ${\displaystyle E^{c}}$  is closed. Choose x ${\displaystyle \in }$  E. Then x ${\displaystyle \notin E^{c}}$ , and so x is not a limit point of ${\displaystyle E^{c}}$ . 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.