# 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
* 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.

## Basic definitions

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 $N_{r}(p)$ 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) $\leq$ r.
* A point p is a limit point of the set E if every neighbourhood of p contains a point q$\neq$p such that q $\in$ E.
* If p $\in$ 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 $\subset$ 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 $\in$ X such that d(p,q) < M for all p $\in$ E.
* 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 = $N_{r}(p)$. 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) $\leq$ 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 $\neq$ 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 $E^{c}$. 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.