Metric induces topology: Difference between revisions

Revision as of 21:14, 19 July 2008

Statement

Suppose $(X, d)$ is a metric space. Then, the collection of subsets:

$B (x, r) : = {y \in X ∣ d (x, y) < r}$

form a basis for a topology on $X$ . These are often called the open balls of $X$ .

Definitions used

Metric space

Further information: metric space

A metric space $(X, d)$ is a set $X$ with a function $d : X \times X \to R$ satisfying the following:

$d (x, y) \geq 0 \forall x, y \in X$ (non-negativity)
$d (x, y) = 0 ⟺ x = y$ (identity of indiscernibles)
$d (x, y) = d (y, x)$ (symmetry)
$d (x, y) + d (y, z) \geq d (x, z) \forall x, y, z \in X$ (triangle inequality)

Basis for a topological space

Further information: Basis for a topological space

A collection of subsets ${U_{i}}_{i \in I}$ of a set $X$ is said to form a basis for a topological space if the following two conditions are satisfied:

$⋂_{i \in I} U_{i} = X$
For any $i, j \in I$ , and any $p \in U_{i} \cap U_{j}$ , there exists $U_{k} \subset U_{i} \cap U_{j}$ such that $p \in U_{k}$ .

Note that this is the definition for a collection of subsets that can form the basis for some topology.

@@ Line 7: / Line 7: @@
 form a [[basis]] for a topology on <math>X</math>. These are often called the ''open balls'' of <math>X</math>.
-==Proof==
+==Definitions used==
-To prove that the subsets form a basis for a topology, we need to prove the following fact: the intersection of two open balls is a union of open balls. Equivalently, given two open balls <math>B(x,r)</math> and <math>B(y,s)</math>, and <math>z \in B(x,r) \cap B(y,s)</math>, then there exists some radius <math>t</math> such that <math>B(z,t) \subset B(x,r) \cap B(y,s)</math>.
+===Metric space===
+{{further|[[metric space]]}}
-It turns out that the following works for <math>t</math>:
+A metric space <math>(X,d)</math> is a set <math>X</math> with a function <math>d:X \times X \to \R</math> satisfying the following:
-<math>t := \min \left( r - d(x,z), r - d(y,z) \right)</math>
+* <math>d(x,y) \ge 0 \ \forall \ x,y \in X</math> (non-negativity)
+* <math>d(x,y) = 0 \iff x = y</math> (identity of indiscernibles)
+* <math>d(x,y) = d(y,x)</math> (symmetry)
+* <math>d(x,y) + d(y,z) \ge d(x,z) \ \forall \ x,y,z \in X</math> (triangle inequality)
-This essentially follows from the triangle inequality.
+===Basis for a topological space===
+{{further|[[Basis for a topological space]]}}
+A collection of subsets <math>\{ U_i \}_{i \in I}</math> of a set <math>X</math> is said to form a '''basis for a topological space''' if the following two conditions are satisfied:
+* <math>\bigcap_{i \in I} U_i = X</math>
+* For any <math>i,j \in I</math>, and any <math>p \in U_i \cap U_j</math>, there exists <math>U_k \subset U_i \cap U_j</math> such that <math>p \in U_k</math>.
+Note that this is the definition for a collection of subsets that can form the basis for ''some'' topology.