Metric induces topology: Difference between revisions

Latest revision as of 21:22, 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.

Proof

It suffices to show the following two things:

The space $X$ is the union of subsets of the form $B (x, r)$
Given two sets $B (x, r)$ and $B (y, s)$ , and any $z \in B (x, r) \cap B (y, s)$ , there exists $t > 0$ such that $B (z, t) \subset B (x, r) \cap B (y, s)$ (this suffices because $z \in B (z, t)$ for any $t$ , since $d (z, z) = 0$ ).

Proof that the union is the whole space

For any $x \in X$ , and any $r > 0$ , we have, because $d (x, x) = 0$ :

$x \in B (x, r)$

Thus, in particular, we have:

$x \in ⋃_{r > 0} B (x, r)$

Taking the union over all $x$ , we get:

$X = ⋃_{x \in X, r > 0} B (x, r)$

Proof for intersection of two

Consider two balls $B (x, r)$ and $B (y, s)$ , where $r, s > 0$ and $x, y \in X$ . (Note that $x, y$ may be equal). Suppose $z \in B (x, r) \cap B (y, s)$ . Then, by definition of the balls, we have:

$d (x, z) < r, d (y, z) < s$

Define:

$t : = min {r - d (x, z), s - d (y, z)}$

Then, $t > 0$ . We want to claim that $B (z, t)$ lies completely inside $B (x, r) \cap B (y, s)$ .

Let's prove this. Suppose $p \in B (z . t)$ . Then:

$d (z, p) < t \leq r - d (x, z)$

By the triangle inequality and an application of the above we have:

$d (x, p) \leq d (x, z) + d (z, p) < d (x, z) + r - d (x, z) = r$

Thus, $p \in B (x . r)$ . Analogously, $p \in B (y, s)$ . This completes the proof.

@@ Line 28: / Line 28: @@
 Note that this is the definition for a collection of subsets that can form the basis for ''some'' topology.
+==Proof==
+It suffices to show the following two things:
+* The space <math>X</math> is the union of subsets of the form <math>B(x,r)</math>
+* Given two sets <math>B(x,r)</math> and <math>B(y,s)</math>, and any <math>z \in B(x,r) \cap B(y,s)</math>, there exists <math>t > 0</math> such that <math>B(z,t) \subset B(x,r) \cap B(y,s)</math> (this suffices because <math>z \in B(z,t)</math> for any <math>t</math>, since <math>d(z,z) = 0</math>).
+===Proof that the union is the whole space===
+For any <math>x \in X</math>, and any <math>r > 0</math>, we have, because <math>d(x,x) = 0</math>:
+<math>x \in B(x,r)</math>
+Thus, in particular, we have:
+<math>x \in \bigcup_{r > 0} B(x,r)</math>
+Taking the union over all <math>x</math>, we get:
+<math>X = \bigcup_{x \in X, r > 0} B(x,r)</math>
+===Proof for intersection of two===
+Consider two balls <math>B(x,r)</math> and <math>B(y,s)</math>, where <math>r,s >0</math> and <math>x,y \in X</math>. (Note that <math>x,y</math> may be equal). Suppose <math>z \in B(x,r) \cap B(y,s)</math>. Then, by definition of the balls, we have:
+<math>d(x,z) < r, d(y,z) < s</math>
+Define:
+<math>t := \min \{ r - d(x,z), s - d(y,z) \}</math>
+Then, <math>t > 0</math>. We want to claim that <math>B(z,t)</math> lies completely inside <math>B(x,r) \cap B(y,s)</math>.
+Let's prove this. Suppose <math>p \in B(z.t)</math>. Then:
+<math>d(z,p) < t \le r - d(x,z)</math>
+By the triangle inequality and an application of the above we have:
+<math>d(x,p) \le d(x,z) + d(z,p) < d(x,z) + r - d(x,z) = r</math>
+Thus, <math>p \in B(x.r)</math>. Analogously, <math>p \in B(y,s)</math>. This completes the proof.