The set of all open balls of a metric space are able to generate a topology and are a basis for that topology

Grade: A

This page is currently being refactored (along with many others)

Please note that this does not mean the content is unreliable. It just means the page doesn't conform to the style of the site (usually due to age) or a better way of presenting the information has been discovered.
The message provided is:

This is one of the oldest pages on the wiki, created in Feb 2015. With a slight revision in Apr 2015 when Template:Theorem was moved to Template:Theorem Of. A very old page indeed!

Statement

Let [ilmath]X[/ilmath] be a set, let [ilmath]d:X\times X\rightarrow\mathbb{R}_{\ge 0} [/ilmath] be a metric on that set and let [ilmath](X,d)[/ilmath] be the resulting metric space. Then we claim:

• [ilmath]\mathcal{B}:\eq\left\{ B_\epsilon(x)\ \vert\ x\in X\wedge \epsilon\in\mathbb{R}_{>0}\right\} [/ilmath] satisfies the condition to generate a topology, for which it is a basis

Proof

Recall the definition of a topology generated by a basis

Let [ilmath]X[/ilmath] be a set and let [ilmath]\mathcal{B}\in\mathcal{P}(\mathcal{P}(X))[/ilmath] be any collection of subsets of [ilmath]X[/ilmath], then:

• [ilmath](X,\{\bigcup\mathcal{A}\ \vert\ \mathcal{A}\in\mathcal{P}(\mathcal{B})\})[/ilmath] is a topological space with [ilmath]\mathcal{B} [/ilmath] being a basis for the topology [ilmath]\{\bigcup\mathcal{A}\ \vert\ \mathcal{A}\in\mathcal{P}(\mathcal{B})\}[/ilmath]

if and only if

• we have both of the following conditions:
  1. [ilmath]\bigcup\mathcal{B}=X[/ilmath] (or equivalently: [ilmath]\forall x\in X\exists B\in\mathcal{B}[x\in B][/ilmath]) and
  2. [ilmath]\forall U,V\in\mathcal{B}\big[U\cap V\neq\emptyset\implies \forall x\in U\cap V\exists B\in\mathcal{B}[x\in W\wedge W\subseteq U\cap V]\big][/ilmath]
     • Caveat:[ilmath]\forall U,V\in\mathcal{B}\ \forall x\in U\cap V\ \exists W\in\mathcal{B}[x\in W\subseteq U\cap V][/ilmath] is commonly said or written; however it is wrong, this is slightly beyond just abuse of notation.

Proof that the requisite conditions are met:

1. [ilmath]\forall x\in X\exists B\in\mathcal{B}[x\in B][/ilmath]
   • Let [ilmath]x\in X[/ilmath] be given
     • Lemma: [ilmath]\forall p\in X\forall\epsilon>0[p\in B_\epsilon(x)][/ilmath]
       • Proof:
         • Let [ilmath]p\in X[/ilmath] be given.
           • Let [ilmath]\epsilon>0[/ilmath] be given (with [ilmath]\epsilon\in\mathbb{R}_{>0} [/ilmath] of course)
             • Recall, by definition of an open ball that [ilmath][u\in B_\delta(v)]\iff[d(u,v)<\delta][/ilmath]
               • Thus [ilmath][p\in B_\epsilon(p)]\iff[d(p,p)<\epsilon][/ilmath]
             • Recall, by definition of a metric that [ilmath][d(u,v)\eq 0]\iff[u\eq v][/ilmath]
               • Thus [ilmath]d(p,p)\eq 0[/ilmath]
             • As [ilmath]\epsilon > 0[/ilmath] we see [ilmath]d(p,p)\eq 0<\epsilon[/ilmath], i.e. [ilmath]d(p,p)<\epsilon[/ilmath] thus [ilmath]p\in B_\epsilon(p)[/ilmath]
           • Since [ilmath]\epsilon > 0[/ilmath] was arbitrary we have shown [ilmath]p\in B_\epsilon(p)[/ilmath] for all [ilmath]\epsilon>0[/ilmath] ([ilmath]\epsilon\in\mathbb{R}_{>0} [/ilmath] of course)
         • Since [ilmath]p\in X[/ilmath] was arbitrary we have shown [ilmath]\forall\epsilon>0[p\in B_\epsilon(p)][/ilmath] for all [ilmath]p\in X[/ilmath]
       • This completes the proof.
     • By using the lemma above we see [ilmath]\forall\epsilon>0[x\in B_\epsilon(x)][/ilmath]
       • In particular we see [ilmath]x\in B_1(x)[/ilmath] - there is nothing special about the choice of [ilmath]\epsilon:\eq 1[/ilmath] - we could have picked any [ilmath]\epsilon\in\mathbb{R}_{>0} [/ilmath]
     • Choose [ilmath]B:\eq B_1(x)[/ilmath]
       • Note that [ilmath]B\in\mathcal{B} [/ilmath] by definition of [ilmath]\mathcal{B} [/ilmath], explicitly: [ilmath][B_r(q)\in\mathcal{B}]\iff[q\in X\wedge r\in\mathbb{R}_{>0}][/ilmath]
       • By our choice of [ilmath]B[/ilmath] (and the lemma) we see [ilmath]x\in B[/ilmath]
     • Our choice of [ilmath]B[/ilmath] satisfies the requirements
   • Since [ilmath]x\in X[/ilmath] was arbitrary we have shown it for all [ilmath]x[/ilmath] - as required. This completes part 1
2. [ilmath]\forall U,V\in\mathcal{B}\big[U\cap V\neq\emptyset\implies \forall x\in U\cap V\exists B\in\mathcal{B}[x\in W\wedge W\subseteq U\cap V]\big][/ilmath]
   • Let [ilmath]U,\ V\in\mathcal{B} [/ilmath] be given.
     • Suppose [ilmath]U[/ilmath] and [ilmath]V[/ilmath] are disjoint. Then the logical implication holds regardless of the RHS, and we've shown the statement to be true in this case.
     • Suppose

OLD PAGE

For a metric space [ilmath](X,d)[/ilmath] there is a topology which the metric induces on [ilmath]x[/ilmath] that is the topology of all sets which are open in the metric sense.

---

TODO: Proof that open sets in the metric space have the topological properties

---