The set of all open balls of a metric space are able to generate a topology and are a basis for that topology
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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
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]
- we have both of the following conditions:
- [ilmath]\bigcup\mathcal{B}=X[/ilmath] (or equivalently: [ilmath]\forall x\in X\exists B\in\mathcal{B}[x\in B][/ilmath]^{[Note 1]}) and
- [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]^{[Note 2]}
- 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.^{[Note 3]}
Proof that the requisite conditions are met:
- [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]
- Recall, by definition of an open ball that [ilmath][u\in B_\delta(v)]\iff[d(u,v)<\delta][/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)
- Let [ilmath]\epsilon>0[/ilmath] be given (with [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]
- Let [ilmath]p\in X[/ilmath] be given.
- This completes the proof.
- 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
- Lemma: [ilmath]\forall p\in X\forall\epsilon>0[p\in B_\epsilon(x)][/ilmath]
- Since [ilmath]x\in X[/ilmath] was arbitrary we have shown it for all [ilmath]x[/ilmath] - as required. This completes part 1
- Let [ilmath]x\in X[/ilmath] be given
- [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 [ilmath]U[/ilmath] and [ilmath]V[/ilmath] have non-empty intersection
- We must now show [ilmath]\forall x\in U\cap V\exists B\in\mathcal{B}[x\in W\wedge W\subseteq U\cap V][/ilmath]
- Let [ilmath]x\in U\cap V[/ilmath] be given
- By If the intersection of two open balls is non-empty then for every point in the intersection there is an open ball containing it in the intersection we see that
- There exists an open ball [ilmath]W[/ilmath] such that [ilmath]x\in W[/ilmath] ([ilmath]W[/ilmath] is in fact centred at [ilmath]x[/ilmath]) and [ilmath]W\subseteq U\cap V[/ilmath]
- As required
- By If the intersection of two open balls is non-empty then for every point in the intersection there is an open ball containing it in the intersection we see that
- Since [ilmath]x\in U\cap V[/ilmath] was arbitrary we have shown it for all
- Let [ilmath]x\in U\cap V[/ilmath] be given
- We've shown it
- We must now show [ilmath]\forall x\in U\cap V\exists B\in\mathcal{B}[x\in W\wedge W\subseteq U\cap V][/ilmath]
- Now in either case the logical implication has been shown to hold
- Since [ilmath]U,V\in\mathcal{B} [/ilmath] were arbitrary we have shown it for all open balls
- Let [ilmath]U,\ V\in\mathcal{B} [/ilmath] be given.
Thus [ilmath]\mathcal{B} [/ilmath] suitable to generate a topology and be a basis for that topology
Discussion of result
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Notes
- ↑ By the implies-subset relation [ilmath]\forall x\in X\exists B\in\mathcal{B}[x\in B][/ilmath] really means [ilmath]X\subseteq\bigcup\mathcal{B} [/ilmath], as we only require that all elements of [ilmath]X[/ilmath] be in the union. Not that all elements of the union are in [ilmath]X[/ilmath]. However:
- [ilmath]\mathcal{B}\in\mathcal{P}(\mathcal{P}(X))[/ilmath] by definition. So clearly (or after some thought) the reader should be happy that [ilmath]\mathcal{B} [/ilmath] contains only subsets of [ilmath]X[/ilmath] and he should see that we cannot as a result have an element in one of these subsets that is not in [ilmath]X[/ilmath].
- We then use Union of subsets is a subset of the union (with [ilmath]B_\alpha:\eq X[/ilmath]) to see that [ilmath]\bigcup\mathcal{B}\subseteq X[/ilmath] - as required.
- ↑ We could of course write:
- [ilmath]\forall U,V\in\mathcal{B}\ \forall x\in \bigcup\mathcal{B}\ \exists W\in\mathcal{B}[(x\in U\cap V)\implies(x\in W\wedge W\subseteq U\cap V)][/ilmath]
- ↑ Suppose that [ilmath]U,V\in\mathcal{B} [/ilmath] are given but disjoint, then there are no [ilmath]x\in U\cap V[/ilmath] to speak of, and [ilmath]x\in W[/ilmath] may be vacuously satisfied by the absence of an [ilmath]X[/ilmath], however:
- [ilmath]x\in W\subseteq U\cap V[/ilmath] is taken to mean [ilmath]x\in W[/ilmath] and [ilmath]W\subseteq U\cap V[/ilmath], so we must still show [ilmath]\exists W\in\mathcal{B}[W\subseteq U\cap V][/ilmath]
- This is not always possible as [ilmath]W[/ilmath] would have to be [ilmath]\emptyset[/ilmath] for this to hold! We do not require [ilmath]\emptyset\in\mathcal{B} [/ilmath] (as for example in the metric topology)
- [ilmath]x\in W\subseteq U\cap V[/ilmath] is taken to mean [ilmath]x\in W[/ilmath] and [ilmath]W\subseteq U\cap V[/ilmath], so we must still show [ilmath]\exists W\in\mathcal{B}[W\subseteq U\cap V][/ilmath]
References
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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