Revision as of 20:38, 16 January 2017

Statement

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

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

Proof

Proof that the requisite conditions are met:

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

Thus $\mathcal{B}$ suitable to generate a topology and be a basis for that topology

Discussion of result

Notes

1. Jump up ↑ By the implies-subset relation $\forall x\in X\exists B\in\mathcal{B}[x\in B]$ really means $X\subseteq\bigcup\mathcal{B}$, as we only require that all elements of $X$ be in the union. Not that all elements of the union are in $X$. However:
   • $\mathcal{B}\in\mathcal{P}(\mathcal{P}(X))$ by definition. So clearly (or after some thought) the reader should be happy that $\mathcal{B}$ contains only subsets of $X$ and he should see that we cannot as a result have an element in one of these subsets that is not in $X$.
   Thus $\forall B\in\mathcal{B}[B\in\mathcal{P}(X)]$ which is the same as (by power-set and subset definitions) $\forall B\in\mathcal{B}[B\subseteq X]$.
   • We then use Union of subsets is a subset of the union (with $B_\alpha:\eq X$) to see that $\bigcup\mathcal{B}\subseteq X$ - as required.
2. Jump up ↑ We could of course write:
   • $\forall U,V\in\mathcal{B}\ \forall x\in \bigcup\mathcal{B}\ \exists W\in\mathcal{B}[(x\in U\cap V)\implies(x\in W\wedge W\subseteq U\cap V)]$
3. Jump up ↑ Suppose that $U,V\in\mathcal{B}$ are given but disjoint, then there are no $x\in U\cap V$ to speak of, and $x\in W$ may be vacuously satisfied by the absence of an $X$, however:
   • $x\in W\subseteq U\cap V$ is taken to mean $x\in W$ and $W\subseteq U\cap V$, so we must still show $\exists W\in\mathcal{B}[W\subseteq U\cap V]$
     • This is not always possible as $W$ would have to be $\emptyset$ for this to hold! We do not require $\emptyset\in\mathcal{B}$ (as for example in the metric topology)

References

OLD PAGE

For a metric space $(X,d)$ there is a topology which the metric induces on $x$ 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