Difference between revisions of "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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(Finishing proof)
m (Alec moved page Topology induced by a metric to The set of all open balls of a metric space are able to generate a topology and are a basis for that topology without leaving a redirect: Better name (this is a theorem, not defining the topology...)
 
(No difference)

Latest revision as of 22:16, 16 January 2017

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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 X be a set, let d:X×XR0 be a metric on that set and let (X,d) be the resulting metric space. Then we claim:

Proof

[Expand]Recall the definition of a topology generated by a basis

Proof that the requisite conditions are met:

  1. xXBB[xB]
    • Let xX be given
      • Lemma: pXϵ>0[pBϵ(x)]
        • Proof:
          • Let pX be given.
            • Let ϵ>0 be given (with ϵR>0 of course)
              • Recall, by definition of an open ball that [uBδ(v)][d(u,v)<δ]
                • Thus [pBϵ(p)][d(p,p)<ϵ]
              • Recall, by definition of a metric that [d(u,v)=0][u=v]
                • Thus d(p,p)=0
              • As ϵ>0 we see d(p,p)=0<ϵ, i.e. d(p,p)<ϵ thus pBϵ(p)
            • Since ϵ>0 was arbitrary we have shown pBϵ(p) for all ϵ>0 (ϵR>0 of course)
          • Since pX was arbitrary we have shown ϵ>0[pBϵ(p)] for all pX
        • This completes the proof.
      • By using the lemma above we see ϵ>0[xBϵ(x)]
        • In particular we see xB1(x) - there is nothing special about the choice of ϵ:=1 - we could have picked any ϵR>0
      • Choose B:=B1(x)
        • Note that BB by definition of B, explicitly: [Br(q)B][qXrR>0]
        • By our choice of B (and the lemma) we see xB
      • Our choice of B satisfies the requirements
    • Since xX was arbitrary we have shown it for all x - as required. This completes part 1
  2. U,VB[UVxUVBB[xWWUV]]

Thus B suitable to generate a topology and be a basis for that topology

Discussion of result

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Now we can link "open" in the metric sense to "open" in the topological sense. At long last.

Notes

  1. Jump up By the implies-subset relation xXBB[xB] really means XB, as we only require that all elements of X be in the union. Not that all elements of the union are in X. However:
    • BP(P(X)) by definition. So clearly (or after some thought) the reader should be happy that 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 BB[BP(X)] which is the same as (by power-set and subset definitions) BB[BX].
  2. Jump up We could of course write:
    • U,VB xB WB[(xUV)(xWWUV)]
  3. Jump up Suppose that U,VB are given but disjoint, then there are no xUV to speak of, and xW may be vacuously satisfied by the absence of an X, however:
    • xWUV is taken to mean xW and WUV, so we must still show WB[WUV]
      • This is not always possible as W would have to be for this to hold! We do not require B (as for example in the metric topology)

References

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Requires them, but is very widely known and borderline implicit!



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