Difference between revisions of "Discrete metric and topology/Summary"

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Revision as of 13:03, 27 July 2015

Property Comment
induced topology discrete topology - which is the topology [ilmath](X,\mathcal{P}(X))[/ilmath] (where [ilmath]\mathcal{P} [/ilmath] denotes power set)
Open ball [ilmath]B_r(x):=\{p\in X\vert\ d(p,x)< r\}=\left\{\begin{array}{lr}\{x\} & \text{if }r\le 1 \\ X & \text{otherwise}\end{array}\right.[/ilmath]
Open sets Every subset of [ilmath]X[/ilmath] is open.
Proof outline: as for a subset [ilmath]A\subseteq X[/ilmath] we can show [ilmath]\forall x\in A\exists r[B_r(x)\subseteq A][/ilmath] by choosing say, that is [ilmath]A[/ilmath] contains an open ball centred at each point in [ilmath]A[/ilmath].
Connected The topology generated by [ilmath](X,d_\text{discrete})[/ilmath] is not connected if [ilmath]X[/ilmath] has more than one point.
Proof outline:
  • Let [ilmath]A[/ilmath] be any non empty subset of [ilmath]X[/ilmath], then define [ilmath]B:=A^c[/ilmath] which is also a subset of [ilmath]X[/ilmath], thus [ilmath]B[/ilmath] is open. Then [ilmath]A\cap B=\emptyset[/ilmath] and [ilmath]A\cup B=X[/ilmath] thus we have found a separation, a partition of non-empty disjoint open sets, that separate the space. Thus it is not connected
  • if [ilmath]X[/ilmath] has only one point then we cannot have a partition of non empty disjoint sets. Thus it cannot be not connected, it is connected.
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