MHB Open Sets in a Discrete Metric Space .... ....

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In a discrete metric space open balls are either singleton sets or the whole space ...

Is the situation the same for open sets or can there be sets of two, three ... elements ... ?

If there can be two, three ... elements ... how would we prove that they exist ... ?

Essentially, given the metric or distance function, I am struggling to see how in forming a set of the union of two (or more) singleton sets you can avoid including other elements of the space ...

Peter
 
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As you say, open balls are either singleton sets or the entire space. But the union of any collection of open sets are open. Since any singleton sets are open balls (so open sets) any union of singleton sets is open. But any set is a union of singleton sets! Therefore every set is open in the discrete metric. (And every set is closed.)
 
HallsofIvy said:
As you say, open balls are either singleton sets or the entire space. But the union of any collection of open sets are open. Since any singleton sets are open balls (so open sets) any union of singleton sets is open. But any set is a union of singleton sets! Therefore every set is open in the discrete metric. (And every set is closed.)
Thanks HallsofIvy ...

Appreciate your help ...

Peter
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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