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
 
I posted this question on math-stackexchange but apparently I asked something stupid and I was downvoted. I still don't have an answer to my question so I hope someone in here can help me or at least explain me why I am asking something stupid. I started studying Complex Analysis and came upon the following theorem which is a direct consequence of the Cauchy-Goursat theorem: Let ##f:D\to\mathbb{C}## be an anlytic function over a simply connected region ##D##. If ##a## and ##z## are part of...