I'm trying to solve exercise 6.3#7 from Sidney A. Morris' "Topology without tears": "Prove that each discrete space [...] is a Polish space."

I started by proving that discrete spaces are always completely metrizable with the discrete metric. But then I got stuck. As far as I know, the only dense subset of a discrete space is the whole space.

But does that not mean that only countable discrete spaces are separable (and therefore Polish) spaces?

  • 4
    $\begingroup$ You're right. Polish spaces are second countable by definition and an uncountable discrete space is not second countable. $\endgroup$ – t.b. Apr 3 '12 at 0:28
  • $\begingroup$ Thanks! Just a typo then. Do you want to write it as an answer so that I can give you credit? $\endgroup$ – jerico Apr 3 '12 at 0:30
  • 2
    $\begingroup$ Better yet, write your own answer and (after the necessary lapse of time) accept it. $\endgroup$ – Brian M. Scott Apr 3 '12 at 0:31

See t.b.'s comment above confirming my assumption.


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