Def: a metric space $X$ is separable if it contains a countable dense subset.

Problem: I want to show that $\mathbb{R}^n$ is separable.

My Attempt: Can I show that $\mathbb{R}$ is separable? and then use (if this is true) the following result:

If $\{ A_i \}_{i \in I} $ are separable, then $\prod_{i \in I} A_i $ is separable?

  • $\begingroup$ $\mathbb R$ is not countable (you probably mean to ask whether $\mathbb R$ is separable, the answer on that is 'yes'). Do you know some countable dense subset of $\mathbb R$? It can help you by finding a countable dense subset of $\mathbb R^n$. $\endgroup$ – drhab Oct 7 '14 at 7:52
  • $\begingroup$ @drhab: "... it can help you to find a countable ...." :-) $\endgroup$ – orangeskid Oct 7 '14 at 7:54
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    $\begingroup$ @orangeskid My English is improving daily here ;) $\endgroup$ – drhab Oct 7 '14 at 7:56
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    $\begingroup$ Lamberto’s answer and @drhab’s comment point you towards a proof that $\Bbb R$ is separable. It’s not generally true that an arbitrary Cartesian product of separable spaces is separable, but it is true when the index set $I$ is finite, and the proof isn’t hard. (It’s actually true whenever $|I|\le\mathfrak{c}$, but the proof of that is hard.) $\endgroup$ – Brian M. Scott Oct 7 '14 at 8:14

$\Bbb R$ is separable: the subset $\Bbb Q \subset\Bbb R $ of rational numbers form, as is well known, a countable dense subset.

Further, $\forall \, n > 1$, $\Bbb Q^n \subset\Bbb R^n$ forms a countable set and yet is still dense.


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