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Is every non-empty open set of a complete metric space uncountable ? If not can anyone please provide some examples of metric spaces (other than $\mathbb R$ with usual metric) in which every non-empty open set is uncountable ?

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  • $\begingroup$ There are some pretty small complete metric spaces. $\endgroup$
    – André Nicolas
    Jun 5, 2014 at 4:10
  • $\begingroup$ What about $(\{x\}, d)$ where $d : \{x\}\times\{x\} \to [0, \infty)$ is given by $d(x, x) = 0$? Such a metric space is complete as the only sequence is the constant sequence $x_n = x$. $\endgroup$
    – Michael Albanese
    Jun 5, 2014 at 4:10
  • $\begingroup$ @user123733: Do you mean an open set that is not uncountable? $\endgroup$
    – user99680
    Jun 5, 2014 at 4:21

2 Answers 2

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Every open set in $\mathbb R^n$ is uncountable: if $U$ is open in $\mathbb R^n$, then for every $x$ in $U$ there is a ball $B(x,r) \subset U ; r>0$ (since the open balls are a basis for the metric topology of $\mathbb R^n$). You can use, e.g., stereographic projection to show that these balls contain uncountably-many points. On the other extreme, in any discrete metric space , singletons are open ( as is any collection of points, which is also closed.) Maybe trivially, same thing applies for $\mathbb C^n$, since $\mathbb C^n$ is homeomorphic to $\mathbb R^{2n}$.

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To answer your first question

Is every non-empty open set of a complete metric space uncountable ?

This is false, there are in fact countable complete metric spaces. For instance the discrete metric on any countable set forms a complete metric space. Also something like $\{0,1,1/2,1/3,1/4,...\}$ with the Euclidean metric is a complete metric space. We can also find uncountable complete metric spaces that have countable non-empty sets by taking again the discrete metric on an uncountable set.

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