> A non-empty complete metric space is NOT the countable union of nowhere-dense closed sets.

I don't understand the following point the proof: we assume that $X$ is the countable union of nowhere-dense closed sets. $$ X=\bigcup_{n=1}^\infty C_n $$ and we can choose $x_1 \in A_1=(C_1)^c \,\Rightarrow\exists\,\, \varepsilon_1 <1 : \overline{B(x_1,\varepsilon_1)}\subset A_1$ because $C_1$ has no interior points. What I don't get is the last "because". Someone can help me in understanding this passage of the proof? Fortunately, the rest of the proof is clear to me.


1 Answer 1


$C_1$ has no interior points means that $X\setminus C_1$ is nonempty, more, every nonempty open set intersects $X\setminus C_1$.

From the non-emptyness, we obtain an $x_1 \in X\setminus C_1$, and since $C_1$ is closed, we have a $\rho_1 > 0$ with $B_{\rho_1}(x_1) \subset X\setminus C_1$. For every $0 < \varepsilon_1 < \rho_1$, we know $\overline{B_{\varepsilon_1}(x_1)} \subset B_{\rho_1}(x_1) \subset X\setminus C_1$.

In the following steps, the fact that $C_{n+1}$ is nowhere dense yields $B_{\varepsilon_n}(x_n) \cap X\setminus C_{n+1} \neq \varnothing$.

  • $\begingroup$ Thank you very much for the anwer! $\endgroup$
    – rusca91
    Dec 21, 2013 at 16:38

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