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Let $A$ is a subset of $\mathbb{R}$ and the cardinality of $A$ is $2^\omega$. The question is this: Does the closure of $A$ in $\mathbb{R}$ have a nonempty interior in $\mathbb{R}$?

Added: Thank you for helps. In fact, I can't understand the proof of the lemma 2.11 in this paper. In line 7 of the proof, I don't know why $cl_R(A_{N,M})$ has a non-empty interior in $R$?

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No, for example the Cantor ternary set is nowhere dense, but has cardinality $2^\omega$. –  martini May 15 '12 at 4:37
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You should have provided that link to begin with, your apparent guess as to why it was true; by stripping the context, you've made the answer and hint useless for your true purpose. –  Arturo Magidin May 15 '12 at 5:08
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The above link does not work for me at the moment. It seems to be the paper: Arhangel'skii, A. V. ; Buzyakova, R. Z. The rank of the diagonal and submetrizability. Commentationes Mathematicae Universitatis Carolinae, vol. 47 (2006), issue 4, pp. 585-597. Link at dml. Or google to find many more copies of that paper online. –  Martin Sleziak May 15 '12 at 6:39

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As far as I believe, for your addition (as Arturo remarks, you should have began with that): $\mathbb R$ is Polish (as a $G_\delta$ in $\mathbb R$) and therefore a Baire space. As we have $\mathbb R \setminus \mathbb Q = \bigcup_{n,m}\mathrm{cl}_{\mathbb R \setminus \mathbb Q} A_{n,m}$ one of this sets has non-empty interior, $\mathrm{cl}_{\mathbb R \setminus \mathbb Q}A_{N,M}$, say. But now $\mathrm{cl}_{\mathrm R} A_{N,M}$ has non-empty interior in $\mathbb R$. For if $(x-\epsilon, x+\epsilon) \cap {\mathbb R \setminus \mathbb Q} \subset \mathrm{cl}_{\mathbb R \setminus \mathbb Q} A_{N,M}$, then $(x-\epsilon, x+\epsilon) \subseteq \mathrm{cl}_{\mathbb R} A_{N,M}$.

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Hint. Have you heard of the Cantor set?

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Thank you for helps. In fact, I can't understand the proof of the lemma 2.11 in this [paper][1]. In line 7 of the proof, I don't know why $cl_R(A_{N,M})$ has a non-empty interior in $R$? –  Paul May 15 '12 at 4:59

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