Let $X$ be complete linear metric space. Is it true that if we remove from a dense subset $A$ of $X$ a subset which has cardinality less then cardinality of $A$ then we obtain dense subset of $X$ ? If not, what about Banach spaces?
Subsets of dense set
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general-topology
functional-analysis
1 Answers
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No. Consider $X=\mathbb R$ and $A=\mathbb Q\cup [0,1]$, and remove from $A$ the subset $\mathbb Q$.
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0@col: Suppose $B$ has cardinality $\lambda$ and $B$ is *not* nowhere dense (and we can assume $\lambda$ is infinite). Then $\overline B$ contains some open subset $U$ of $X$. If $x$ is in $U$, then $X=\cup_{n=1}^\infty n(U-x)$, so $\cup_{n=1}^\infty n(B-x)$ is dense in $X$ and has cardinality $\lambda$. – 2011-08-15