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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?

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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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    Sorry about my previous comment. I interpreted the question this way, which makes more sense to me (and should be true, at least for Banach spaces): Suppose $X$ is a complete linear metric space and $\kappa$ is the smallest cardinality of a dense set $D$ (the density character). Is it true that removing $\lambda \lt \kappa$ elements from a dense set $D$ still leaves us with a dense set? This [answer from Bill Johnson on MO](http://mathoverflow.net/questions/72750/density-character-and-cardinality/72824#72824) is somewhat related and interesting.2011-08-15
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    @Theo: I was thinking about the same thing. I figured read again the question and then I saw that Jonas made sense :-)2011-08-15
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    Theo: Thanks for clarifying. That sounds true to me, too, because a set with cardinality $\lambda$ would be nowhere dense.2011-08-15
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    Right, that's it. Thanks!2011-08-15
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    Thanks. How to see that if $A$ has the smallest cardinality κ then set with cardinality λ<κ will be nowhere dense?2011-08-15
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    @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