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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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up vote 5 down vote accepted

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 is somewhat related and interesting. – t.b. Aug 15 '11 at 11:34
@Theo: I was thinking about the same thing. I figured read again the question and then I saw that Jonas made sense :-) – Asaf Karagila Aug 15 '11 at 11:35
Theo: Thanks for clarifying. That sounds true to me, too, because a set with cardinality $\lambda$ would be nowhere dense. – Jonas Meyer Aug 15 '11 at 11:36
Right, that's it. Thanks! – t.b. Aug 15 '11 at 11:40
Thanks. How to see that if $A$ has the smallest cardinality κ then set with cardinality λ<κ will be nowhere dense? – col Aug 15 '11 at 12:00

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