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Let $\langle M_i:i<\theta\rangle$ be an increasing chain of Banach spaces, where each $M_i$ has density character $\mu$ (i.e.,the mininum cardinality of a dense subset of $M_i$ is $\mu$). Let $B_i\subset M_i$ be a dense subset of $M_i$ of cardinality $B_i$. Notice that $\bigcup_{i<\theta}B_i$ is a dense subset of $\overline{\bigcup_{i<\theta}M_i}=:M$, so $density-character(M)\le \mu$. Is it possible to prove that $density-character(M)=\mu$? Thank you.

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closed as too localized by Bill Johnson, Andreas Blass, George Lowther, Matthew Daws, Ryan Budney Nov 3 '11 at 22:05

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I think you probably mean "Let $B_i\subset M_i$ be a dense subset of $M_i$ of cardinality $\mu$" and "... so $density-character(M)\geq \mu$"; if this is what you meant, then the answer is very easily seen to be no. For this, take an increasing chain of separable closed subspaces in $\ell_2(\omega_1)$ whose union is all of $\ell_2(\omega_1)$ (you don't even need to take the closure). –  Philip Brooker Nov 1 '11 at 0:59
@Philip: Perhaps it would hold if $\theta<\mu$? –  Ilya Bogdanov Nov 1 '11 at 6:51
@Ilya: Indeed, the answer is yes under the additional assumption that $\vert \theta \vert \leq \mu$). –  Philip Brooker Nov 1 '11 at 12:00
Well, actually I mean that since $M_i$ has density character $\mu$, let $B_i\subset M_i$ be a dense subset of $M_i$ with such minimal cardinality $\mu$, it is straightforward to see that $\bigcup_{i<\theta}B_i$ is a dense subset of $M$ of size $\mu$ -since $|\theta|\le \mu$- and then $dc(M):=\min \{\lambda:$ there exists a dense subset $A$ of $M$ of size $\lambda\}\le \mu$. Even it holds if $|\theta|\le \omega$ in that example? –  Peter Nov 1 '11 at 15:08
I see that even $dc(M)$ could decrease under these assumptions. My question is, if $dc(M)=\mu$ in general. –  Peter Nov 1 '11 at 15:12

1 Answer 1

up vote 3 down vote accepted

If $X$ is any metric space and $Y$ is any subspace of $X$ then $dc(Y) \leq dc(X)$.

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Take $X:=\mathbb{R}$ and $Y:=\mathbb{I}$ with the usual metric. Notice that $dc(X)=\aleph_0$ and $dc(Y)=2^{\aleph_0}$. Does your claim hold if both $X$ and $Y$ are complete in the metric sense? –  Peter Nov 4 '11 at 20:00
Peter $dc(\mathbb{I})=\aleph_0$ since it is in fact second countable. The claim says "any metric space" and "any subspace". –  Ramiro de la Vega Nov 5 '11 at 14:01
Where can I find a reference about it? Or better, how do you prove that fact? (I mean, $Y\le X$ implies $dc(Y)\le dc(X)$)? –  Peter Nov 6 '11 at 20:56

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