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Nov 5, 2011 at 13:38 vote accept Peter
Nov 5, 2011 at 13:38 vote accept Peter
Nov 5, 2011 at 13:38
Nov 3, 2011 at 22:05 history closed Bill Johnson
Andreas Blass
George Lowther
Matthew Daws
Ryan Budney
too localized
Nov 3, 2011 at 21:28 answer added Ramiro de la Vega timeline score: 3
Nov 1, 2011 at 20:12 comment added Ilya Bogdanov Well, in this case for $b\in B$ you can simply take a point $b'\in M_i$ such that $\|b-b'\|<2\rho(b,M_i)$. Now, for every $m\in M_i$ there exists $b\in B$ such that $\|m-b\|<\varepsilon$; then $\|m-b'\|<3\varepsilon$.
Nov 1, 2011 at 17:32 comment added Peter What does an almost orthogonal projection in a Banach space mean? Thank you.
Nov 1, 2011 at 16:09 comment added Ilya Bogdanov So, if we add the condition $|\theta|\leq \mu$ then the claim is true. Actually, you may take a dense set $B$ in $M$ and project it to $M_i$ almost orthogonally to obtain a dense set in $M_i$ of the same cardinality.
Nov 1, 2011 at 15:12 comment added Peter I see that even $dc(M)$ could decrease under these assumptions. My question is, if $dc(M)=\mu$ in general.
Nov 1, 2011 at 15:08 comment added Peter 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?
Nov 1, 2011 at 12:00 comment added Philip Brooker @Ilya: Indeed, the answer is yes under the additional assumption that $\vert \theta \vert \leq \mu$).
Nov 1, 2011 at 6:51 comment added Ilya Bogdanov @Philip: Perhaps it would hold if $\theta<\mu$?
Nov 1, 2011 at 0:59 comment added Philip Brooker 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).
Oct 31, 2011 at 22:51 history asked Peter CC BY-SA 3.0