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May 31, 2011 at 15:43 comment added Philip Brooker and we have $Y_\alpha\cap Y_\beta=\{0\}$ when $\alpha\neq\beta$. Moreover, the closed span of all the $Y_\alpha$s is all of $C([0,\omega_1])$. Well, I am pretty sure all that is true - let me know if I have made a mistake! I haven't checked it closely but am pretty sure it is correct.
May 31, 2011 at 15:39 comment added Philip Brooker Tomek, let $\psi$ be a bijection of $\omega_1\times\omega_1$ onto the countable successor ordinals, $\phi$ a bijection of $\omega_1$ onto the countable limit ordinals, for $0<\alpha<\omega_1$ let $Y_\alpha$ be the closed span of $\{\chi_{\{\psi((\alpha,\eta))\}}\mid \eta<\omega_1\}\cup\{\chi_{(\alpha',\phi(\alpha)]}\}$, where $\alpha'=\min\{\mu\mid\exists\nu\mbox{ such that }\phi(\alpha)=\mu+\omega^\nu\}$, and let $Y_0$ be the closed span of $\{\chi_{\{\psi((0,\eta))\}}\mid \eta<\omega_1\}\cup\{\chi_{\{0\}}\}\cup\{\chi_{[0,\omega_1]}\}$. Then each $Y_\alpha$ is isomorphic to $c_0(\omega_1)$...
May 31, 2011 at 9:28 comment added Tomasz Kania I am wondering what is the answer if we take $K=[0,\omega_1]$. Can $Y$ (following the notation) be isomorphic to the whole space for some $Y_\alpha$s?
May 31, 2011 at 0:51 comment added Bill Johnson No problem with the tags, but the question is too vague.
May 31, 2011 at 0:22 comment added Zen Harper I added the Functional Analysis tag; apologies if you think it's not appropriate.
May 31, 2011 at 0:22 history edited Zen Harper
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May 30, 2011 at 13:55 history asked Wiktor Jaszak CC BY-SA 3.0