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Sep 14, 2017 at 20:01 vote accept Matthias Ludewig
Sep 13, 2017 at 7:06 answer added user43326 timeline score: 4
Sep 12, 2017 at 20:22 answer added Anton Fetisov timeline score: 11
Sep 12, 2017 at 18:37 history edited Matthias Ludewig CC BY-SA 3.0
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Sep 12, 2017 at 18:15 comment added Matthias Ludewig @Anton: Why would I require the delooping to be $(n-1)$-connected? I can surely take the $n$-fold loop space of a space that is not $(n-1)$-fold connected? Maybe this is the issue: If one makes this requirement, then one gets connective spectra, otherwise one gets any spectrum?
Sep 12, 2017 at 17:07 comment added Anton Fetisov You've got your definition of an $\infty$-loop space wrong. An $\infty$-loop space is a space admitting all deloopings (strictly speaking, equipped with a specific choice of them). An n-fold delooping of an n-fold loop space is by definition an $(n-1)$-connected basepointed space, so in your sequence each $Y_i$ must be $(i-1)$-connected, unlike the case of general spectra.
Sep 12, 2017 at 17:03 comment added Phil Tosteson Truncate $X_j$ to have no homotopy groups above degree $j$, and you get a connective spectrum with 0th space $Y_0$. The point is that there is a unique connective spectrum with $Y_0$ as it's zeroth space
Sep 12, 2017 at 16:38 history asked Matthias Ludewig CC BY-SA 3.0