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Jun 4, 2018 at 23:06 comment added skd The last part of the last sentence in my comment above is not exactly correct: one also has $\langle I_\mathbf{Z} \rangle \leq \bigvee_n \langle K(n) \rangle$.
Jun 4, 2018 at 17:12 vote accept skd
Jun 4, 2018 at 16:19 history edited Drew Heard CC BY-SA 4.0
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Jun 4, 2018 at 12:38 comment added skd Awesome, thanks for the references! This means that the question is probably really hard to solve; (3) is an infinite height analogue of the telescope conjecture. Indeed, one equivalent way to rephrase the telescope conjecture is that if the thick subcategory of $E$-acyclics consists of finite spectra of type $n+1$, then $\langle E \rangle \geq \langle K(n) \rangle$. Moreover, $\langle I \rangle \leq \bigvee_n \langle K(n) \rangle$, and I think this is the only class which is less than or equal to $\bigvee_n \langle K(n) \rangle$.
Jun 4, 2018 at 8:44 comment added Drew Heard I just noticed your comments to the question - it seems you have already noticed that (1) is equivalent to (3)!
Jun 4, 2018 at 8:43 history answered Drew Heard CC BY-SA 4.0