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Apr 23, 2019 at 11:47 vote accept ar.grig
Apr 23, 2019 at 11:17 comment added ar.grig @EmilJeřábek: if the answer is negative then there exists α-sequence with $\alpha>\omega_1$
Apr 23, 2019 at 11:05 history edited ar.grig CC BY-SA 4.0
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Apr 23, 2019 at 10:59 comment added ar.grig @Wojowu: infinitesimal $\omega$-sequence is the sequence with countably many members which converges to zero.
Apr 23, 2019 at 10:35 review Close votes
Apr 23, 2019 at 18:22
Apr 23, 2019 at 10:34 comment added Asaf Karagila Do you know about Hausdorff gaps?
Apr 23, 2019 at 10:12 answer added Wojowu timeline score: 2
Apr 23, 2019 at 9:58 comment added Johannes Schürz I believe infinitesimal sequence refers to the field-extension of $\mathbb{R}$ adding non-standard reals $r$ , i.e. $\forall q \in \mathbb{Q}^+ \colon 0< r<q$. The way you do this, you consider $\mathbb{Q}^\omega$ modulo eventually <.
Apr 23, 2019 at 9:51 comment added Emil Jeřábek A straightforward diagonalization argument shows that if $f$ is any sequence of rationals, and $G$ is a countable set of rational sequences such that $f\ll g$ for each $g\in G$ (meaning $\lim_n f(n)/g(n)=0$), then there exists a rational sequence $h$ such that $f\ll h$ and $h\ll g$ for all $g\in G$. Thus, for any fixed $f$, one can construct by induction a sequence $\{g_\alpha:\alpha<\omega_1\}$ such that $\alpha<\beta\implies f\ll g_\beta\ll g_\alpha$. This is likely a counterexample to the question, except that I don’t know what exactly “infinitesimal sequence” means.
Apr 23, 2019 at 9:39 answer added Johannes Schürz timeline score: 1
Apr 23, 2019 at 9:33 comment added Emil Jeřábek Is the title a typo? The question seems to only speak about $\alpha<\omega_1$.
Apr 23, 2019 at 9:27 comment added Wojowu What is an infinitesimal $\omega$-sequence?
Apr 23, 2019 at 8:53 history asked ar.grig CC BY-SA 4.0