Weltschmerz
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 Apr17 accepted Smooth maps (between manifolds) are continuous (comment in Barrett O'Neill's textbook) Apr14 awarded Enlightened Apr14 awarded Nice Answer Feb24 awarded Popular Question Feb15 comment Every collection of disjoint non-empty open subsets of $\mathbb{R}$ is countable? @Fardad I did use the Axiom of Choice. Namely, when I say, pick a rational number from each set. Feb13 revised $G/Z(G)$ is cyclic then is abelian? added 13 characters in body Feb13 revised $G/Z(G)$ is cyclic then is abelian? added 6 characters in body Feb13 answered $G/Z(G)$ is cyclic then is abelian? Feb11 revised a countable dense subset of Lipschitz functions added 106 characters in body Feb11 comment a countable dense subset of Lipschitz functions That's very good, thanks. Do you reckon it'd be true if $X$ were compact? Feb10 comment a countable dense subset of Lipschitz functions $q_2$ is a rational number between 0 and 1. Feb10 comment a countable dense subset of Lipschitz functions Isn't $q_1\leq h_{q_1,q_2,k,y}(x)\leq k< 1 \ \forall x$ by definition, implying the functions in $\mathcal{D}$ are bounded? Feb10 revised a countable dense subset of Lipschitz functions added 2 characters in body Feb10 comment a countable dense subset of Lipschitz functions @copper.hat I've rephrased the question. I don't know if it's true. Feb10 revised a countable dense subset of Lipschitz functions added 38 characters in body Feb10 comment a countable dense subset of Lipschitz functions @copper.hat I've edited the question, thanks. Feb10 revised a countable dense subset of Lipschitz functions added 152 characters in body Feb10 revised a countable dense subset of Lipschitz functions added 20 characters in body Feb10 asked a countable dense subset of Lipschitz functions Jan13 comment Normalizer of a subgroup generated by a cycle. @D_S From the way you define $r$ and $s$ it seems $r=s$ because the number of elements in the same conjugacy class as $(1234)$ is exactly the number of elements with the same cycle type of $(1234)$ (in $S_4$).