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Apr
17
accepted Smooth maps (between manifolds) are continuous (comment in Barrett O'Neill's textbook)
Apr
14
awarded  Enlightened
Apr
14
awarded  Nice Answer
Feb
24
awarded  Popular Question
Feb
15
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.
Feb
13
revised $G/Z(G)$ is cyclic then is abelian?
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Feb
13
revised $G/Z(G)$ is cyclic then is abelian?
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Feb
13
answered $G/Z(G)$ is cyclic then is abelian?
Feb
11
revised a countable dense subset of Lipschitz functions
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Feb
11
comment a countable dense subset of Lipschitz functions
That's very good, thanks. Do you reckon it'd be true if $X$ were compact?
Feb
10
comment a countable dense subset of Lipschitz functions
$q_2$ is a rational number between 0 and 1.
Feb
10
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?
Feb
10
revised a countable dense subset of Lipschitz functions
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Feb
10
comment a countable dense subset of Lipschitz functions
@copper.hat I've rephrased the question. I don't know if it's true.
Feb
10
revised a countable dense subset of Lipschitz functions
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Feb
10
comment a countable dense subset of Lipschitz functions
@copper.hat I've edited the question, thanks.
Feb
10
revised a countable dense subset of Lipschitz functions
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Feb
10
revised a countable dense subset of Lipschitz functions
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Feb
10
asked a countable dense subset of Lipschitz functions
Jan
13
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 $S_4$).