DonAntonio
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102,075
98/100 score
 Feb23 awarded Enlightened Feb23 awarded Nice Answer Sep30 awarded Refiner Sep30 awarded Explainer Sep28 awarded Nice Answer Sep24 awarded Autobiographer Sep21 awarded Good Answer Sep14 awarded induction Aug14 awarded Nice Answer Aug4 awarded Nice Answer Jul8 awarded Necromancer Jul7 awarded Good Answer Jul2 awarded Curious Jun29 awarded Nice Answer Jun25 awarded Revival Jun18 revised FC groups with infinite derived subgroup which are not constructed by direct product of finite groups. edited body Jun18 comment Any Help would thanks : Product of continuous and Riemann integrable function Yes @user....but remember that first you have to prove $\;f\;$ is Riemann integrable (it is almost trivial). Jun18 comment Why must a field with a cyclic group of units be finite? I think I got it now, @JyrkiLahtonen: "since otherwise" means "if $\;char\neq 2\;$ then $\;-1\neq 1\;$ and ...etc. ". Thanks. Jun18 comment Integral domain (rings and fields) Perhaps the easiest and shortest way: we have that $\;\Bbb Z[i]\le\Bbb C\;$ , and since the last one is a field we're done. Jun17 comment Prove the cartesian product $K \times L \subset \mathbb R^{m+n}$ is also a compact set "bounded", not limited.