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 Mar23 awarded Good Question Feb5 comment Prove this alternative formula for derivative $f'(x)$ You can calculate the limit of each term seperately and in the second computation you replace $h$ with $-h$ (both go to $0$ as $h$ goes to $0$) Dec16 comment Complex projective space and its dual are homeomorphic? I believe you can find this theorem in Milnor & Stasheff characteristic classes book chapter 5. Dec15 awarded Informed Dec14 awarded Caucus Nov16 accepted Good textbooks on combinatorics for self-study Nov8 awarded Disciplined Jul14 awarded Notable Question Jul2 awarded Curious Jun18 awarded Yearling Apr12 comment $k[x]/(x^n) \otimes_{k[x]} k[y]/(y^m)$ Well start by changing the name of the variable in one side from $x$ to $y$ so it becomes $k[x]\otimes k[y]$ Apr12 comment One to one mapping from $A(S_1)$ into $A(S_2)$ Elements of $f\in A(S^1)$ are 1-1 functions from $f:S^1\to S^1$ and $\phi:S^1\to S^2$ is 1-1, what can you say about their composition $\phi\circ f$ Mar26 comment Why is the fibre of each point compact? The fiber is a closed subset of a compact space hence compact. It is closed because it is the inverse image of the closed set $\{x\}$ by a continuous function. Mar25 comment Trivialization of the normal bundle of a knot @Lepanais Yes the Mobius band is the unique non orientable line bundle over $S^1$. Mar24 revised Trivialization of the normal bundle of a knot added 8 characters in body Mar24 revised Trivialization of the normal bundle of a knot edited body Mar24 answered Trivialization of the normal bundle of a knot Mar17 revised Prove that $\frac{2^n}{n!}$ converges 0. added 3 characters in body Mar16 comment Example of a non-measurable set of a particular kind I see now what is the problem, for the a proof that every Riemann integrable function is measurable check folland's book. I think he gives a proof by constructing measurable functions such that $f$ is their pointwise limit. Mar16 comment Example of a non-measurable set of a particular kind This is not possible because there is a theorem called Lebesgue integrability criterion tells you that the set of discontinuities of $f$ are of measure 0 (hence any subset of it is measurable)