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Dec
16
comment Complex projective space and its dual are homeomorphic?
I believe you can find this theorem in Milnor & Stasheff characteristic classes book chapter 5.
Dec
15
awarded  Informed
Dec
14
awarded  Caucus
Nov
16
accepted Good textbooks on combinatorics for self-study
Nov
8
awarded  Disciplined
Jul
14
awarded  Notable Question
Jul
2
awarded  Curious
Jun
18
awarded  Yearling
Apr
12
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 $k[x]\otimes k[y]$
Apr
12
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 $\phi\circ f$
Mar
26
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.
Mar
25
comment Trivialization of the normal bundle of a knot
@Lepanais Yes the Mobius band is the unique non orientable line bundle over $S^1$.
Mar
24
revised Trivialization of the normal bundle of a knot
added 8 characters in body
Mar
24
revised Trivialization of the normal bundle of a knot
edited body
Mar
24
answered Trivialization of the normal bundle of a knot
Mar
17
revised Prove that $\frac{2^n}{n!}$ converges 0.
added 3 characters in body
Mar
16
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.
Mar
16
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)
Mar
12
answered Lie group, Lie algebra and surjectivity
Mar
4
comment Limit of a function, not using L'Hospital's Rule
This is just the derivative of $f(x)=x^x$ at $a$, to evaluate it write $f(x)=e^{x\log x}.$