HerrWarum
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Recall that an equivalence relation $\sim$ on a topological space X is said to be $\textit{open}$ if for every open subset $A$ of $X$, the set $$[A] := \{x \in X | x \sim a \text{ for some a} \in A\}$$...
Answer to question 1: Observe that since $$a_1 \geq a_2 \geq a_3 \cdots \geq 0,$$we have $$a_3 + a_4 \geq a_4 + a_4,$$ and similarly $\underbrace{(a_{2^{k-1}+1}+...+a_{2^k})}_{\text{$2^{k-1}$terms}}... View answer 2 votes In the non abelian group that you wrote down, we have $$a_1 a_2 = a_3$$ and $$a_1 a_3 = a_4.$$ Using the first relation in the second gives $$a_1 a_1 a_2 = a_4$$ i.e, $$a_2 = a_4$$ which contradicts ... View answer Accepted answer 0 votes For Q1,$\Delta$is non-vanishing on the upper half plane: indeed, since$\Delta$is a cusp form of weight$12$, using the valence formula (aka the$k/12$formula) we get$ord_{\infty}(\Delta) = 1$, ... View answer 0 votes Yes. The boundary of a Jordan measurable set is a measure zero set. Consider a closed rectangle$R$that contains the given Jordan measurable set$A\$ (boundedness is assumed to be part of the ...