Assume that
$$
\sum_{k=1}^\infty\frac{\sin(H_k)}{kH_k}\tag{1}
$$
converges. Then
$$
\begin{align}
&\sum_{k=1}^\infty\frac{\sin(H_k)}{kH_k}\\
&=\sum_{k=1}^\infty\frac{\sin(H_k)}{H_k}(H_k-H_{k-1})\\
&=\sum_{k=1}^\infty\left(\frac{\sin(H_k)}{H_k}-\frac{\sin(H_{k+1})}{H_{k+1}}\right)H_k\\
&=\sum_{k=1}^\infty\frac{\left(H_k+\frac1{k+1}\right)\sin(H_k)-H_k\left(\sin(H_k)\cos\left(\frac1{k+1}\right)+\cos(H_k)\sin\left(\frac1{k+1}\right)\right)}{H_{k+1}}\\
&=\sum_{k=1}^\infty\frac{H_k\sin(H_k)\left(1-\cos\left(\frac1{k+1}\right)\right)+\frac1{k+1}\sin(H_k)-H_k\cos(H_k)\sin\left(\frac1{k+1}\right)}{…

Question: Is following subset of $\mathbb{R}^{3}$ also a subspace. Explain why and why not.
$$U=\left \{ (x,y,z)\in \mathbb{R}^{3}|z^{2}=x^{2}+y^{2} \right \}$$

I proved that the subset is not empty as it contains $\vec{0}$ and it satisfies the condition since 0^{2}=0^{2}+0^{2}
I also proved that the subset is closed under scalar multiplication since $\alpha ^{2}z^{2}=\alpha ^{2}\left ( x^{2}+y^{2} \right )\therefore z^{2}=x^{2}+y^{2},\alpha \in \mathbb{R}$

However, I am not sure whether it is closed under vector addition or not.

Suppose we have $f: A \backslash g(B)$ and $g: B \backslash f(A)$
then by complement def we have $f: A $ and $g:B$. Anything in B can't be retrieved by applying $f$ to elements of $A$. $f$ acts on $A$ and $G$ acts on $b$. Now suppose we have $g^{-1}: g(B) \backslash g \circ f(A)$ then $g^{-1}: g(b)$ We can use the $g^{-1}$ map to retrieve the elements of B that can't be gathered in A
$g(B \backslash f(A)) =g(B) - g \circ f(A)$ Similarly, we can use $f$ to map from $A \backslash g(B)$ which produces $f(A \backslash g(B) ) = f(A) - f \circ g(B)$. the elements in A can't be retrieved by applyi…