I am trying to prove that the limit of a sequence along a non-principal ultrafilter on $\mathbb{N}$ is multiplicative and I am getting stuck. This is what I have until now.
Let $U$ be a non-principal ultrafilter on $\mathbb{N}$ and let $\boldsymbol{x}=(x_n)_n$ and $\boldsymbol{y}=(y_n)_n$ be two elements in $\ell^{\infty}$. Define a map $f$ such that $f(\boldsymbol{x})=U-\lim\limits_{n}x_n$. Then
$$
f(\boldsymbol{x})\cdot f(\boldsymbol{y})=U-\lim\limits_{n}x_n \cdot U-\lim\limits_{n}y_n=L_1 L_2,
$$
where $L_1,L_2$ are such that for every $\varepsilon >0$ $\{n\in \mathbb{N} : |x_n-L_1|< \varepsilon \}\subset U$ and $\{n\in \mathbb{N} : |y_n-L_2|< \varepsilon \}\subset U$.
Now,
$$
f(\boldsymbol{x}\boldsymbol{y})=U-\lim\limits_{n}x_ny_n=L,
$$
where $L$ is such that for every $\varepsilon >0$ $\{n\in \mathbb{N} : |x_ny_n-L|< \varepsilon \}\subset U$.
How do I conclude that $L_1 L_2 =L$?
Best Answer
HINT: Fix $\epsilon>0$. You know by the continuity of multiplication that there is a $\delta>0$ such that $|x_ny_n-L_1L_2|<\epsilon$ whenever $|x_n-L_1|<\delta$ and $|y_n-L_2|<\delta$. Let
$$u_x=\{n\in\Bbb N:|x_n-L_1|<\delta\}$$
and
$$u_y=\{n\in\Bbb N:|y_n-L_2|<\delta\}\,,$$
and consider $u=u_x\cap u_y$.