$1_{A}$, the characteristic function of the set $A \subset \mathbb{R}$, is defined by:
$$ 1_{A}(x)=\begin{cases} 1,& \text{ for } x \in A\\ 0,& \text{ for } x \notin A \end{cases} $$ Show that $\sideset{^*}{}(1_{A}) = 1_{ \sideset{^*}{}{A}}$
This is exercise (3.36) from Sousa Pinto's Infinitesimal Methods for Mathematical Analysis.
I was trying to prove the equality, but I found out that: $$ (1,0,1,0,1,...) \in \sideset{^*}{}(1_{A})(\sideset{^*}{}{\mathbb{R}})\\ (1,0,1,0,1,...) \notin 1_{ \sideset{^*}{}{A}}(\sideset{^*}{}{\mathbb{R}}) $$ because
$$ 1_{ \sideset{^*}{}{A}} = \begin{cases} (1,1,1,...),& \text{ for } \sideset{^*}{}{x} \in \sideset{^*}{}{A} \\ (0,0,0,...), & \text{ for } \sideset{^*}{}{x} \notin \sideset{^*}{}{A} \end{cases} $$ and $(1,0,1,0,1,...)\neq(1,1,1,...) \land (1,0,1,0,1,...)\neq(0,0,0,...)$, is it right?
This is incorrect, assuming these sequences are supposed to be representing elements of an ultrapower. Either the set of even numbers or the set of odd numbers is in your ultrafilter. In the first case, $(1,0,1,0,1,\dots)=(0,0,0,\dots)$ and in the second case $(1,0,1,0,1,\dots)=(1,1,1,\dots)$.