A slightly different argument using the sequence $x=(1,0,1,0,1,0,\dots)$.$\newcommand{\I}{\mathcal I}\newcommand{\Ilim}{\operatorname{\I-lim}}$

This sequence is almost convergent to $1/2$. At the same time, it is not difficult to show that if this sequence has $\I$-limit of some ideal $\I$, then the $\I$-limit can only be $0$ or $1$.

 * We can use the fact that $\I$-limit of a sequence is a [cluster point](https://en.wikipedia.org/wiki/Limit_point#For_Sequences_and_Nets) of that sequence. (This holds for any admissible ideal, i.e., for any ideal which contains all finite sets. If we allow also non-admissible ideals, then we can get cluster points or *terms of the sequence* as limits.)
 * For $\I$-convergence we have multiplicativity, i.e., $\Ilim (x_ny_n)=\Ilim x_n\cdot\Ilim y_n$. In particular, for our sequence $x$ we have $x^2=x$. Consequently, if $L$ is an $\I$-limit, then we get $L^2=L$.