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

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 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 $L^2=L$.

This is basically just a reformulation of Lorenz's criterion for almost convergent sequence, but since you're looking at connection between almost convergent ideals, I'll mention that a sequence is almost convergent to $L$ if and only if $$\Flim_n \Glim_k \frac{x_k+\dots+x_{k+n-1}}n=L$$ for any free ultrafilters $\mathcal F$, $\mathcal G$.

You can check the section on almost convergent sequences of the paper Jerison, Meyer, The set of all generalized limits of bounded sequences, Can. J. Math. 9, 79-89 (1957). ZBL0077.31004, MR83697. I have notes on some part of the paper where the results are expressed using ultrafilters (rather than using nets in Stone-Čech compactification) available here: notes, slides.

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

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 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 $L^2=L$.

This is basically just a reformulation of Lorenz's criterion for almost convergent sequence, but since you're looking at connection between almost convergent ideals, I'll mention that a sequence is almost convergent to $L$ if and only if $$\Flim_n \Glim_k \frac{x_k+\dots+x_{k+n-1}}n=L$$ for any free ultrafilters $\mathcal F$, $\mathcal G$.

You can check the section on almost convergent sequences of the paper Jerison, Meyer, The set of all generalized limits of bounded sequences, Can. J. Math. 9, 79-89 (1957). ZBL0077.31004, MR83697. I have notes on some part of the paper where the results are expressed using ultrafilters (rather than using nets in Stone-Čech compactification) available here: notes (Wayback Machine), slides (Wayback Machine).

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

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 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 $L^2=L$.

This is basically just a reformulation of Lorenz's criterion for almost convergent sequence, but since you're looking at connection between almost convergent ideals, I'll mention that a sequence is almost convergent to $L$ if and only if $$\Flim_n \Glim_k \frac{x_k+\dots+x_{k+n-1}}n=L$$ for any free filters $\mathcal F$, $\mathcal G$.

You can check the section on almost convergent sequences of the paper Jerison, Meyer, The set of all generalized limits of bounded sequences, Can. J. Math. 9, 79-89 (1957). ZBL0077.31004, MR83697. I have notes on some part of the paper where the results are expressed using ultrafilters rather than using nets available here: notes, slides.

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

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 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 $L^2=L$.

This is basically just a reformulation of Lorenz's criterion for almost convergent sequence, but since you're looking at connection between almost convergent ideals, I'll mention that a sequence is almost convergent to $L$ if and only if $$\Flim_n \Glim_k \frac{x_k+\dots+x_{k+n-1}}n=L$$ for any free ultrafilters $\mathcal F$, $\mathcal G$.

You can check the section on almost convergent sequences of the paper Jerison, Meyer, The set of all generalized limits of bounded sequences, Can. J. Math. 9, 79-89 (1957). ZBL0077.31004, MR83697. I have notes on some part of the paper where the results are expressed using ultrafilters (rather than using nets in Stone-Čech compactification) available here: notes, slides.

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 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 $L^2=L$.

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

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 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 $L^2=L$.

This is basically just a reformulation of Lorenz's criterion for almost convergent sequence, but since you're looking at connection between almost convergent ideals, I'll mention that a sequence is almost convergent to $L$ if and only if $$\Flim_n \Glim_k \frac{x_k+\dots+x_{k+n-1}}n=L$$ for any free filters $\mathcal F$, $\mathcal G$.

You can check the section on almost convergent sequences of the paper Jerison, Meyer, The set of all generalized limits of bounded sequences, Can. J. Math. 9, 79-89 (1957). ZBL0077.31004, MR83697. I have notes on some part of the paper where the results are expressed using ultrafilters rather than using nets available here: notes, slides.