Assume that $ \omega$ is a selective nonprinciple ultrafilter. Fix $\omega$ : If $S$ is infinite subset in $\mathbb{N}$ then $\omega(S)=1$ And if $S$ is a finite subset in $\mathbb{N}$, then $\omega(S)=0$.
Def : If $X$ is a metric space and $x_n\in X$, then $$ \omega\ \bigg\{ n\bigg| |x_\omega -x_n|<\epsilon \bigg\} =1$$ for any $\epsilon$ iff $$x_\omega =\lim_{n\rightarrow \omega}\ x_n$$
Def : Define $X_\omega$ to be set of equivalence classes of sequence $(x_n),\ x_n\in X_n$ where $X_n$ is a metric space. And we have a metric $d_\omega$ on $X_\omega$ $$ (x_n)\sim (y_n)\Leftrightarrow \lim_{n\rightarrow \omega}\ |x_n-y_n|=0 $$
Question : Here I have a confusing. Assume that $X_n=\{p,q\}$ with $d(p,q)=1$. If $x_{2n}=p,\ x_{2n+1}=q$, then consider three classes $(x_n),\ (p),\ (q)$ Then $$\lim_{n\rightarrow \omega}\ |x_n-p|=\lim_{n\rightarrow \omega}\ |x_n-q|=0$$
since $$ \omega\ \{n||x_{n}-p|=0\}=1$$
And $$ \lim_{n\rightarrow \omega}\ |p-q|=1$$
So $$ 0=d_\omega((x_n),(p) ) + d_\omega ((x_n),(q)) \geq d_\omega ((p),(q))=1 $$
Am I miss something ? Thank you in advance.
The problem is that your $\omega$ is not an ultrafilter. An ultrafilter $\omega$ must satisfy $\omega(A\cap B)=\omega(A)\omega(B)$ for any $A,B\subseteq\mathbb{N}$. This is false for your $\omega$, since you can take $A$ and $B$ to be disjoint infinite sets and then $\omega(A)=\omega(B)=1$ but $\omega(A\cap B)=0$.