Ultralimits vs limits of subsequences Let $(x_n)_{n \geq 1} \subset \mathbb{R}$ be a sequence of real numbers, and let $\omega \subset \mathcal{P}(\mathbb{N})$ be a non-principal ultrafilter. We say that $x$ is an ultralimit of $(x_n)_{n \geq 1}$ along $\omega$ if for all $\varepsilon > 0$ we have $\{ n \geq 1 : |x - x_n| < \varepsilon \} \in \omega$.
Suppose that there exists a set $\{ n_k : k \geq 1 \} \in \omega$ such that the subsequence $(x_{n_k})_{k \geq 1}$ has converges to $x$ (in the standard sense). Then it follows that $x$ is the ultralimit $(x_n)_{n \geq 1}$ along $\omega$, since $\omega$ is non-principal and so taking away finitely many elements from $(n_k)_{k \geq 1}$ we are still in $\omega$.
Conversely, if $x$ is the ultralimit of $(x_n)_{n \geq 1}$ along $\omega$, we can find a subsequence $(x_{n_k})_{k \geq 1}$ converging to $x$, for instance we can choose inductively $n_k > n_{k-1}$ such that $n_k \in \{ n \geq 1 : |x_n - x| < 1/k \}$. Although this subsequence is constructed using sets in $\omega$, it is not clear to me that the resulting set $\{ n_k : k \geq 1 \}$ should be in $\omega$.
So my question is: if $x$ is the ultralimit of $(x_n)_{n \geq 1}$ along $\omega$, can we always choose a subsequence $(x_{n_k})_{k \geq 1}$ converging to $x$ so that $\{ n_k : k \geq 1 \} \in \omega$?
 A: Consider the following counterexample: Let $(N_n)_n$ be a sequence in $\mathcal{P}(\mathbb{N})$ such that
$$ \mathbb{N} = N_1 \supseteq N_2 \supseteq N_3 \supseteq \dots $$
and for every $n \in \mathbb{N}$ we have
$$ | N_n \setminus N_{n+1} | = \aleph_0, \quad | N_n | = \aleph_0 \quad \text{ and } \quad \forall m \in \mathbb{N} \; \exists n \in \mathbb{N}: m \not \in N_n. $$
(Such a sequence can easily be obtained, for example by inductively throwing away half of the elements of $N_n$ starting from $N_1$ in a systematic way.)
Set $\Delta_n := N_n \setminus N_{n+1}$ for every $n \in \mathbb{N}$ and consider the family
$$ U := \{V \subseteq \mathbb{N}: \exists N \in \mathbb{N} \; \forall n \geq N: | \Delta_n \setminus V | < \aleph_0 \}. $$
Intuitively, $U$ contains all $V \subseteq \mathbb{N}$ such that for almost all $n \in \mathbb{N}$ the set $V$ contains almost all points in $\Delta_n$. It is clear that $U$ has the finite intersection property since the intersection of sets $V_1, \dots, V_m$, which for almost all $n \in \mathbb{N}$ contain almost all $k \in \Delta_n$, still satisfies this property. Thus, by a general theorem there is an ultrafilter $\mathcal{U} \subseteq \mathcal{P}(\mathbb{N})$ extending $U$.
Now, define the sequence $(x_n)_n \subseteq \mathbb{R}$ such that
$$ x_n = \frac{1}{k} \quad \text{if } n \in \Delta_k. $$
First of all it is to be shown that $0$ is an ultralimit of $(x_n)_n$ with respect to $\mathcal{U}$. For that let $\epsilon > 0$ and choose $n \in \mathbb{N}$ with $\frac{1}{n} < \epsilon$. Then
$$  \{n \in\mathbb{N}: |x_n| = x_n \leq \epsilon\} \supseteq 
\left\{n \in \mathbb{N}: x_n \leq \frac{1}{n}\right\} = N_n \in \mathcal{U}, $$
where $N_n \in \mathcal{U}$ holds since for all $m \geq n$ the set $N_n$ contains every element from $\Delta_m = N_m \setminus N_{m+1}$ as $N_n \supseteq N_m$.
We show that for every subsequence $(x_{n_k})_k$ converging to $0$ in the standard sense we have
$$ S := \{n_k: k \in \mathbb{N}\} \not \in \mathcal{U}. $$
In fact for every $n \in \mathbb{N}$ there are only finitely many $k \in \mathbb{N}$ such that
$$ \frac{1}{n+1} < x_{n_k} \leq \frac{1}{n} \quad \Leftrightarrow \quad n_k \in \Delta_n. $$
Hence for $S^C := \mathbb{N} \setminus S$ we find that for all $n \in \mathbb{N}$ the set $\Delta_n \setminus S^C = \Delta_n \cap S$ is finite, thus $S^C \in U \subseteq \mathcal{U}$. It follows that $S \not \in \mathcal{U}$.
