Would anyone happen to have a proof that the following use of $\equiv$ is an equivalence relation:
Let $\mathcal{F}(\mathbb{N})$ be an ultrafilter on $\mathbb{N}$, constructed by choosing subsets of $\mathbb{N}$ with large cardinality. Suppose $\lbrace s_n\rbrace$ and $\lbrace r_n\rbrace$ are real-valued sequences. Now, let $A$ be the set of all $n\in\mathbb{N}$ such that $r_n=s_n$. Then, $\lbrace s_n\rbrace$ and $\lbrace r_n\rbrace$ satisfy $\lbrace s_n\rbrace\equiv\lbrace r_n\rbrace$ if and only if $A\in\mathcal{F}(\mathbb{N})$.
Thank you in advance.
"constructed by chooising subsets of large cardinality" is somewhat inexact and misleading. You probably mean to say it is a so-called free ultrafilter, which is not of the form $\mathcal{F}_n=\{A\subseteq \mathbb{N}: n \in A\}$, for some $n \in \mathbb{N}$, which can be "constructed" (not really) by extending the filter of co-finite subsets (all subsets of $\mathbb{N}$ with finite complement) to an ultrafilter (which uses a form of the axiom of choice, and is far from constructive).
So let $\mathcal{F}$ be non-empty a(n ultra)filter (free or not, it does not matter for the proof) and define $(a_n)_n \equiv (b_n)_n$ iff $\{n : a_n = b_n\} \in \mathcal{F}$.
This is easily an equivalence relation: $(a_n) \equiv (a_n)_n$ follows from $\mathbb{N} \in \mathcal{F}$ for all non-empty filters (from closedness under supersets). Symmetry is clear from symmetry of $=$ itself:
$$\{n: a_n = b_n\} = \{n: b_n = a_n\}$$
And moreover if $\{n: a_n = b_n \} \in \mathcal{F}$ and $\{n: b_n = c_n\} \in \mathcal{F}$ then
$$\{n: a_n = b_n \} \cap \{n: b_n = c_n \} \subseteq \{n: a_n = c_n \}$$
so again the latter set is in the filter too. This shows transitivity.