I do not know if this is a known problem on here; excuse me if it is. It does not have an easy way to search for it.
Consider $X = \mathbb{R}^n$ as a normed vector space and $D$ a subset of $X$ which is also a subgroup with respect to $+$, so $v + w$ and $-v$ are in $D$ whenever $v,w \in D$. In particular, also $0 \in D$. This does not necessarily have to be a vector subspace: take for example $X = \mathbb{R}$ and $D = \mathbb{Q}$. We define a point $x \in X$ to be a limit point of a sequence $(x_n) = (x_1, x_2, \ldots)$ in $X$ if there exists a subsequence of $(x_n)$ converging to $x$.
Define a relation on the set of all sequences on $X$ as follows: $$(x_n) \sim (y_n) \iff \text{$(x_n - y_n)$ is bounded and all of its limit points are in $D$.}$$ Prove that $\sim$ is an equivalence relation.
I will answer the question myself. I post it because I believe it to be a good exercise for beginners in real analysis (change $X$ with $\mathbb{R}$ and $D$ with $\mathbb{Q}$, for example). Any comments are always welcome.
We prove the relation is an equivalence relation by proving it is reflexive, symmetric and transitive.
Reflexivity and symmetry are easy. $(x_n - x_n)$ is the constant zero sequence $0,0, \ldots$ which trivially converges to $0$ and so is bounded with only limit point $0 \in D$. Further, $(x_n - y_n)$ has a limit point $d \in D$ if and only if $(y_n - x_n) = -(x_n - y_n)$ has limit point $-d \in D$ because $$\lvert| (x_n - y_n) - d\rvert| = \lvert| (y_n - x_n) - (-d) \rvert|,$$ and if $\lvert| (x_n - y_n) \rvert| < M$ is bounded, then of course also $\lvert| (y_n - x_n) \rvert| < M$ by the same constant.
Transitivity requires a bit more work. Suppose $(x_n - y_n)$ and $(y_n - z_n)$ have their limit points in $D$ and they are both bounded, say in norm by $M_1$ and $M_2 > 0$. Because $x_n - z_n = (x_n - y_n) + (y_n - z_n)$, it is easily with the triangle inequality that $(x_n - z_n)$ is bounded in norm by $M_1 + M_2$. We know prove that $(x_n - z_n)$ has all its limit points in $D$.
For ease of notation, we let $a_n = x_n - y_n$, $b_n = y_n - z_n$ and $c_n = a_n + b_n$. By the previous observations, the sequences $(a_n)$, $(b_n)$ and $(c_n)$ are bounded and $(a_n)$ and $(b_n)$ are known to have their limit points in $D$. Let $c$ be a limit point of $(c_n)$. So, there exists a subsequence $( c_{n_k} )$ of $(c_n)$ which converges to $c$. Now boundedness is crucial; the subsequences of $(a_n)$ and $(b_n)$ are still bounded, so by the Bolzano-Weierstrass theorem we know that there exists a subsequence of $(a_{n_k})$ , say $(a_{n_{k_j}})$ which converges to some $a \in D$. And similarly $(b_{n_{k_j}})$ has a subsequence $b_{n_{k_{j_i}}} =: b_{m_i}$ which converges to some $b \in D$. Then the subsequences $(a_{m_i})$ and $(c_{m_i})$ still converge to $a$ and $c$ and crucially $a_{m_i} + b_{m_i} = c_{m_i}$. Taking $i \to \infty$ yields $a + b = c \in D$, which concludes the proof.