Let $A$ and $B$ be two subsets of $\mathbb{R}$ in the standard topology. Is it true that
$$\overline{A+B} = \overline{A} + \overline{B}$$
where $\overline{A}$ denotes the closure of $A$.
I see that it is true that $\overline{A} + \overline{B} \subset \overline{A+B}$ because standard $\mathbb{R}$ is a topological vector space. And I also see that the converse is generally not true.
But the counter example given in that previous answer uses $\mathbb{R}^2$. I kind of struggle to come up with a counter example or a proof.
If $A$ and $B$ are bounded, I proved it with the following.
For any $x \in \overline{A + B}$, consider the sequence $(a_n + b_n)_{n \in \mathbb{N}}$ such that $a_n + b_n \rightarrow x$ and $a_n \in A$ and $b_n \in B$.
By sequential compactness, find a convergent subsequence $(a_{n_k})_{k \in \mathbb{N}}$ which converges of $a$, and then another convergent subsequence of the subsequence above $(b_{n_{k_m}})_{m \in \mathbb{N}}$ which converges to $b$. Now,
$$a_{n_{k_m}} + b_{n_{k_m}} \rightarrow a + b = x$$
since sum of convergent sequence converges and the limit is unique. Thus, $x = a + b \in \overline{A} + \overline{B}$.
This proof obviously falls apart for unbounded $A$ and $B$. I am not very sure how to proceed.
