I think it can be shown for the Euclidean metric in an $\mathbb{R}^n$ set, since $\|x-y\|=\|(x+a)-(y+a)\|$ but is it true for any metric space? Let's say the discrete metric space?
2026-02-22 21:04:11.1771794251
If the set $A$ is open in $X$, is the set $\{x+y : x\in A \}$ also open for a given $y \in X$ under any metric space?
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Note that addition is not defined on an arbitrary metric space.
It is true for any normed vectors space because $$B(x,r) + y = B(x+y, r)$$
so if for $x\in A$ we have $B(x,r) \subseteq A$ then likewise for $x+y\in A+y$ we have $B(x+y, r) \subseteq A+y$.