Translation of Open Ball in Invariant Pseudometric on Vector Space

Theorem

Let $K$ be a field.

Let $X$ be a vector space over $K$.

Let $d$ be an invariant pseudometric on $X$.

For $x \in X$ and $\epsilon > 0$, let $\map {B_\epsilon} x$ be the open ball centered at $x$ with radius $\epsilon$.

Let $y \in X$.

Then:

$\map {B_\epsilon} x + y = \map {B_\epsilon} {x + y}$

Proof

Let $z \in X$.

Then we have $z \in \map {B_\epsilon} x + y$ if and only if $z = u + y$ for $u \in \map {B_\epsilon} x$.

That is, if and only if $z - y \in \map {B_\epsilon} x$.

This is equivalent to:

$\map d {z - y, x} < \epsilon$

Since $d$ is invariant, this is equivalent to:

$\map d {z, x + y} < \epsilon$

So, we have $z \in \map {B_\epsilon} x + y$ if and only if $z \in \map {B_\epsilon} {x + y}$.

Hence $\map {B_\epsilon} x + y = \map {B_\epsilon} {x + y}$.

$\blacksquare$