Translation of Open Ball in Invariant Pseudometric on Vector Space
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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 $\map {B_\epsilon} x + y = \map {B_\epsilon} {x + y}$.
$\blacksquare$