# Result from a metric linear space

#### dray

##### Member
Let X be a metric linear space, d a translationally invariant metric defining the metric topology on X, and S(x,r) the open d-ball of radius r centred at the point x in X. How do you prove that S(x,r)=S(0,r) + x ?

Furthermore, why is this an important result?

#### CaptainBlack

##### Well-known member
Let X be a metric linear space, d a translationally invariant metric defining the metric topology on X, and S(x,r) the open d-ball of radius r centred at the point x in X. How do you prove that S(x,r)=S(0,r) + x ?
The usual method: let $$y \in S(x,r)$$ then show that it is also in $$S(0,r)+x$$, then let $$y \in S(0,r)+x$$ and show that it is also in $$S(x,r)$$.

CB

#### dray

##### Member
OK.

Let y in S(x,r) so that d(x,y)< r .... and now I have no idea!

#### Opalg

##### MHB Oldtimer
Staff member
OK.

Let y in S(x,r) so that d(x,y)< r .... and now I have no idea!
The translational invariance means that $d(x,y) = d(x-x,y-x)$.

#### dray

##### Member
Thanks.

Does this look right?

Let y in S(x, r). Then d(x, y) = d(x-x, y-x) = d(0, y-x)< r, so that y in S(0,r) + x.

Now assume that y in S(0,r) + x. Then d(y,0)+d(x-0) >= d(y,x) = d(x,y)< r, so that y in S(0, r).

Since y was chosen arbitrarily from X, it follows that S(x, r)=S(0,r)+x.