Define metrics $d$ and $e$ on $ \mathbb R^{\oplus }$ as follows (where $ \mathbb R^{\oplus }$ refers to the set of non negative real numbers):
$d(a,b) = |a-b|+|~[a]-[b]~|$ where $[x]$ refers to the greatest integer function.
$e(a,b) = |a-b|+\dfrac {|~[2a]-[2b]~|}{2}$
Prove that $U=[\dfrac{1}{2},1)$ is open in $(\mathbb R^{ \oplus },e)$ but not open in $(\mathbb R^ \oplus,d)$.
Attempt: If we prove that $\dfrac {1}{2} $ is not a boundary point of the set $U=[\dfrac{1}{2},1)$ in $(\mathbb R^{ \oplus },e)$ but is a boundary point in $(\mathbb R^ \oplus,d)$, then we obtain the desired result.
For that, we must prove that $dist_e(\dfrac {1}{2},U^c ) > 0$ but $dist_d(\dfrac {1}{2},U^c ) = 0$ where $dist_e$ is the distance function = $\inf \{e(\dfrac{1}{2},w )~|~w \in U^c \}.$ Similarily for $d$.
$U^c = [0,\dfrac{1}{2})~ \bigcup ~[1, \infty)$.
$\forall w \in U^c: ~d(\dfrac{1}{2},w) = |\dfrac{1}{2}-w|+|~[\dfrac{1}{2}]-[w]~|=|\dfrac{1}{2}-w|+ [w] $.
For $w$ sufficiently close to $\dfrac{1}{2},$ we have $\inf d(\dfrac{1}{2},w) = 0$ which means: $U$ is not open in $d$
Similarily: $\forall w \in U^c: ~e(\dfrac{1}{2},w) = |\dfrac{1}{2}-w|+|~[1]-[w]~|=|\dfrac{1}{2}-w|+ |~1-[w]~| >0 $.
Hence, $U$ is open w.r.t $e$.
Is it Correct? Thanks!