Open ball with respect to the Euclidean metric is open ball with respect to advance French railroad metric

general-topologymetric-spaces

Let $x\neq0$ in $\mathbb{R}^2$, and consider the open ball (i.e. open disk) of radius $r$ around $x$, with respect to the Euclidean metric. Then is it true that this is an open set with respect to the advanced French railroad metric? I think it is, because for every point $y$ in the disk, we can define an advanced French open ball of appropriate radius $l$ (i.e. a line segment of length $l$ with $y$ as its centre, on the straight line passing from $y$ and the origin) such that the line segment is completely contained in the Euclidean disk. Is this correct? Sorry if the description is a bit confusing, but I do not have a picture available atm. Thanks in advance!

Edit: the advanced French railway metric is, per my notes, defined by: $d(x,y)=|x-y|$, if $x=ay$, for some $a\in \mathbb{R}$, and $|x|+|y|$ otherwise.

Best Answer

As for the question in the title, the French-railroad open ball $B_F(x,r)$ is:

  1. the open euclidean ball $B(x,r)$ if $x=0$;
  2. the segment $x_{(r)}:=\left\{\frac t{\lVert x\rVert}\,:\,t\in(\lVert x\rVert-r,\lVert x\rVert+r)\right\}$ if $x\ne 0$ and $r\le\lVert x\rVert$;
  3. $B(0,r-\lVert x\rVert)\cup x_{(r)}$ if $r>\lVert x\rVert$ and $x\ne0$.

Therefore I would contend that an Euclidean open ball $B$ is a French-railroad open ball if and only if $0$ is the center of $B$. It is clear that an Euclidean ball does not fall into the case (1) unless its center is $0$, and it can never fall into case (2) or (3) because they aren't Euclidean open (case (3) always has an "antenna" protruding out of the ball in the direction of $x$).

As for $B$ being French-railroad open, this is always the case because you can see it as union of open segments and possibly a small Euclidean ball centered in $0$, and all of those can be chosen to be French-railroad balls.

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