[Math] Open set with respect to french railway metric

general-topologymetric-spaces

Can't seem to tackle this question,

Consider $A = (0,1)\times \mathbb{R}$. Is A open w.r.t the topology induced by the French railway metric in $\mathbb{R}^2$?

Also consider $B =(-1,1)\times \mathbb{R}$. Is B open w.r.t the topology induced by the French railway metric on $\mathbb{R}^2$?

where

$$d(x,y) = \begin{cases} \|x-y\|, & \text{if $x,y,0$ are collinear;} \\
\|x\| +\|y\|, & \text{otherwise} \end{cases}$$

And i have to sketch for $A = \{x \in \mathbb{R}^2 \colon d(x,(0,1)) = 2\}$

also $B = \{x \in \mathbb{R}^2 \colon d(x,(2,1)) \leq 1\}$

To be honest, i dont truly understand what it means for a topology to be 'induced' by a metric, but i could tell you what open, topology and metric mean etc…

Best Answer

A set is open if it is a neighborhood of each point $x\in U$, i.e. if for every $x\in U$ there is an $\varepsilon>0$ such that $B_ε(x)⊆U$. It's the same as saying that $U$ is a union of open balls $B_\delta(y)$ for various $y$ and $\delta>0$. Intuitively, an open set is a set without a boundary. From every element in the open set you can move a bit in any direction and stay within $U$.

Consider $A=(0,1)\times\Bbb R$. If $x\in A$, then $x=(s,t)$ where $0<s<1$ and $t\in\Bbb R$. Then $||x||=r>0$ since $x\ne(0,0)$. What can you say about $B_r(x)$? In particular, what does $B_r(x)$ "look like" ? If $B_r(x)$ is not small enough, can you find a smaller value depending only on $s$ ?

What about $B=(-1,1)\times\Bbb R$? Does the same method work for every point in $B$?

Here are some examples you could check for openness/closedness if you like:

  • $C=\{0\}\times(-1,1)$
  • $D=\{0\}\times[0,1]$
  • $D=\{0\}\times(0,1)$
  • $F=\{1\}\times (1,2)$
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