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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...

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    $\begingroup$ What's the French Railway metric? $\endgroup$
    – voldemort
    Commented Feb 3, 2014 at 23:25
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    $\begingroup$ @voldemort it's a sad reality. Where ever you are going by train, you're always moving radially away or towards Paris. $\endgroup$
    – Olivier Bégassat
    Commented Feb 3, 2014 at 23:28
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    $\begingroup$ Sorry, edited. I was just being lazy! $\endgroup$
    – user65972
    Commented Feb 3, 2014 at 23:29
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    $\begingroup$ @OlivierBégassat: Ah- that's why I prefer flying to the muggle methods ;-) $\endgroup$
    – voldemort
    Commented Feb 3, 2014 at 23:33
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    $\begingroup$ The topology induced by a metric is the one whose open sets are unions of open balls around a point, i.e. of sets $B_{x,r}=\{y\in X:d(x,y)<r\}$. So for instance a set $S$ is open iff around each of its point there exists an open ball contained entirely in $S$. $\endgroup$
    – Kevin Carlson
    Commented Feb 3, 2014 at 23:34

1 Answer 1

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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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  • $\begingroup$ The part that confuses me, is that there are two cases. I understand that I have to define all the of open sets of the railway metric on the set A to find the topology induced by it, but the fact that i have two cases for the metric confuses me. $\endgroup$
    – user65972
    Commented Feb 4, 2014 at 10:32
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    $\begingroup$ You don't have to define the open sets of the railway metric, they are determined by that metric. It's rather important to look at the open balls $B_ε(x)$ as these are the basic "building blocks" for the open sets, which are just arbitrary union of these balls. It's not a good idea to find a formula for the general open sets. Instead, given any set $A$, in order to figure out if it's open or not, you should rather consider an arbitrary point $x\in A$ and try to find an open ball which is contained in $A$. In other word, you should check for the openness "locally". $\endgroup$
    – Stefan Hamcke
    Commented Feb 4, 2014 at 13:29
  • $\begingroup$ Can you say what a set $B_r(x)$ looks like if $||x||=r$ ? $\endgroup$
    – Stefan Hamcke
    Commented Feb 4, 2014 at 13:31

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