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Is it true that every open set in Euclidean space $\left(\mathbb{R}^2, \lVert\cdot \lVert_{2} \right)$, is also open in space $\left(\mathbb{R}^2, d_{r} \right)$, where $d_{r}$ is jungle river metric?

I may have found an example proving that it's not true and I would like to know if it's correct.

I took into consideration examples of open balls $B((0,0),1)$ (diamond and circle). But there is a point $x \in B((0,0),1)$ in $\left(\mathbb{R}^2, \lVert\cdot \lVert_{2} \right)$, which doesn't belongs to diamond and we can't find the ball $B_{r}(x,\epsilon)$ in $\left(\mathbb{R}^2, d_{r} \right)$, which is a subset of diamond. So $B((0,0),1)$, which is open in Euclidean space, is not open in $\left(\mathbb{R}^2, d_{r} \right)$.

I will be grateful for any help and hints, because topology is still a hard topic for me.

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2 Answers 2

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Your reasoning doesn't work. It goes wrong at the end of

But there is a point $x \in B((0,0),1)$ in $\left(\mathbb{R}^2, \lVert\cdot \lVert_{2} \right)$, which doesn't belongs to diamond and we can't find the ball $B_{r}(x,\epsilon)$ in $\left(\mathbb{R}^2, d_{r} \right)$, which is a subset of diamond

In order for the ball to be open in $d_r$ all you need is to find a $B_r(x,\varepsilon)$ that is a subset of the ball -- it doesn't need to be a subset of the diamond $B_r(0,1)$.


In fact it is true that every open set according to $\|{\cdot}\|_2$ is also open according to $d_r$.

This is because we always have $d_r(x,y) \ge \|x-y\|_2$, and therefore $B_r(x,\varepsilon) \subseteq B_2(x,\varepsilon)$.

Now suppose $A\subseteq \mathbb R^2$ is open according to $\|{\cdot}\|_2$. We want to prove it is open according to $d_r$. To do this, by definition, we must take an arbitrary $x\in A$ and find a $\varepsilon>0$ such that $B_r(x,\varepsilon)\subseteq A$. However, since $A$ is open in $\|{\cdot}\|_2$ there is a $\varepsilon>0$ such that $B_2(x,\varepsilon)\subseteq A$, and therefore $$ B_r(x,\varepsilon) \subseteq B_2(x,\varepsilon) \subseteq A$$

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  • $\begingroup$ thank you for your answer too! I notice my mistake... $\endgroup$
    – Pumpkin
    May 24, 2021 at 18:48
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If $B((p,q),r)$ is any Euclidean open ball, consider $(p',q')$ in it. If $q' \neq 0$, then $B_r((p',q'), |q'|)$ only contains points on the line $x=p'$ and so for a small enough $r' \le |q'|$ we will stay inside $B((p,q),r)$ with a ball $B_r((p',q'), r')$.

If $q'=0$ we have a $\|\cdot\|_1$ ball around $(p',q')$ that sits inside $B((p,q),r)$ and that is also te river-metric ball around such a point.

So all points of $B((p,q),r)$ are river-metric interior points, so the ball is open in the river metric topology.

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  • $\begingroup$ thank you for your answer! now it's more clear for me $\endgroup$
    – Pumpkin
    May 24, 2021 at 18:48

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