Every open set in Euclidean space is also open in space with Jungle River metric

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.

• How is "jungle river metric" defined? May 23, 2021 at 12:00
• @Troposphere This question has the definition May 23, 2021 at 12:09
• About jungle metric that Pumpkin hasn't defined: geogebra.org/m/FddDg7Ew May 23, 2021 at 14:08
• An interesting explanation for people like me who are not Tarzan: thestudentroom.co.uk/showthread.php?t=2239843 May 23, 2021 at 14:12

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 $$B_r(x,\varepsilon) \subseteq B_2(x,\varepsilon) \subseteq A$$

• thank you for your answer too! I notice my mistake... May 24, 2021 at 18:48

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.

• thank you for your answer! now it's more clear for me May 24, 2021 at 18:48