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I want to prove that Jungle River metric is indeed a metric space, and determine it is open and closed balls. Firstly, i know that the metric is given by $x,y\in \mathbb{R}^2$, such that $x=(x_1,x_2), y=(y_1,y_2),$ and $$ d(x,y) = \begin{cases} |x_2-y_2|, & \text{if } x_1 = y_1,\\ |x_2| + |y_2| + |x_1-y_1|, & \text{if } x_1 \neq y_1. \end{cases}$$

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    $\begingroup$ Also $d(x,y)\geq 0$ since absolute value is non-negative, and $d(x,y)=d(y,x)$ since all the experession on RHS are symmetrical. $\endgroup$ – Aysha A. Jan 6 '15 at 8:59
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    $\begingroup$ Geometrically speaking $d(x,y)+d(y,z) \geq d(x, z)$ since you can draw this paths and equality will be only if $y$ lies on the path between $x$ and $z$. $\endgroup$ – Jihad Jan 6 '15 at 9:04
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If $d(x,y)=0,$ then $x_1=x_2$ (why?) and therefore $|y_1-y_2|=0$ which then yields $y_1=y_2.$ Also, if $x=y,$ then $(x_1,x_2)=(y_1,y_2),$ so that $$d(x,y)=|x_2-y_2|=0.$$ To show that $d(x,y)\leq d(x,z)+d(z,y), z:=(z_1,z_2)$; Consider the following four cases: (a) $x_1=z_1;$ (b) $x_1,x_2,z_1$ are pairwise different; (c) $x_1=x_2$ but $z_1$ is different (d) $x_1$ is different from $$x_2=z_1.$$ Note that: Drawing some balls in this metric you just seem to get a diamonds of some size along the x-axis and then potentially a vertical line extending from its centre that passes through the tip.

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    $\begingroup$ Pls., this is not clear, could you elaborate? $\endgroup$ – Aysha A. Jan 6 '15 at 17:04
  • $\begingroup$ @AyshaA. What parts don't you understand exactly? $\endgroup$ – Viktor Glombik Mar 13 at 15:51
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I'm also struggling a bit with proving the triangle inequality with the cases proving in @Yusuf's answer and hope to revive this post a little by positing what I've got so far: Let $x,y,z \in \mathbb{R}^2$.

Case 1: $x_1 = z_1$. Then, if (1a) $x_1 \neq y_1$ and $y_1 \neq z_1$ we have \begin{align*} d(x,z) & = | x_2 - z_2 | \le | x_2 | + | z_2 | + | x_1 - z_1 | \\ & \le | x_2 | + | z_2 | + 2 | y_2 | + | x_1 - y_1 + y_1 - z_1 | \\ & \le | x_2 | + | y_2 | + | x_1 - y_1 | + | z_2 | + | y_2 | + | y_1 - z_1 | = d(x,y) + d(y,z). \end{align*} and if (1b) $x_1 = y_1 = z_1$ \begin{align*} d(x,z) = | x_2 - z_2 | \le |x_2 - y_2 + y_2 - z_2 | \le |x_2 - y_2 | +| y_2 - z_2 |. \end{align*}

Case 2: $x_1 \neq x_2 \neq z_1$. Then for (2a) $y_1 \neq z_1$ we have \begin{equation*} d(x,z) = | x_2 | + | z_2 | + | x_1 - z_1 | \overset{\triangle \neq }{\le} | x_2 | + |y_2 | + | x_1 - y_1 | + | y_2 | + | z_2 | + | y_1 - z_1 | \end{equation*} For (2b) $y_1 = z_1 \neq x_2$ we would have to show \begin{equation*} d(x,z) = | x_2 | + | z_2 | + | x_1 - z_1 | \le | x_2 | + | y_2 | + | x_1 - y_1 | + | z_2 - y_2 |, \end{equation*} which I couldn't manage.

Case 3: $x_1 = x_2 \neq z_1$ If also $z_1 \neq y_1$ we have Case (1a). For $z_1 = y_1$ we need to show \begin{align*} d(x,z) & = | x_2 | + | z_2 | + | x_1 - z_1 | = | x_2 | + | z_2 | + | x_2 - z_1 | \\ & \overset{(\star)}{\le} | x_2 | + | y_2 | + | x_2 - y_1| + | y_2 - z_2 | \\ & = | x_2 | + | y_2 | + | x_1 - y_1| + | y_2 - z_2 |, \end{align*} but I haven't been able to show $(\star)$, which is very similar to (2b)

Case 4: $x_1 \neq x_2 = z_1$. For (4a) $y_1 = z_1$ we would have to show \begin{align*} d(x,z) & = | x_2 | + | z_2 | + | x_1 - z_1 | = | z_1 | + | z_2 | + | x_1 - z_1 | \\ & \overset{(\star)}{\le} | z_1 | + | y_2 | + | x_1 - y_1 | + |y_2 - z_2|, \end{align*} but I haven't been able to show $(\star)$, which is very similar to (2b)

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