# How to show the Jungle River Metric is Complete

Currently, I'm struggling to prove that the Jungle-River Metric is complete over $R^2$. More generally, however, I'm having trouble grasping the approach that one should take when prooving completeness.

Specific Question: Prove the Jungle River Metric, defined as $$d((x,y),(x',y')) = |y|+|y'|+|x-x'|$$ if $x \neq x'$. and $$d((x,y),(x',y')) =|y-y'|$$ if $x = x'$.

My approach so far: So, I know that a metric space is complete if every Cauchy space in the sequence converges. From what I understand, I want to define a general Cauchy Sequence $(x_n,y_n)_{n \in N}$ and then prove that the limit of this Cauchy sequence is inside of my metric space. However, I'm not too sure how to proceed from there.

I've tried drawing open balls in my metric space in hopes of some inspiration. I found that the open balls in this metric space are diamonds centered around the x-axis, with possible vertical lines extending upwards and downwards depending on the radius of the ball. However, I don't see how this can help me. I feel like, please let me know if I'm wrong, but that I understand what it means to be complete, but I don't understand how specifically to begin proving that what I know needs to be true, is true. Any tips or advice would be greatly appreciated. Thank you!

Hints: First it probably helps to understand the metric in the following way: it is the length of the shortest path given that we can only travel vertically at any $x$ value, or horizontally along $y=0$ (I guess we are supposed to imagine a horizontal river along $y=0$ with many vertical tributaries).

Note that $d((x,y),(x',y'))\geq d_1((x,y),(x',y'))$ where $$d_1((x,y),(x',y'))=|x-x'|+|y-y'|$$ is the metric induced by the norm $\|\cdot\|_1$. Assuming you already know $\mathbb R^2$ is complete wrt $d_1$, this means a sequence which is Cauchy wrt $d$ is also Cauchy wrt $d_1$, and therefore converges to some $(x,y)$ wrt $d_1$. If $y\neq0$, you should be able to show that $x_n=x$ for $n$ sufficiently large.

There are a few tricks we can employ to solve this problem. Firstly, since this metric seems to incorporate the standard Euclidean metric on $\mathbb{R}$ (that is, the distance between $a,b\in\mathbb{R}$ is $|a-b|$), we should use the fact that $\mathbb{R}$ is complete with respect to this metric. This is helpful because we can look at the sequences $(x_{n})_{n\in\mathbb{N}}$ and $(y_{n})_{n\in\mathbb{N}}$ in $\mathbb{R}$ and use their limits to find a limit for $\{(x_n,y_n)\}_{n \in \mathbb{N}}$ in $(\mathbb{R}^{2},d)$.

Another trick which is useful in general metric spaces is the fact that if a subsequence of a Cauchy sequence converges, then the whole Cauchy sequence converges (to the same limit as the subsequence). For a proof of this see here. This is useful for this metric $d$ since we can consider two cases which come from the two scenarios given in the definition of $d$. You don't have to use this trick, but it does make splitting the problem up into cases a bit easier.

Suppose $S=\{(x_n,y_n)\}_{n \in \mathbb{N}}$ is a Cauchy sequence in $\mathbb{R}^{2}$ with respect to $d$. We only have to show that a subsequence of $S$ converges.

Suppose first that we can find $x\in\mathbb{R}$ such that $x_{n}=x$ for infinitely many $n\in\mathbb{N}$. Then we can find a subsequence $S'=\{(x_{n_{k}},y_{n_{k}})\}_{k \in \mathbb{N}}$ of $S$ with $x_{n_{k}}=x$ for all $k$. As $S$ is Cauchy, so is $S'$. Thus, for each $\varepsilon>0$, we can find $N_{\varepsilon}\in\mathbb{N}$ such that $$d((x_{n_{k}},y_{n_{k}}),(x_{n_{k'}},y_{n_{k'}})) =|y_{n_{k}}-y_{n_{k'}}|<\varepsilon$$ holds for all $k,k'\geqslant N_{\varepsilon}$. This shows that the sequence $(y_{n_{k}})_{k\in\mathbb{N}}$ is a Cauchy sequence in $\mathbb{R}$ (with respect to the usual Euclidean metric). As $\mathbb{R}$ is complete, we can find $y\in\mathbb{R}$ such that $(y_{n_{k}})_{k\in\mathbb{N}}$ converges to $y$ in $\mathbb{R}$. It is now easy to show that $(x,y)$ is the limit in $(\mathbb{R}^{2},d)$ for $S'=\{(x_{n_{k}},y_{n_{k}})\}_{k \in \mathbb{N}}$.

Now let's look at the case where such a subsequence of $S$ doesn't exist. In other words, for each $x\in\mathbb{R}$, there are only finitely many $n\in\mathbb{N}$ such that $x_{n}=x$. By replacing $S$ with a subsequence, we can therefore assume without loss of generality that the $x_{n}$ are all distinct. As before, for each $\varepsilon>0$, we can find $N_{\varepsilon}\in\mathbb{N}$ such that $$d((x_{m},y_{m}),(x_{n},y_{n})) =|y_{m}|+|y_{n}|+|x_{m}-x_{n}|<\varepsilon$$ holds for all $m,n\geqslant N_{\varepsilon}$.

Observe from this inequality that $y_{n}\rightarrow0$ as $n\rightarrow\infty$. Also, this inequality shows that $(x_{n})_{n\in\mathbb{N}}$ is a Cauchy sequence in $\mathbb{R}$, and so converges to some limit $x\in\mathbb{R}$. It is then easy to show that $(x,0)$ is the limit in $(\mathbb{R}^{2},d)$ of $S$ (take care to check both cases in the definition of $d$).

• I don't know if this will shorten your argument, but, notice that that $|x_m-x_n| \le d((x_m,y_m),(x_n,y_n))$. The sequence $x_n$ is therefore Cauchy in the usual metric on $\mathbb{R}$, hence it converges. Oct 10, 2017 at 2:42