# Sequence of points in a metric space

I have to check if the set of points $$p_n = (\frac{1}{n},\frac{1}{n})$$ is closed in the metric space $$(\mathbb{R}^2, d_R)$$, where $$d_R$$ is the river metric.

Intuition tells me that this is an open set, but I don't know how to show that.

Edit: the river metric is defined as follows:

if $$x_1 = x_2$$

$$d_R((x_1,y_1),(x_2,y_2)) = |y_1 - y_2|$$

else $$d_R((x_1,y_1),(x_2,y_2)) = |x_1 - x_2| + |y_1| + |y_2|$$

• What is your definition of the river metric? – upanddownintegrate Jan 12 at 18:04
• Note that $d_R(p_n,(0,0))=\frac{1}{n}+0+\left|\frac{1}{n}-0\right|\to0$ as $n\to\infty$. Therefore, $p_n\to (0,0)$ as $n\to\infty$, while $(0,0)$ is not a point of the form $\left(\frac{1}{n},\frac{1}{n}\right)$. – MoonLightSyzygy Jan 12 at 18:11
• I've edited the question to add my definition of the river metric – schrader2 Jan 12 at 18:52
• Is is equivalent to the standard metric? – orientablesurface Jan 12 at 18:54
• @topologicalmagician No, for example the sequence $q_n=(1/n,1)$ is not Cauchy for $d_R$, since $d_R((1/n,1),(1/m,1))=\left|\frac{1}{n}-\frac{1}{m}\right|+2\geq2$, while it is Cauchy for $d((1/n,1),(1/m,1))=\sqrt{(1/n-1/m)^2+(1-1)^2}$. – MoonLightSyzygy Jan 12 at 19:02

The set (a sequence really) has $$(0,0)$$ as a limit point, so is not closed. And it has no interior points, as open balls are uncountable.