Closure of product with an particular metric

Question

Let $(\mathbb{R}^2,d)$ a metric space, with $d(x,y)=\|x\|+\|y\|$ when $x\not=y$ and $d(x,y)=0$ when $x=y$. Find the closure of $(0,1)\times(0,1)$. My intuition says that the closure is $[0,1]\times[0,1]$ but since we are working with that metric, I don't know if the closure changes, and if it doesn't, I don't know how to formally prove that $[0,1]\times[0,1]$ is the closure. I would appreciate any help.

Answer 1

The first thing that one has to notice here is that for all $x\neq 0$, the set $\{x\}$ is in fact open. Let’s prove this:

Let $x\neq 0$. Then, we choose $\varepsilon=\lVert x\rVert>0$, and we have that $d(y,x) = \lVert y\rVert + \lVert x\rVert\geq\lVert x\rVert = \varepsilon$ for all $y\neq x$, so $B(x,\varepsilon)=\{y\in\mathbb R^2 : d(y,x)<\varepsilon\} = \{x\}$.

This implies that any set $U$ such that $0\notin U$ is open: $$U = \bigcup_{x\in U} \{x\}.$$

However, for $0$ we have that $B(0,\varepsilon)$ is the regular ball: $$B(0,\varepsilon) = \{x\in\mathbb R^2 : d(0,x) = \lVert x\rVert<\varepsilon\}.$$

We now ask: is $U=(0,1)\times(0,1)$ closed? Well, no, because its compliment is not open: all the balls centered at $0$ have nonempty intersection with $U$. However, the set $U\cup\{0\}$ is closed, since its compliment does not include $0$, hence it is open. We conclude that $U\cup\{0\}$ is the smallest closed set that contains $U$, so $\overline U = U\cup\{0\}.$

Answer 2

Observe that, if $y\neq 0$ and $y$ is not in your set (call it $A$) you cannot have a sequence $x_n\in A$ approaching $y$, since $d(x_n,y)\ge \|y\|>0$. But $0$ is approached by the sequence $(1/n,1/n)$ say. So the closure is $A\cup \{0\}$.