3

Ex. 1.2.65. Let $d$ be the Euclidean metric on $\Bbb R^2$. Define $$\delta(p,q):=\begin{cases} d(p,0)+d(q,0), & p\ne q \\ 0, & p = q, \end{cases}$$ for $p, q \in \Bbb R^2$. Show that $\delta$ is a metric on $\Bbb R^2$. What are $B_\delta(0,\varepsilon)$ for $\varepsilon > 0$? Can you describe the $\delta$-open sets?

I've shown for $\varepsilon>0,~p\in\mathbb R^2$ $$B_\delta(p,\varepsilon)=\begin{cases} \{ p \} & \text{if }p\ne0,\; d(p,0)\ge\varepsilon\\ B_d(0,\varepsilon-d(p,0))\cup\{p\} & \text{if }p\ne0,\;d(p,0)<\varepsilon\\ B_d(0,\varepsilon) & \text{if }p=0 \end{cases}$$

Thus all the subset of $\mathbb R^2$ not containning $0$ is an open set and so do each open ball of $(\mathbb R^2,d)$ centered at the origin.

Is it correct to describe the family $\tau$ of all open sets of $(\mathbb R^2,\delta)$ as [assuming $B_\delta(0,0)=\emptyset$] $$\tau=\{X\cup B_d(0,\varepsilon):\varepsilon\ge 0, \,0\notin X\subset\mathbb R^2\}$$ Text: Topology of Metric Spaces - S. Kumaresan

dfeuer
  • 9,069
Sriti Mallick
  • 6,137
  • 3
  • 30
  • 64
  • Yes, that is correct. – Daniel Fischer Jul 18 '13 at 04:52
  • $B(p, \varepsilon)$ should only really be defined for $\varepsilon > 0$, so you can throw that problem away immediately. Otherwise, yes, you're right. All $\delta$-open balls are a singleton set (but not ${0}$), or a Euclidean open ball around zero, or a union of the two. So any set that doesn't contain zero is open; and any set that contains zero and a small Euclidean open ball around it is open. A non-open set would be, for example, ${(x,y) : y \leq x^2}$. – Billy Jul 18 '13 at 04:54
  • 1
    @Billy Why should $B(p, \varepsilon)$ only be defined for strictly positive $\varepsilon$? If you go by the most basic definition, $$ B(p, \varepsilon) = {x : |: d(x, p) < \varepsilon} $$ then setting $\varepsilon$ to be zero or even negative would make perfect sense. The result would just be the empty set. – Arthur Jul 18 '13 at 05:05

1 Answers1

2

The solution was confirmed to be correct in the comments. I'll add two remarks:

  1. I think that $X\subseteq \mathbb R^2\setminus\{0\}$ is easier to understand (and is more standard) than $0\notin X\subset \mathbb R^2$.

  2. The metric happens to be an ultrametric: it satisfies $d(p,q)\le \max(d(p,r), d(q,r))$ for all $p,q,r$.

40 votes
  • 9,736