1

I could not understand this question on my textbook. It's in the chapter of Metric Spaces, and I do think it's refering to a metric induced by a function $f$.

What is the maximum number of points that a subspace $X \subset \mathbb{R}^2$ can have so that $\mathbb{R}^2$ induces the discrete metric in $X$?

José Carlos Santos
  • 427,504
MrBr
  • 367
  • Do you mean that you don't know how to answer the question, or do you mean that you literally can't understand the question? – bof Oct 08 '17 at 05:48
  • Hahaha. I literally did not understand the question. But now i think i get it.Thanks to José Carlos. Suposse that there are the following Metric Space $(X,d_1)$, with $X \subset R²$ and $(R²,d_2)$ where $d_1$ is the discrete metric and $d_2$ is the euclidian metric. If $(X,d_1)$ is induced by $R²$ then there is a injective function $f$ where $d_1(x,y) = d_2(f(x),f(y))$ for any $x,y \in X $. There can be only 3 points in $(R²,d_2)$ where the distance between any two of them is 1. Therefore, because of the injectivity of $f$, there can be only 3 points in $X$. Am i correct? – MrBr Oct 08 '17 at 19:30

2 Answers2

2

Take three points in $A,B,C\in\mathbb{R}^2$ such that the distance between any two of them is $1$. For instance, you can take $A=(0,0)$, $B=(1,0)$, and $C=\left(\frac12,\frac{\sqrt3}2\right)$. Could there be a fourth point $D\in\mathbb{R}^2$ whose distance to the other three is also $1$? No, because then it would belong to the circle centered at $A$ with radius $1$ and also to the circle centered at $B$ with radius $1$. There are only two points at the intersection of these circles. One of them is $C$ and the distance from the other one to $C$ is $\sqrt3$, which is not $1$.

Therefore, the answer is three.

José Carlos Santos
  • 427,504
1

Suppose we have such an $X$. Note that $\mathbb{R}^2$ has a countable base, so every subspace too. And a discrete second countable space has at most countably many points as all sets $\{x\}$ must be in any base for $X$.

Directly: for every $x \in X$ pick rationals $q_1,q_2,r_1, r_2$ such that $\{x\} = ((q_1,q_2) \times (r_1, r_2)) \cap X$, which can be done as $\{x\}$ is open in the subspace. Then this defines an injection of $X$ into the countable set $\mathbb{Q}^4$.

As David Hartley noted we want the discrete metric on $X$ in which case the answer is $3$, a equilateral triangle. We cannot have $4$ by geometric reasons. (We can have a regular tetrahedron in $\mathbb{R}^3$ of course).

Henno Brandsma
  • 242,131