Discrete space

In topology, a discrete space is a particularly simple example of a topological space or similar structure, one in which the points form a discontinuous sequence, meaning they are isolated from each other in a certain sense. The discrete topology is the finest topology that can be given on a set. Every subset is open in the discrete topology so that in particular, every singleton subset is an open set in the discrete topology.

Given a set ${\displaystyle X}$:

A metric space ${\displaystyle (E,d)}$ is said to be uniformly discrete if there exists a packing radius ${\displaystyle r>0}$ such that, for any ${\displaystyle x,y\in E,}$ one has either ${\displaystyle x=y}$ or ${\displaystyle d(x,y)>r.}$ The topology underlying a metric space can be discrete, without the metric being uniformly discrete: for example the usual metric on the set ${\displaystyle \left\{2^{-n}:n\in \mathbb {N} _{0}\right\}.}$

Let ${\textstyle X=\left\{2^{-n}:n\in \mathbb {N} _{0}\right\}=\left\{1,{\frac {1}{2}},{\frac {1}{4}},{\frac {1}{8}},\dots \right\},}$ consider this set using the usual metric on the real numbers. Then, ${\displaystyle X}$ is a discrete space, since for each point ${\displaystyle x_{n}=2^{-n}\in X,}$ we can surround it with the open interval ${\displaystyle (x_{n}-\varepsilon ,x_{n}+\varepsilon ),}$ where ${\displaystyle \varepsilon ={\tfrac {1}{2}}\left(x_{n}-x_{n+1}\right)=2^{-(n+2)}.}$ The intersection ${\displaystyle \left(x_{n}-\varepsilon ,x_{n}+\varepsilon \right)\cap X}$ is therefore trivially the singleton ${\displaystyle \{x_{n}\}.}$ Since the intersection of an open set of the real numbers and ${\displaystyle X}$ is open for the induced topology, it follows that ${\displaystyle \{x_{n}\}}$ is open so singletons are open and ${\displaystyle X}$ is a discrete space.

However, ${\displaystyle X}$ cannot be uniformly discrete. To see why, suppose there exists an ${\displaystyle r>0}$ such that ${\displaystyle d(x,y)>r}$ whenever ${\displaystyle x\neq y.}$ It suffices to show that there are at least two points ${\displaystyle x}$ and ${\displaystyle y}$ in ${\displaystyle X}$ that are closer to each other than ${\displaystyle r.}$ Since the distance between adjacent points ${\displaystyle x_{n}}$ and ${\displaystyle x_{n+1}}$ is ${\displaystyle 2^{-(n+1)},}$ we need to find an ${\displaystyle n}$ that satisfies this inequality: $${\displaystyle {\begin{aligned}2^{-(n+1)}&<r\\1&<2^{n+1}r\\r^{-1}&<2^{n+1}\\\log _{2}\left(r^{-1}\right)&<n+1\\-\log _{2}(r)&<n+1\\-1-\log _{2}(r)&<n\end{aligned}}}$$

Since there is always an ${\displaystyle n}$ bigger than any given real number, it follows that there will always be at least two points in ${\displaystyle X}$ that are closer to each other than any positive ${\displaystyle r,}$ therefore ${\displaystyle X}$ is not uniformly discrete.

The underlying uniformity on a discrete metric space is the discrete uniformity, and the underlying topology on a discrete uniform space is the discrete topology. Thus, the different notions of discrete space are compatible with one another. On the other hand, the underlying topology of a non-discrete uniform or metric space can be discrete; an example is the metric space ${\displaystyle X=\{n^{-1}:n\in \mathbb {N} \}}$ (with metric inherited from the real line and given by ${\displaystyle d(x,y)=\left|x-y\right|}$). This is not the discrete metric; also, this space is not complete and hence not discrete as a uniform space. Nevertheless, it is discrete as a topological space. We say that ${\displaystyle X}$ is topologically discrete but not uniformly discrete or metrically discrete.

Additionally: