Editing Discrete space (section)

== Definitions ==
Given a set <math>X</math>:{{unordered list
| the '''{{visible anchor|discrete topology}}''' on <math>X</math> is defined by letting every [[subset]] of <math>X</math> be [[Open set|open]] (and hence also [[Closed set|closed]]), and <math>X</math> is a '''{{visible anchor|discrete topological space}}''' if it is equipped with its discrete topology;
| the '''{{visible anchor|discrete uniformity|discrete [[Uniform space|uniformity]]}}''' on <math>X</math> is defined by letting every [[superset]] of the diagonal <math>\{(x,x) : x \in X\}</math> in <math>X \times X</math> be an [[Entourage (topology)#Entourage definition|entourage]], and <math>X</math> is a '''{{visible anchor|discrete uniform space}}''' if it is equipped with its discrete uniformity.
| the '''{{visible anchor|discrete metric|text=discrete [[Metric space|metric]]}}''' <math>\rho</math> on <math>X</math> is defined by <math display=block>\rho(x,y) = \begin{cases} 
1 &\text{if}\ x\neq y , \\
0 &\text{if}\ x = y
\end{cases}</math> for any <math>x,y \in X.</math> In this case <math>(X,\rho)</math> is called a '''{{visible anchor|discrete metric space}}''' or a '''space of [[isolated point]]s'''.
| a '''{{visible anchor|discrete subspace}}''' of some given topological space <math>(Y, \tau)</math> refers to a [[topological subspace]] of <math>(Y, \tau)</math> (a subset of <math>Y</math> together with the [[subspace topology]] that <math>(Y, \tau)</math> induces on it) whose topology is equal to the discrete topology. For example, if <math>Y := \R</math> has its usual [[Euclidean topology]] then <math>S = \left\{\tfrac{1}{2}, \tfrac{1}{3}, \tfrac{1}{4}, \ldots\right\}</math> (endowed with the subspace topology) is a discrete subspace of <math>\R</math> but <math>S \cup \{0\}</math> is not. 
| a [[Set (mathematics)|set]] <math>S</math> is '''discrete''' in a [[metric space]] <math>(X,d),</math> for <math>S \subseteq X,</math> if for every <math>x \in S,</math> there exists some <math>\delta > 0</math> (depending on <math>x</math>) such that <math>d(x,y) > \delta</math> for all <math>y \in S\setminus\{x\}</math>; such a set consists of [[isolated point]]s.  A set <math>S</math> is '''uniformly discrete''' in the [[metric space]] <math>(X,d),</math> for <math>S \subseteq X,</math> if there exists <math>\varepsilon > 0</math> such that for any two distinct <math>x, y \in S, d(x, y) > \varepsilon.</math>
}}

A metric space <math>(E,d)</math> is said to be ''[[Uniformly discrete set|uniformly discrete]]'' if there exists a ''{{visible anchor|packing radius}}'' <math>r > 0</math> such that, for any <math>x,y \in E,</math> one has either <math>x = y</math> or <math>d(x,y) > r.</math><ref>{{cite book | zbl=0982.52018 | last=Pleasants | first=Peter A.B. | chapter=Designer quasicrystals: Cut-and-project sets with pre-assigned properties | editor1-last=Baake | editor1-first=Michael | title=Directions in mathematical quasicrystals | location=Providence, RI | publisher=[[American Mathematical Society]] | series=CRM Monograph Series | volume=13 | pages=95–141 | year=2000 | isbn=0-8218-2629-8 }}</ref>  The topology underlying a metric space can be discrete, without the metric being uniformly discrete: for example the usual metric on the set <math>\left\{2^{-n} : n \in \N_0\right\}.</math>

{{math proof|title=Proof that a discrete space is not necessarily uniformly discrete | proof =
Let <math display="inline">X = \left\{2^{-n} : n \in \N_0 \right\} = \left\{1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \dots\right\},</math> consider this set using the usual metric on the real numbers. Then, <math>X</math> is a discrete space, since for each point <math>x_n = 2^{-n} \in X,</math> we can surround it with the open interval <math>(x_n - \varepsilon, x_n + \varepsilon),</math> where <math>\varepsilon = \tfrac{1}{2} \left(x_n - x_{n+1}\right) = 2^{-(n+2)}.</math> The intersection <math>\left(x_n - \varepsilon, x_n + \varepsilon\right) \cap X</math> is therefore trivially the singleton <math>\{x_n\}.</math> Since the intersection of an open set of the real numbers and <math>X</math> is open for the induced topology, it follows that <math>\{x_n\}</math> is open so singletons are open and <math>X</math> is a discrete space.

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

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