Discrete metric and topology

Metric space definition

Let $X$ be a set. The discrete metric, or trivial metric is the metric defined as follows:

• $$d:X\times X\rightarrow \mathbb{R}_{\ge 0}$$ with $$d:(x,y)\mapsto\left\{\begin{array}{lr}0 & \text{if }x=y \\1 & \text{otherwise}\end{array}\right.$$

However any strictly positive value will do for the $x\ne y$ case. For example we could define $d$ as:

• $$d:(x,y)\mapsto\left\{\begin{array}{lr}0 & \text{if }x=y \\v & \text{otherwise}\end{array}\right.$$
  • Where $v$ is some arbitrary member of $\mathbb{R}_{> 0}$ - traditionally (as mentioned) $v=1$ is used.
    

Note: however in proofs we shall always use the case $v=1$ for simplicity

Metric summary

Property	Comment
induced topology	discrete topology - which is the topology $(X,\mathcal{P}(X))$ (where $\mathcal{P}$ denotes power set)
Open ball	$B_r(x):=\{p\in X\vert\ d(p,x)< r\}=\left\{\begin{array}{lr}\{x\} & \text{if }r\le 1 \\ X & \text{otherwise}\end{array}\right.$
Open sets	Every subset of $X$ is open. Proof outline: as for a subset $A\subseteq X$ we can show $\forall x\in A\exists r[B_r(x)\subseteq A]$ by choosing say, that is $A$ contains an open ball centred at each point in $A$.
Connected	The topology generated by $(X,d_\text{discrete})$ is not connected if $X$ has more than one point. Proof outline: Let $A$ be any non empty subset of $X$, then define $B:=A^c$ which is also a subset of $X$, thus $B$ is open. Then $A\cap B=\emptyset$ and $A\cup B=X$ thus we have found a separation, a partition of non-empty disjoint open sets, that separate the space. Thus it is not connected if $X$ has only one point then we cannot have a partition of non empty disjoint sets. Thus it cannot be not connected, it is connected.

Metric objects

Open balls

The open balls of $X$ with the discrete topology are entirely $X$ or a single point, that is:

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• $$B_r(x):=\{p\in X\vert\ d(x,p)<r\}=\left\{\begin{array}{lr}\{x\} & \text{for }r\le 1\\ X & \text{otherwise}\end{array}\right.$$

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By definition $$B_r(x):=\{p\in X\vert\ d(x,p)<r\}$$ note that for:

• $r\le 1$ we have $B_r(x)=\{x\}$ as
  • $d(x,p)< r \le 1$ so $d(x,p)<1$ only when $x=p$, as if $x\ne p$ then $d(x,p)=1\not<1$ (proof by contrapositive)
• $r> 1$ we have $B_r(X)=X$ as
  • $d(x,y)\le 1$ always, so we have $\forall x,y\in X[d(x,y)\le 1< r]$ so $\forall x,y\in X[d(x,y)< r]$ thus the ball contains every point in $X$

This completes the proof

Open sets

The open sets of $(X,d_\text{discrete})$ consist of every subset of $X$ (the power set of $X$) - this is how the topology induced by the metric may be denoted $(X,\mathcal{P}(X))$

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Every subset of $X$ is an open set

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Let $A$ be a subset of $X$, we will show that $\forall x\in A\exists r>0[B_r(x)\subseteq A]$

• Let $x\in A$ be given
  • Choose $r=\tfrac{1}{2}$

    Now $B_r(x)=\{x\}\subseteq A$

    • This must be true as we know already that $x\in A$ (to show this formally use the implies-subset relation)

  • We have shown that given an $x\in A$ we can find an open ball about $x$ entirely contained within $A$
• We have shown that for any $x\in A$ we can find an open ball about $x$ entirely contained within $A$

This completes the proof

Discrete topology

The discrete topology on $X$ is the topology that considers every subset to be open. We may write $X$ imbued with the discrete topology as:

• $(X,\mathcal{P}(X))$ where $\mathcal{P}$ denotes power set

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TODO: find reference - even though it is obvious as I show above that every subset is open

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Related theorems

• Every map from a space with the discrete topology is continuous
• Every continuous map from a non-empty connected space to a discrete space is constant