What is meant be the equivalence relation generated by $0\sim 1$?

Question

I came across a couple occurrences of this notation in my notes. Here’s one example:

Ex. The quotient space X/~ where $X=\lbrack 0,1 \rbrack$ with the Euclidean topology, and $\sim$ the equivalence relation generated by $0\sim 1$.

What does that mean? How do I determine the other pairs in that relation?

Answer 1

In this case, it means $a\sim b$ iff $a=b$ or $a,b\in\{0,1\}.$ That is, it is the smallest equivalence relationship on $[0,1]$ such that $0\sim 1.$

You'll see this as a short form of defining equivalence relations a lot, such as the definition of the Möbius strips as $X/~$ where $X=[0,1]\times[0,1]$ where $(x,0)\sim(1-x,1)$ for all $x\in[0,1].$

Answer 2

The intersection of a set $\mathcal{R}$ of equivalence relations on $X$ (as subsets of $X^2$) is again an equivalence relation. This is a simple matter of checking the axioms: e.g. set $R_0:=\bigcap \mathcal{R}$. Then let $x \in X$; for all $R \in \mathcal{R}$ we know that $xRx$ (as $R$ is an equivalence relation), or equivalently $(x,x) \in R$ and so $(x,x) \in R_0$ and $xR_0x$ holds. $x$ was arbitrary so $R_0$ is reflexive; the other two axioms are similar.

So if we specify a relation like $0 \sim 1$ we actually specify that $(0,1) \in R$. So the equivalence relation generated by $0 \sim 1$ is the intersection of all equivalence relations that contain $(0,1)$ (and $(1,0)$). This is the minimal equivalence relation we get when we know this relation plus the axioms for an equivalence relation, so in practice:

$0 \sim 1$ so $1 \sim 0$ by symmetry. Always $x \sim x$ for all $x \in [0,1]$. But transitivity gives no new forced relations, as $0 \sim 1$ plus $1 \sim 0$ gives $0 \sim 0$ which was not new, etc.

This gives a more constructive (rather than the abstract intersection idea) way of looking at it: the generated equivalence relation by a set of pairs $A$ is basically all $(x,y)$ such that there is $n \ge 1$ and a finite set $a_1, a_2,\ldots a_n$ such that $a_1=x, a_n=y$ and for all $1 \le i \le n-1$, $(a_i, a_{i+1}) \in A$ or $(a_{i+1},a_i) \in A$, so basically all pairs for which we can build a "path" between with pairs we want to be in it the relation (the generating set). See Wikipedia for a similar exposition.