Corollary

Let R be an equivalence relation in $\mathbf{I}$. The following conditions are equivalent:

(i) Every equivalence class according to R is a closed interval of $\mathbf{I}$, distinct from $\mathbf{I}$;

(ii) There exists an increasing surjective mapping $u:\mathbf{I}\rightarrow \mathbf{I}$ such that R is the equivalence relation associated with $u$.

Such a mapping $u$, when it exists, is continuous and defines by passing to the quotient a homeomorphism of $\mathbf{I}/R$ onto $\mathbf{I}$.

We shall denote by $p:\mathbf{I}\rightarrow \mathbf{I}/R$ the canonical surjection.

Suppose condition (i) is satisfied. For A and B equivalence classes following R, write $A<B$ if one has $a < b$ for every $a\in A$ and every $b\in B$. In $\mathbf{I}/R$, the relation « A = B or $A<B$ » is an order relation. Indeed, it is reflexive; it is antisymmetric since one cannot have simultaneously $A<B$ and $B<A$; it is transitive since the relations $A<B$ and $B<C$ imply $A<C$. If A and B are distinct elements of $\mathbf{I}/R$, they are disjoint closed intervals of $\mathbf{I}$, and one then has either $A<B$, or $B<A$. Endowed with the order relation thus defined, $\mathbf{I}/R$ is therefore totally ordered. The mapping $p:\mathbf{I}\rightarrow \mathbf{I}/R$ is increasing.

The set $\mathbf{I}/R$ is not reduced to one element, by virtue of (i).

Let F be a subset of $\mathbf{I}/R$. Let us prove that F has a least upper bound in $\mathbf{I}/R$. Put $F'=\overset{-1}{p}(F)$; denote by $a$ the least upper bound of $F'$ in $\mathbf{I}$ and by A the equivalence class of $a$ following R. Since $a$ is an upper bound of $F'$ in $\mathbf{I}$, A is an upper bound of F in $\mathbf{I}/R$. Conversely, let $B\in \mathbf{I}/R$ be an upper bound of F; put $b=$ sup(B). Every element of $F'$ is then bounded above by $b$. We therefore have $a\leqslant b$. Since the equivalence classes following R are closed intervals, $a$ belongs to A and $b$ to B. Consequently we have $A =p(a)\leqslant p(b) = B$. This proves that A is the least upper bound of F.

Let A and B be elements of $\mathbf{I}/R$ such that $A<B$. Let $a$ be the greatest lower bound of A and $b$ the least upper bound of B. Since $a\in A$ and $b\in B$, we have $a < b$. The equivalence class following R of any element of $]a, b[$ is an element of the interval $]A,B[$ of $\mathbf{I}/R$. Since $\mathbf{I}\cap \mathbf{Q}$ meets $]a, b[$, its image by $p$ meets $]A,B[$.

We have thus proved that the totally ordered set $\mathbf{I}/R$ satisfies the hypotheses of prop. 16. There therefore exists an isomorphism of ordered sets $f:\mathbf{I}/R\rightarrow \mathbf{I}$. The mapping $u=f\circ p$ is a surjective and increasing mapping of $\mathbf{I}$ onto $\mathbf{I}$ and the equivalence relation associated with $u$ is the relation R; this proves that condition (ii) is satisfied.

Conversely, suppose that condition (ii) is satisfied. Let $u:\mathbf{I}\rightarrow \mathbf{I}$ be an increasing and surjective mapping such that R is the equivalence relation associated with $u$.

Let $a$ be a point of $\mathbf{I}$. The set $A =\overset{-1}{u}(a)$ is an interval of $\mathbf{I}$, since the mapping $u$ is increasing. Since $u$ is surjective, A is neither empty nor equal to $\mathbf{I}$. Let $b$ be the greatest lower bound of A in $\mathbf{I}$. We have $u(b)\geqslant a$. If $u(b)> a$, there exist $c\in ]a, u(b)[$ and $d\in \mathbf{I}$ such that $u(d) =c$ since $u$ is surjective. Since $u$ is increasing and $a < u(d)< u(b),d$ majorizes every element of A and we have $d < b$, which contradicts the assumption that $b$ is the greatest lower bound of A. We therefore have $u(b) =a$, that is to say $b\in A$. Analogously one proves that A contains its greatest lower bound. The set A is therefore a closed interval of $\mathbf{I}$, distinct from $\mathbf{I}$. This proves that condition (i) is satisfied.

Let us now prove that the mapping $u$ is continuous. It suffices for this to prove that, for every $a\in \mathbf{I}$, the sets $\overset{-1}{u}(]a,\rightarrow [)$ and $\overset{-1}{u}(]\leftarrow , a[)$ are open. Let $b$ be the greatest lower bound of $\overset{-1}{u}(a)$; we have $u(b) =a$. If $x\in \mathbf{I}$ satisfies $u(x)> a$, we necessarily have $x > b$; conversely, if $x > b$, we have $u(x)\geqslant a$ and $u(x)=\not a$ since $b$ is the greatest lower bound of $\overset{-1}{u}(a)$. Consequently $\overset{-1}{u}(]a,\rightarrow [) = ]b,\rightarrow [$, which proves that the inverse image by $u$ of the interval $]a,\rightarrow [$ is open. Analogously one proves that that of $]\leftarrow , a[$ is open. It follows that the mapping $u$ is continuous.

The mapping $v:\mathbf{I}/R\rightarrow \mathbf{I}$ deduced from $u$ by passing to the quotient is then continuous and bijective. Since $\mathbf{I}$ is compact, $\mathbf{I}/R$ is quasi-compact (TG, I, p. 62, th. 2) and $v$ is a homeomorphism (TG, I, p. 63, cor. 2).