Lemma 8

The set $\mathscr{U}$, ordered by inclusion, is inductive.

Let $\mathscr{V}$ be a totally ordered subset of $\mathscr{U}$. Let us prove that the union V of the sets belonging to $\mathscr{V}$ is an element of $\mathscr{U}$, that is to say that, for every connected component $]u, v[$ of V, one has $f(u) =f(v)$.

Let $x$ be a point of $]u, v[$. Let $\mathscr{V}_x$ be the set of the $U\in \mathscr{V}$ such that $x\in U$; for such a U, denote by $]u_U, v_U[$ the connected component of $x$ in U. One has $u_{U'}\leqslant u_U< v_U\leqslant v_{U'}$ if U and $U'$ are elements of $\mathscr{V}_x$ such that $U\subset U'$. Since the union for $U\in \mathscr{U}_x$ of the $]u_U, v_U[$ is equal to $]u, v[$ according to Lemma 7 of III, p. 280, one has $u=$ lim $u_U$ and $v=$ lim $v_U$, where the limits are taken following the filtered set $\mathscr{V}_x$. Since the mapping $f$ is continuous and the topological space X is separated, the equalities $f(u_U) =f(v_U)$ for every $U\in \mathscr{V}_x$ imply that $f(u) =f(v)$, which was to be proved.

By virtue of E, III, p. 20, Th. 2, there exists an open subset U belonging to $\mathscr{U}$ which is maximal for the relation of inclusion.

Let $g:\mathbf{I}\rightarrow X$ be the mapping defined as follows. If $t\notin U$, we put $g(t) =f(t)$; otherwise, denoting by $]u, v[$ the connected component of $t$ in U, we put $g(t) =f(u) =f(v)$, so that the mapping $g|[u, v]$ is constant with image $f(u)$.

Prop. 18 follows from the following lemma.