Lemma 1

Let X be a connected and locally connected metric space, let $a$ and $b$ be points of X and let $\varepsilon$ be a real number $>0$. There exists in X a train of width $\leqslant \varepsilon$ joining $a$ to $b$.

More precisely, every train T joining $a$ to $b$ possesses a refinement $(T', f)$, where $T'$ is a train of width $\leqslant \varepsilon$ joining $a$ to $b$.

The relation “there exists a train of width $\leqslant \varepsilon$ joining $x$ to $y$” is an equivalence relation between $x$ and $y$ in X. The equivalence class of a point $x$ contains every connected open neighbourhood of $x$ of diameter $\leqslant \varepsilon$, and $x$ possesses such a neighbourhood since X is locally connected. The equivalence classes for this relation are therefore open, and consequently also closed. There is at most one of them, since X is connected. This proves the first assertion.

Let $T = (W_i)_{1\leqslant i\leqslant n}$ be a train in X joining $a$ to $b$. Put $x_0=a$, $x_n=b$ and choose for $1\leqslant i\leqslant n-1$ a point $x_i$ in the nonempty set $W_i\cap W_{i+1}$. For $1\leqslant i\leqslant n$, the open set $W_i$ is connected and locally connected and $x_{i-1},x_i$ are two of its points; there exists by the preceding paragraph a train $(W_{i,k})_{1\leqslant k\leqslant m_i}$ in $W_i$, of width $\leqslant \varepsilon$, joining $x_{i-1}$ to $x_i$.

Put $m=m_1+\cdots +m_n$. For $1\leqslant j\leqslant m$, put $W'_j= W_{i,k}$, where $(i, k)$ is the unique pair of integers such that $1\leqslant i\leqslant n,1\leqslant k\leqslant m_i$ and $j=m_1+\cdots +m_{i-1}+k$. For $0\leqslant i\leqslant n$, put $f(i) =m_1+\cdots +m_i$. Then $T'= (W'_j)_{1\leqslant j\leqslant m}$ is a train of width $\leqslant \varepsilon$ in X joining $a$ to $b$, and $(T', f)$ is a refinement of T.

Let us now prove Proposition 11. Equip the connected and locally connected space X with a distance $d$, compatible with its topology, for which it is complete. Let $a$ and $b$ be points of X. Lemma 1 permits us to construct by induction sequences $(T_s)_{s\geqslant 0}$ and $(f_s)_{s\geqslant 1}$ such that, for every $s\geqslant 0$, $T_s= (W_{s,i})_{1\leqslant i\leqslant n_s}$ is a train of width $\leqslant 2^{-s}$ joining $a$ to $b$ and $(T_{s+1}, f_{s+1})$ is a refinement of $T_s$.

We can choose by induction, for $s\geqslant 0$, a strictly increasing mapping $g_s:\{0,1, . . . , n_s\} \rightarrow \mathbf{I}$ such that $g_s(0) = 0$ and $g_s(n_s) = 1$, in such a way that $g_{s+1}\circ f_s=g_s$. Define, for $s\geqslant 0$, a subset $A_s$ of $\mathbf{I}\times X$ by putting

The sequence $(A_s)_{s\geqslant 0}$ is decreasing: indeed, for every integer $j\in \{1, . . . , n_{s+1}\}$, there exists a unique integer $i\in \{1, . . . , n_s\}$ such that $f_s(i-1)< j\leqslant f_s(i)$, and one has

Let $t\in \mathbf{I}$. For every $s\geqslant 0$, denote by $A_s(t)$ the set of the $x\in X$ such that $(t, x)\in A_s$. The set $A_s(t)$ is either one of the cars, or the union of two consecutive cars of the train $T_s$; it is therefore a nonempty subset of X, of diameter $\leqslant 2^{1-s}$. The sequence $(A_s(t))_{s\geqslant 0}$ is decreasing. The set of its terms is a Cauchy filter basis. This one converges to a point $c(t)$ since the metric space X is complete.

Since $a$ belongs to each of the sets $A_s(0) = W_{s,1}$, we have $c(0) =a$; analogously $c(1) =b$. Let $t\in \mathbf{I}$ and let $s$ be an integer $\geqslant 0$. The point $t$ possesses in $\mathbf{I}$ a neighbourhood V of one of the following forms: $[g_s(0), g_s(1)[$, $]g_s(i-1), g_s(i+ 1)[$ for an integer $i$ such that $1\leqslant i\leqslant$ $n_s-1$, or $]g_s(n_s-1), g_s(n_s)]$. According to the case, the set $c(V)$ is contained in the closure of the first car, of the union of two consecutive cars, or of the last car of the train $T_s$. It is therefore of diameter $\leqslant 2^{1-s}$, and we have $d(c(t), c(t'))\leqslant 2^{1-s}$ for $t'\in V$. This proves that the mapping $c:\mathbf{I}\rightarrow$ X is continuous. Thus $c$ is a path in X joining $a$ to $b$, and X is arcwise connected.