Lemma 5

Let X and Y be uniform spaces and $f$ a mapping of X into Y. Let $\mathscr{F}$ be a set of closed subsets of X covering X and possessing the following properties:

(i) There exists a set $F_0\in \mathscr{F}$ which meets all the sets $F\in \mathscr{F}$;

(ii) For every entourage U of the uniform structure of X, there are only a finite number of sets $F\in \mathscr{F}$ which are not small of order U;

(iii) For every entourage V of the uniform structure of Y, there are only a finite number of sets $F\in \mathscr{F}$ such that $f(F)$ is not small of order V.

Thus, if the restriction of $f$ to each of the sets $F\in \mathscr{F}$ is continuous, $f$ is continuous.

Let $x$ be a point of X. Let us prove that $f$ is continuous at $x$. There exists a set $F_1\in \mathscr{F}$ such that $x\in F_1$. The restriction of $f$ to $F_0\cup F_1$ is continuous (TG, I, p. 19, prop. 4). By replacing $F_0$ by $F_0\cup F_1$, we reduce to the case where $x\in F_0$.

Let V be an entourage of the uniform structure of Y. Choose

an entourage $V'$ of this same uniform structure such that $\overset{2}{V'}\subset V$. Since the restriction of $f$ to $F_0$ is continuous, there exists an entourage U of the uniform structure of X such that $f(z)\in V'(f(x))$ for every $z\in F_0\cap U(x)$. Let $U'$ be an entourage of the uniform structure of X

such that $\overset{2}{U'}\subset U$. Let A denote the union of $F_0$, of the sets $F\in \mathscr{F}$ which are not small of order $U'$ and of those such that $f(F)$ is not small of order $V'$. By assumption, A is the union of a finite number of sets belonging to $\mathscr{F}$, and the restriction of $f$ to A is continuous (loc. cit.). There therefore exists a neighbourhood W of $x$ in X, contained in $U'(x)$, such that $f(y)\in V(f(x))$ for $y\in A\cap W$. To conclude, it will suffice to prove that one also has $f(y)\in V(f(x))$ for every point $y\in (X-A)\cap W$. Let $y$ be such a point. Let F be an element of $\mathscr{F}$ such that $y\in F$. By definition of A, F is small of order $U'$ and $f(F)$ is small of order $V'$. By assumption, F meets $F_0$. Let $z\in F\cap F_0$. We have $z\in U'(y)$ since F is small of order $U'$ and $y\in U'(x)$ since W

is contained in $U'(x)$, whence $z\in \overset{2}{U'}(x)$ and fortiori $z\in U(x)$. But then, since $z$ belongs to $F_0$, we have $f(z)\in V'(f(x))$. Moreover $f(F)$ is small of order $V'$, whence $f(y)\in V'(f(z))$. It follows that we have

$f(y)\in \overset{2}{V'}(f(x))$ and finally $f(y)\in V(f(x))$. This concludes the proof of lemma 5.

Let us now prove theorem 1. Let X be a compact nonempty, connected and locally connected metric space. Such a space is connected by arcs and locally connected by arcs (III, p. 267, corollary 2). According to the corollary and the example of III, p. 270, there exists a continuous and surjective mapping $f$ from the triadic Cantor set K (TG, IV, p. 9, example) into X. We shall construct a continuous extension $g$ of $f$ to $[0,1]$, which will prove theorem 1.

Let $\varepsilon$ be a real number $>0$. The open and arcwise connected subsets of X of diameter $\leqslant \varepsilon$ cover X. Let $\mathscr{R}$ denote the set of their inverse images by $f$: it is a set of open sets of K covering K. According to Lemma 4 of III, p. 272, there exists a real number $\alpha >0$ such that every closed ball of K of radius $\alpha$ is contained in an element of $\mathscr{R}$. In particular, if $t$ and $t'$ are points of K such that $|t-t'|\leqslant \alpha$, there exists a path in X joining $f(t)$ to $f(t')$ whose image has diameter $\leqslant \varepsilon$.

The preceding paragraph makes it possible to construct by induction a strictly increasing sequence $(n_k)_{k\geqslant 0}$ of integers $\geqslant$ 0 possessing the following property: for every integer $k\geqslant 0$ and every pair $(t, t')$ of points of K such that $|t-t'|\leqslant 3^{-n_k}$, there exists a path in X joining $f(t)$ to $f(t')$ whose image has diameter $\leqslant 2^{-k}$. The complement of K in $[0,1]$ is the union of a family $(I_{n,p})$ of pairwise disjoint open intervals, where $n$ runs through the set of integers $\geqslant 0$ and $p$ the set of integers between 1 and $2^n$ (TG, IV, p. 9, example). Consider one of these intervals $I_{n,p}$ and write it $]a, b[$. The points $a$ and $b$ belong to K and we have $b-a= 3^{-n-1}$. The function $g$ is defined on the interval $I_{n,p}$ in the following way: one chooses a path $c$ in X joining $f(a)$ to $f(b)$, whose image has diameter $\leqslant 2^{-k}$ if $n_k\leqslant n < n_{k+1}$, and one sets $g(t) =c(\frac{t-a}{b-a})$ for $t\in I_{n,p}$. The function $g: [0,1]\rightarrow X$ thus defined extends $f$. It is continuous on K as well as on each of the closed intervals $\overline{I_{n,p}}$. The latter meet K. Moreover, for every real number $\varepsilon >0$, there are only finitely many intervals $\overline{I_{n,p}}$ of length $> \varepsilon$, and there are only finitely many intervals $\overline{I_{n,p}}$ whose images by $g$ have diameter $> \varepsilon$. According to Lemma 5, the mapping $g$ is continuous.