Theorem 2

Let X be a compact connected topological space possessing a countable everywhere dense subset. Let $a,b$ be distinct points of X. The following conditions are equivalent:

(i) There exists a homeomorphism $f:\mathbf{I}\rightarrow X$ such that $f(0) =a$ and $f(1) =b$;

(ii) Every connected subset of X which contains $\{a, b\}$ is equal to X ;

(iii) The space X is locally connected and every connected compact subset of X which contains $\{a, b\}$ is equal to X ;

(iv) For every $x\in X-\{a, b\}$, the space X $-\{x\}$ is not connected.

Assertion (i) implies all the others.

Let $x$ be a point of X $-\{a, b\}$. Since X $-\{x\}$ contains $\{a, b\}$, assertion (ii) implies that it is not a connected subset of X, so that (ii) implies (iv).

Let us show that (iii) implies (iv). Suppose the assumptions of (iii) are satisfied. Let $x\in X-\{a, b\}$ and suppose that X$-\{x\}$ is connected. Let T be the space X $-\{x\}$ and let $\mathscr{U}$ be the set of subsets of T of the form X-V, where V is a compact neighbourhood of $x$. By Lemma 7, there exists a compact neighbourhood V of $x$ such that $a$ and $b$ belong to the same connected component of X-V. They belong in particular to the same connected component of $X-\mathring{V}$; this is a compact connected set of X, distinct from X, which contradicts the assumption (iii).

It remains to prove that assertion (iv) implies condition (i).

Let $x$ be a point of X $-\{a, b\}$ and let U, V be open and closed, disjoint and nonempty subsets of X$-\{x\}$, such that X$-\{x\}= U\cup V$. According to III, p. 278, prop. 17, c), there exists a point of U (resp. of V) whose complement in X is connected. Since X admits only two such points, $a$ and $b$, one of them, say $a$, is contained in U and the other in V. According to loc. cit., a nonempty open and closed subset, $U'$ of U, contains a point of $\{a, b\}$, hence contains $a$. Applying this to $U-U'$, it follows that $U = U'$. Consequently, U is connected; it is the connected component of $a$ in X$-\{x\}$. Analogously, V is the connected component of $b$ in X $-\{x\}$.

For every $x\in X-\{a, b\}$, denote thus by $U_x$ and $V_x$ the connected components of $a$ and $b$ in X $-\{x\}$. From what precedes, they are open and closed in X $-\{x\}$, disjoint, and their union is equal to X $-\{x\}$.

Denote by $\preccurlyeq$ the relation in X defined in the following way: on the one hand, $a\preccurlyeq x$ and $x\preccurlyeq b$ for every $x\in X$, on the other hand, if $x$ and $y$ belong to X $-\{a, b\}$, then $x\preccurlyeq y$ if $x\in \overline{U_y}$. For $x$ and $y$ in X $-\{a, b\}$, the subsets $V_x$ and $V_y$ have the point $b$ in common, the relation $x\preccurlyeq y$ is in fact equivalent to the assertion $\overline{U_x}\subset \overline{U_y}($III, p. 278, prop. 17, b)). It follows that the relation $\preccurlyeq$ is an order relation in X.

Let $x$ and $y$ be points of X such that one does not have $x\preccurlyeq y$. Necessarily, $x=\not a$ and $y=\not b$, and $x\in V_y$. If $x=b$ or $y=a$, one has $y\preccurlyeq x$. Suppose then that $x$ and $y$ are distinct from $a$ and $b$. Since the parts $U_x$ and $U_y$ have the point $a$ in common, one has the inclusion $\overline{V_x}\subset \overline{V_y}($loc. cit.), whence, taking the closures of the complements, $\overline{U_y}\subset \overline{U_x}$ and, fortiori, $y\preccurlyeq x$. The relation $\preccurlyeq$ in the space X is therefore a total order relation. For every $x\in X-\{a, b\}$, one moreover has $U_x= ]\leftarrow , x[$ and $V_x= ]x,\rightarrow [$; for $x, y\in X-\{a, b\}$, one has $]x, y[ = U_y\cap V_x$. When $x, y$ run through the points of X $-\{a, b\}$, the sets $U_y\cap V_x$, the sets $U_y$ and the sets $V_x$ form a basis of a topology on X. Let $\widetilde{X}$ be the topological space corresponding to it and let $i: X\rightarrow \widetilde{X}$ be the identical mapping of X. Since, for every $x\in X$, the sets $U_x$ and $V_x$ are open in X, the mapping $i$ is continuous.

The space $\widetilde{X}$ is separated. Indeed, let $x$ and $y$ be distinct points of $\widetilde{X}$ such that $x\preccurlyeq y$. The parts $]\leftarrow , y[$ and $]x,\rightarrow [$ are open neighbourhoods of $x$ and $y$; if they have a common point $z$, $]\leftarrow , z[$ and $]z,\rightarrow [$ are then disjoint open neighbourhoods of $x$ and $y$. Since X is compact, the mapping $i$ is therefore a homeomorphism (TG, I, p. 63, cor. 2).

Consequently, the image by $i$ of a countable everywhere dense subset of X meets every nonempty open interval of the ordered set $(X,\preccurlyeq )$. It follows then from Prop. 16 of III, p. 276 that there exists an isomorphism $c$ of the ordered set $\mathbf{I}$ onto $(X,\prec )$. According to the remark following this proposition, this isomorphism is a homeomorphism of $\mathbf{I}$ onto the topological space $\widetilde{X}$. The mapping $f=i^{-1}\circ c$ is then a homeomorphism of $\mathbf{I}$ onto X which maps 0 onto $a$ and 1 onto $b$.