Theorem 4

A subset $A$ of $\mathbf{R}$ is connected if and only if $A$ is an interval.

i) The condition is necessary. Suppose that $A$ is connected: if $A$ consists of a single point, it is an interval. If $A$ has more than one point, let $a$ and $b$ be two points of $A$ such that $a < b$; by Proposition 1 of no. 4 it is enough to show that every $x$ such that $a < x < b$ belongs to $A$. Now, if $x \notin A$ we should have $A \subset C\{x\}$; but $C\{x\}$ is the union of two disjoint open sets ]$\leftarrow, x[$ and $]x, \rightarrow[$, each of which meets $A$, and therefore $A$ would not be connected, contrary to hypothesis.

2) The condition is sufficient. Let us show first that every compact interval $[a, b]$ is connected. For each integer $n > 0$, let $V_{1/n}$ be the entourage consisting of all pairs $(x, y)$ such that $|x - y| \leq 1/n$; by Proposition 6 of Chapter II, § 4, no. 4, it is enough to show that every pair of points $x, y$ of $[a, b]$ can be joined by a $V_{1/n}$-chain. Let $p$ be the greatest integer such that $p/n \leq x$ and let $q$ be the greatest integer such that $q/n \leq y$ ($p$ and $q$ exist by reason of Theorem 1 of no. 1); then $p \leq q$. If $p = q$ then $y - x < 1/n$ and the points $x$ and $y$ form a $V_{1/n}$-chain. If $q > p$, put $x_i = (p + i)/n$ ($i = 1, 2, \ldots, q - p$); we have $x_1 - x \leq 1/n$, $y - x_{q-p} \leq 1/n$ and $x_{i+1} - x_i = 1/n$, hence the points $x, x_1, x_2, \ldots, x_{q-p}, y$ form a $V_{1/n}$-chain joining $x$ and $y$.

If now $I$ is any interval not consisting of a single point, and if $a$ and $b$ are any two points of $I$ such that $a < b$, then the interval $[a, b]$ is contained in $I$ and is connected, and hence $I$ is connected.