There exist two rational numbers $p/q$ and $r/s$ such that $0 < p/q < x$ and $y < r/s$, since the open intervals ]$0, x$ and ]$y, \to$ are not empty ($\S$ I, no. 4, Proposition 4); take $n$ such that $nps > qr$, and we have $y < nx$.

1) The condition is necessary. Let $A$ be a compact subset of $\mathbf{R}$ and let $a$ be a real number $> 0$. The set $A$ is closed (Chapter I, $\S$ 9, no. 3, Proposition 4) and there exists a finite number of points $x_i$ ($1 \leq i \leq n$) of $\mathbf{R}$ such that $A$ is contained in the union of the neighbourhoods $[x_i - a, x_i + a]$ (Chapter I, $\S$ 9, no. 3). Let $b$ be the maximum of the numbers $|x_i|$; then we have $A \subset [-b - a, b + a]$.

2) The condition is sufficient. It is enough to show that every interval $[ -a, +a ]$ ($a > 0$) is compact, and since this interval is a closed subset of a complete uniform space, it is enough to show that, for each $b > 0$, we can cover $[ -a, +a ]$ by a finite number of intervals of the form $[x - b, x + b]$ (Chapter II, $\S$ 4, no. 2, Corollary to Theorem 3). Now, let $n$ be an integer $> 0$ such that $a < nb$; if $x \in [ -a, +a ]$ and if $m$ is the largest integer (positive or negative) such that $mb \leq x$, then we have $-n \leq m \leq n$ and $mb \leq x \leq (m + 1)b$. Hence the $2n + 1$ intervals $[ (k - 1)b, (k + 1)b ]$ ($-n \leq k \leq n$) form a covering of the required type.

Let $A$ be a non-empty subset of $\mathbf{R}$, bounded above, and let $b$ be an upper bound of $A$, so that $A \subset ]\leftarrow, b]$. For each $x \in A$ consider the set $A_x$ of numbers $\geqslant x$ which belong to $A$; the sets $A_x$ form a filter base $\mathfrak{B}$ on $\mathbf{R}$, since $A_y \subset A_x$ if $y \geqslant x$. Let $a$ be a point of $A$. For each $x \geqslant a$ belonging to $A$, $A_x$ is contained in the compact interval $[a, b]$ and thus the filter base $\mathfrak{B}$ has a cluster point $c$. Since the intervals $[x, \rightarrow[$ are closed, $c$ belongs to their intersection and therefore $c$ is an upper bound of $A$. But, on the other hand, every upper bound $z$ of $A$ is $\geqslant c$, otherwise the neighbourhood $]z, \rightarrow[$ of $c$ would not contain any point of $A$. Hence $c$ is the least upper bound of $A$.

We can argue similarly for a non-empty set $B$ bounded below, or else simply remark that $-B$ is non-empty and bounded above, and that if $c$ is the least upper bound of $-B$, then $-c$ is the greatest lower bound of $B$.

The least upper bound $c$ of $A$ can be characterized by the following two properties: (i) For each $x \in A, x \leqslant c_*$. (ii) For each $a < c$, there exists $x \in A$ such that $a < x \leqslant c$.

The least upper bound of a closed set (non-empty and bounded above) belongs to the set and is its greatest element; and the least upper bound of any non-empty subset $A$ of $\mathbf{R}$ which is bounded above may be defined as the largest real number in the closure of $A$.

The condition is clearly necessary. Conversely, suppose that it is satisfied. If $A$ is neither bounded above nor below it must be the whole of $\mathbf{R}$, for if $x$ is any point of $\mathbf{R}$ there are then two points $a, b$ of $A$ such that $a < x < b$. If $A$ is bounded above but not below, let $k$ be its least upper bound; then for any $x < k$ there exist $a$ and $b$ in $A$ such that $a < x < b \leqslant k$, hence $x \in A$, and therefore $A$ can only be one of the two intervals $]\leftarrow, k], ]\leftarrow, k[$. The argument is similar in the other cases.

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.

By Theorem 4, a subset of $\mathbf{R}$ which does not contain any interval consisting of more than one point is totally disconnected; this is true, for example, of the set $\mathbf{Q}$ of rational numbers, since the set $C\mathbf{Q}$ of irrational numbers is dense in $\mathbf{R}$.

Let $A$ be a non-empty open set in $\mathbf{R}$. Since $\mathbf{R}$ is locally connected, every component of $A$ is a connected open set (Chapter I, § 11, no. 6, Proposition 11) and therefore an open interval by Theorem 4. Any two of these open intervals are disjoint; on the other hand, each of them contains a rational number; hence the set of these intervals has a power less than or equal to that of $\mathbf{Q}$, i.e. is countable.

It follows that every closed set in $\mathbf{R}$ is the complement of the union of a (finite or infinite) sequence $(I_n)$ of mutually disjoint open intervals. These intervals are said to be contiguous to the closed set under consideration. Conversely, given such a sequence of intervals, the complement of their union is a closed set to which these intervals are contiguous.

1) The condition is necessary. Let $a$ and $b$ be two points of I such that $a < b$, and suppose for example that $f(a) < f(b)$. Let us show that $f$ is strictly increasing on I. First, if $a < c < b$ then we must have $f(a) < f(c) < f(b)$; if, for example, we had $f(a) < f(b) < f(c)$ then the image of the interval $[a, c]$ under $f$ would be a connected set (Chapter I, , no. 2, Proposition 4) and would therefore contain the interval $[f(a), f(c)]$; hence there would exist $x \in [a, c]$ such that $f(x) = f(b)$, contrary to the hypothesis that $f$ is injective

It follows that if $x$ and $y$ are any two points of I such that $x < y$, then $f(x) < f(y)$; for we have $f(a) < f(x) < f(b)$ if $a < x < b$, $f(a) < f(b) < f(x)$ if $b < x$, and $f(x) < f(a) < f(b)$ if $x < a$; repeating the argument with $a, x, y$ in place of $a, b, x$ respectively, we see that $f(x) < f(y)$.

2) The condition is sufficient. Suppose that $f$ is continuous and strictly monotonic on I (say, strictly increasing): $f(I)$ is connected and is therefore an interval, and since $f$ is strictly increasing, $f$ is a bijective mapping of I onto $f(I)$. Moreover, the image under $f$ of an open interval in I is an open interval in $f(I)$, and therefore (§ 1, no, 4, Proposition 5) $f$ is a homeomorphism of I onto $f(I)$.