Proposition 3

As $r$ runs through the set of rational numbers $> 0$, the intervals $s_r = [-r, +r]$ in $\mathbf{R}$ form a fundamental system of neighbourhoods of 0.

By Proposition 7 of Chapter III, § 3, no. 4 we obtain a fundamental system of neighbourhoods of 0 in $\mathbf{R}$ by taking the closures in $\mathbf{R}$ of the intervals

S_r \cap \mathbf{Q} = [-r, +r] \textit{of} \mathbf{Q}. The proof will be complete if we show that S_r is the closure of S_r \cap \mathbf{Q}. Now S_r is closed in \mathbf{R}, and we need therefore only prove that, if x is a real number such that -r < x < r, then x is in the closure of S_r \cap \mathbf{Q}. The interval ]-r, +r[ is an open set in \mathbf{R} and therefore for all sufficiently small neighbourhoods V of o in \mathbf{R} we have x + V \subset ]-r, +r[; but \mathbf{Q} being dense in \mathbf{R}, there is a rational number r' \in x + V, so that -r < r' < r and therefore r' \in S_r \cap \mathbf{Q}.