Proposition 2

Every closed (resp. open) interval in $\mathbf{R}$ is a closed (resp. open) set in $\mathbf{R}$.

The sets $[a, \to[ = a + \mathbf{R}_+$ and $]\leftarrow, a] = a - \mathbf{R}_+$ are obtained by translation from $\mathbf{R}_+$ and $-\mathbf{R}_+$ respectively and are therefore closed (Chapter III, § 1, no. 1); the sets $]\leftarrow, a[$ and $]a, \to[,$ which are their complements, are open; finally, the closed interval $[a, b]$ (resp. the open interval $]a, b[$) is the intersection of $[a, \to[$ and $]\leftarrow, b]$ (resp. of $]a, \to[$ and $]\leftarrow, b[$) and is therefore a closed (resp. open) set.

The closed intervals $[-a, +a]$ ($a > 0$) in $\mathbf{R}$ are therefore neighbourhoods of 0. Let us show that they form a fundamental system of neighbourhoods of 0 as $a$ runs through $\mathbf{R}_+^*$. For this it is enough to establish the following proposition: