Proposition 3

Every non-discrete closed subgroup of $\mathbf{R}^n$ contains a line passing through $o$.

Let $(x_p)_{p \in \mathbf{N}}$ be an infinite sequence of points of $G$ such that $x_p \neq o$ and $\lim_{p \to \infty} x_p = o$; by hypothesis such a sequence exists. Let $P$ be an open cube with centre $o$, containing the $x_p$. Let $k_p$ denote the largest integer $h > 0$ such that $hx_p \in P$ (since $P$ is a bounded box and $x_p \neq o$, the existence of $k_p$ follows from Archimedes' axiom). The points $k_p x_p$ lie in a compact set $\overline{P}$, hence the sequence $(k_p x_p)_{p \in \mathbf{N}}$ has a cluster point $a \in \overline{P}$. If $\| (k_p + 1)x_p - a \| \leq \varepsilon$, we have

and since $\lim_{p \to \infty} x_p = o$, it follows that $a$ is also a cluster point of the sequence $((k_p + 1)x_p)$, whose points belong to the closed set $\mathbf{C}P$, by definition of $k_p$; hence $a \in \overline{P} \cap \mathbf{C}P$ (the frontier of $P$, Fig. 5), which implies $a \neq o$. Moreover, since $G$ is closed, we have $a \in G$. Let $t$ be any real number; since $|tk_p - [tk_p]| < 1$, the relation $\| k_p x_p - a \| \leq \varepsilon$ implies that $\| [tk_p]x_p - ta \| \leq |t| \varepsilon + \| x_p \|$; since $\lim_{p \to \infty} x_p = o$, $ta$ is a

!Figure 5 Figure 5.

!Figure 6 Figure 6.

cluster point of the sequence ([tk_p]x_p); but the points of this sequence belong to G, and consequently ta ∈ G, since G is closed. This completes the proof.