Since every translation is a homeomorphism of $E$, it is sufficient to demonstrate the proposition for a vector subspace $M$ of $E$, and in this case, the proposition has been proved in I, p. 4.

In fact, the closure of a homogeneous hyperplane $H$ can only be $H$ or the whole space $E$, since it is a vector subspace containing $H$ (prop. 1).

A hyperplane $H$ is closed in $E$ if, and only if, $\complement H$ contains an interior point.

The vector subspace $M$ generated by a set $A$, in a topological vector space $E$, is the set of linear combinations of points of $A$ (A, II, § 1.7, prop. 9); the closure of $M$ in $E$ is, by prop. 1, the smallest closed vector subspace containing $A$; we say that this is the closed vector subspace generated by $A$.

As the mapping $\xi \mapsto \xi a$ of $K_s$ on E is bijective and continuous (I, p. 1, def. 1), it is sufficient to show that it is bicontinuous. Let $\alpha$ be a real number $> 0$, we show that there exists a neighbourhood V of 0 in E such that if $\xi a \in V$ then $|\xi| < \alpha$. As K is not discrete, there exists an element $\xi_0 \in K$ such that $0 < |\xi_0| < \alpha$; but, as E is Hausdorff, there is a neighbourhood V of 0 such that $\xi_0 a$ does not belong to V. We can suppose that V is balanced (I, p. 7, prop. 4). But then if $\xi a \in V$ and $|\xi| \geq |\xi_0|$ we have $|\xi_0 \xi^{-1}| \leq 1$, and $\xi_0 a = (\xi_0 \xi^{-1}) (\xi a) \in V$; since this last statement is false we see that $\xi a \in V$ implies $|\xi| < |\xi_0| < \alpha$. This completes the proof.

In D, the set $\{0\}$ is closed, being the intersection of D and the closed set H; D is therefore Hausdorff. But as E/H is also Hausdorff, the canonical mapping of D on E/H, which is linear, is also an isomorphism by prop. 2, from which the conclusion follows (GT, III, § 6.2).

The condition is evidently sufficient (GT, I, § 2.1, th. 1); we show that it is necessary. We can suppose that H is a closed homogeneous hyperplane with the equation $f(x) = 0$. The quotient space E/H is then a Hausdorff topological vector space of dimension 1 on K. We can write $f = g \circ \phi$, where $\phi$ is the canonical mapping of E on E/H and g is a linear mapping of E/H on $K_s$; from prop. 2, g is continuous, thus the same is true of f.

Proposition 2 of I, p. 12, implies that th. 2 is true for $n = 1$; we argue by induction on $n$. Let $H$ be the vector subspace of $E$ generated by $e_1, e_2, ..., e_{n-1}$; the induction hypothesis is that the mapping $(\xi_i)_{1 \leq i \leq n-1} \mapsto \sum_{i=1}^{n-1} \xi_i e_i$ is an isomorphism of $K_s^{n-1}$ on $H$. The subspace $H$, being isomorphic to a product of complete spaces, is complete (GT, II, § 3.5, prop. 10); hence it is closed in $E$ (GT, II, § 3.4, prop. 8). Let $D$ be the subspace $Ke_n$ complementary to $H$ in $E$; $E$ is the topological direct sum of $H$ and $D$ (I, p. 13, cor. 2), therefore the mapping

of $K_s^{n-1} \times K_s$ on $E$ is an isomorphism.

When $n > 1$ the hypothesis that $K$ is complete is essential for the validity of theorem 2. In fact, let $K$ be a non-complete valued division ring, and let $\hat{K}$ be its completion : for each $a \neq 0$ of $\hat{K}$ the set $K.a$ is everywhere dense in $\hat{K}$, since $x \mapsto xa$ is a homeomorphism of $\hat{K}$ on itself. If $a \notin K$, the subspace $K + Ka$ of the topological vector space $\hat{K}$ on $K$ is of dimension 2 on $K$, but it is not isomorphic to $K_s^2$ since every subspace of dimension 1 in $K + Ka$ is dense in $K + Ka$.

We note that, if $M$ and $N$ are any two closed vector subspaces in a Hausdorff topological vector space $E$, then $M + N$ is not necessarily closed in $E$, \* even if $E$ is a Hilbert space \* (cf. IV, p. 64, exerc. 13, d)).

In $N$, the set $\{0\}$ is closed, since it is the intersection of $N$ and the set $M$ which is closed in $E$; thus $N$ is Hausdorff. As $E/M$ is also Hausdorff, the canonical mapping of $N$ on $E/M$, which is linear and bijective, is bicontinuous (I, p. 14, cor. 2), from which the proposition follows.

In proving the first assertion, we need consider only the case when $E$ is complete; for $E$ is an everywhere dense subspace of its completion $\hat{E}$, and the closure $\overline{V}$ of $V$ in $\hat{E}$ is compact and is a neighbourhood of 0 in $\hat{E}$ (GT, III, § 3.4, prop. 7).

We can suppose then that there is a compact neighbourhood $V$ of 0 in $E$. Let $\alpha \in K$ be such that $0 < |\alpha| < 1$; then there are finitely many points $a_i \in V$ such that

Let $M$ be the finite dimensional subspace of $E$ generated by the $a_i$; it is closed in $E$ (I, p. 14, cor. 1). In the Hausdorff topological vector space $E/M$ the canonical image of $V$ is a compact neighbourhood $W$ of 0, such that $W \subset \alpha W$; hence $\alpha^{-1} W \subset W$, and, by induction on $n$, $\alpha^{-n} W \subset W$ for every positive integer $n$. As $W$ is absorbent, we conclude that $W = E/M$; and thus $E/M$ is compact. To complete the proof of the first assertion in the theorem, it is sufficient, therefore, to establish the following lemma.

Since E is complete we can suppose K is complete (I, p. 6). If E $\neq \{0\}$ then E contains a line that is closed in E (I, p. 14, cor. 1) and therefore compact. This line is isomorphic to $K_s$ (I, p. 12, prop. 2) and hence K must be compact. Now the mapping $\xi \mapsto |\xi|$ of K in $\mathbf{R}$ is continuous and thus the image of K must be bounded, on the other hand there exists $\gamma \in K$ with $|\gamma| > 1$, and the set $|\gamma^n| = |\gamma|^n, n \in \mathbf{N}$, is unbounded. This contradiction shows that $E = \{0\}$.

To prove the second assertion in the theorem, if $E \neq \{0\}$ then from the first part of the theorem E is isomorphic to $K_s^n$ with $n > 0$; now K is complete, hence so is E, and thus E is locally compact. But $K_s$ is isomorphic to a line in E (I, p. 12, prop. 2) which is necessarily closed in E (I, p. 14, cor. 1); it follows that K is locally compact.