Theorem 2 (Approximation Theorem for Absolute Values)

Let $f_i$ ($1 \leq i \leq n$) be absolute values on the same field $K$ which are not improper and no two of which are equivalent. Let $a_i \ (1 \leq i \leq n)$ be elements of $K$ and $\varepsilon$ a real number $> 0$. Then there exists $x \in K$ such that $f_i(x - a_i) \leq \varepsilon$ for all $i$.

Let $K_i$ denote the field K with the topology defined by $f_i$. The result to be proved is equivalent to the following: in the product $P = K_1 \times \cdots \times K_n$, the closure $\overline{D}$ of the diagonal D is equal to P. This is obvious for $n = 1$. Suppose that this point has been established in the case of k absolute values for $k < n$.

We show first that there exists, for $2 \leq h \leq n$, an element $x_h$ of K such that $f_1(x_h) < 1, f_2(x_h) > 1$ and $f_i(x_h) \neq 1$ for $3 \leq i \leq h$. We argue by induction on $h$. If $h = 2$, this follows from the fact that $f_1$ and $f_2$ are not equivalent. We therefore suppose that the existence of $x_{h-1}$ has been shown and prove that of $x_h$. If $f_h(x_{h-1}) \neq 1$, we may take $x_h = x_{h-1}$; if $f_h(x_{h-1}) = 1$, we choose $z \in K^*$ such that $f_h(z) \neq 1$ and $x_h = z(x_{h-1})^s$ solves the problem for s sufficiently large. We have thus proved the existence of $x_h$.

As the integer $q$ tends to infinity, $f_1(x_n^q)$ tends to 0, $f_2(x_n^q)$ tends to $+\infty$ and $f_i(x_n^q)$ tends to 0 or $+\infty$ for $i \geq 3$. Writing $y_q = x_n^q(1 + x_n^q)^{-1}$,

hence the sequence $(y_q)$ tends to 0 in $K_1$, to 1 in $K_2$ and to 0 or 1 in $K_i$ for $i \geq 3$. By changing the numbering of the $K_i$, it may therefore be assumed that there exists an integer $r$ ($1 \leq r < n$) such that $\overline{D}$ contains the point $(e_1, \ldots, e_n)$ where $e_i = 1$ for $1 \leq i \leq r$ and $e_i = 0$ for $r + 1 \leq i \leq n$. Now, $\overline{D}$ is a vector sub-K-space of P. Hence $\overline{D}$ contains the diagonals D' and D'' of and $P'' = K_{r+1} \times \cdots \times K_n$. By the induction hypothesis, $P' = \overline{D}'$ and $P'' = \overline{D}''$. Hence $\overline{D} = P$.