Example 2

\* The topology of the rational line $Q$ (resp. the real line $R$) is compatible with the field structure of $R$ (resp. $R$) (see Chapter IV, § 3). \*

Definition 4 shows that, if $K$ is a topological division ring, then the topology induced by that of $K$ on the multiplicative group $K^*$ is compatible with the group structure of $K^*$.

If $a \neq 0$, the homotheties $x \to ax$ and $x \to xa$ are homeomorphisms of $K$ onto itself; and so is the mapping $x \to ax + b$ for all $b \in K$. Note that the homotheties $x \to ax$ and $x \to xa$ are automorphisms of the (topological) additive group of $K$ if $a \neq 0$. If $V$ is any neighbourhood of $o$ in $K$, it follows therefore that $aV$ and $Va$ are neighbourhoods of $o$ for all $a \neq 0$.

Let $X$ be a topological space, and let $f$ be a mapping of $X$ into a topological division ring $K$. If $f$ is continuous at a point $x_0 \in X$ and if $f(x_0) \neq 0$, then $f^{-1}$ is continuous at $x_0$. In particular, if $K$ is a topological field, then every rational function in $n$ variables with coefficients in $K$ is continuous at every point of $K^n$ where its denominator does not vanish.

Likewise, if $f$ is a mapping of a set $X$, filtered by a filter $\mathfrak{F}$, into a Hausdorff topological division ring $K$, and if $\lim_{\mathfrak{F}} f$ exists and is not $0$, then $\lim_{\mathfrak{F}} f^{-1}$ exists and we have

If $H$ is a division subring of a topological division ring $K$, then the topology induced on $H$ by the topology of $K$ is compatible with the division ring structure of $H$. The structure of a topological division ring thus defined on $H$ is said to be induced by that of $K$. Furthermore, $\overline{H}$ is also a division subring of $K$ (the proof is analogous to that of Proposition 5).

In a topological division ring $K$, the closure of the set $\{0\}$ is a two-sided ideal, by Proposition 5, and hence must be either $\{0\}$ or $K$. In other words, if the topology of $K$ is not the coarsest topology (Chapter I, § 2, no. 2) then it is Hausdorff (§ 1, no. 2, Proposition 2).