The additive group of an ordered field along with the order topology is a topological group

Question

Let $(\mathbb{K},+, \cdot, \leq)$ be an ordered field. I'm trying to prove without success that the sum $+$ is continuous with respect to the order topology.

The problem is that such a space need not be metrizable - When is an ordered field a metric space?, so the arguments for $\mathbb{R}$ may not be reproduced.

Could you give me any hint?

Any help would be appreciated.

Answer 1

You just need to prove that the map $f:\mathbb{K}\times\mathbb{K}\to\mathbb{K}$, $(x,y)\mapsto x+y$ is continuous with respect to the topologies involved (the product of the order topologies on $\mathbb{K}\times\mathbb{K}$ and the order topology on $\mathbb{K}$). You could do it as follows:

1. Since the open intervals $\{(a,b) : a<b\}$ form a basis of the order topology, take an interval $(a,b)$.
2. Think about what is the inverse image under $f$ of your interval $(a,b)$: $$U=f^{-1}(a,b) = \{(x,y)\in\mathbb{K}\times\mathbb{K} : a < x+y < b\}$$
3. Prove that $U$ is open. For that, just take any $(x^*,y^*)\in U$ (i.e., such that $a<x^*+y^*<b$) and then try to find some neighborhood $N=(x_1,x_2)\times(y_1,y_2)\ni(x^*,y^*)$ such that $N\subseteq U$. Observe that, for all $(x,y)\in N$, $x_1+y_1<x+y<x_2+y_2$, so you just really have to find $x_1<x^*<x_2$ and $y_1<y^*<y_2$ such that $a<x_1+y_1<x_2+y_2<b$.

Hint for the last part: Calling $d=x^*+y^*-a>0$, you could take $x_1=x^*-d/4<x^*$, $y_1=y^*-d/4<y^*$, so that $x_1+y_1 = x^*+y^*-d/2>x^*+y^*-d=a$. Can you find suitable $x_2,y_2$ and conclude the argument?