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

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.

• You need, given an open intervals $(a,b)\in \mathbb K,$ the inverse image $A=\{(x,y)\in \mathbb K^2\mid x+y\in(a,b)\}$ to be open in $\mathbb K\times \mathbb K.$ That means for each $(x,y)\in A,$ you need a pair open open intervals $x\in (x_1,x_2),$ $y\in(y_1,y_2)$ such that $x_1+y_1,x_2+y_2\in (a,b).$ Jul 9, 2023 at 0:36

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 $$\{(a,b) : a form a basis of the order topology, take an interval $$(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) 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 $$(x,y)\in N$$, $$x_1+y_1, so you just really have to find $$x_1 and $$y_1 such that $$a.
Hint for the last part: Calling $$d=x^*+y^*-a>0$$, you could take $$x_1=x^*-d/4, $$y_1=y^*-d/4, 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?