The question is based on Proposition $5.8$ in 'Infinite dimensional Lie algebras' by Victor G. Kac. The Proposition describes the closure of the tits cone in the $\textit{metric topology}$ of $\mathfrak{h}_{\mathbb{R}}$ (defined in Section $2.7$) of a Kac-Moody Lie algebra $\mathfrak{g}=\mathfrak{g}(A).$ I am really confused about the term $\textit{metric topology}$ here because of the following fact:
If $A$ if of finite type, then the restriction of the standard bilinear form on $\mathfrak{h}_{\mathbb{R}}$ is positive definite and hence there is a topology generated by the inner product. But even if affine case the restriction of the form to $\mathfrak{h}_{\mathbb{R}}$ is not an inner product since $(\delta,\delta)=0,$ $\delta$ being the basic imaginary root. So what metric topology the author is referring to?
There is a chance that we can declare a basis of $\mathfrak{h}_{\mathbb{R}}$ (of course with all the requirement fullfilled, e.g. all the $\alpha_i^\vee$'s are included etc) orthonormal and work with the topology generated by that since as an affine space $\mathfrak{h}_{\mathbb{R}}$ is same as $\mathbb{R}^n$ for some $n\in \mathbb{N}.$ But to prove the Proposition $8.3(c),$ he is using the fact that any open ball containing an element $h$ such that $(\alpha_i,h)\in \mathbb{Z}$ which has the property that there are only finitely many positive imaginary roots $\gamma$ with $(\gamma,h)<0$ intersects the tits cone $X$ which can be described by (Proposition $3.12$) $$X=\{h\in \mathfrak{h}_{\mathbb{R}}:\text{ there are only finitely many positive imaginary roots }\gamma \text{ with }(\gamma,h)<0\}.$$ Hence the set of integral linear combinations of $\alpha_i^\vee$ is dense in $X$ which I don't see why it should be true because $\mathbb{Z}^n$ is discrete in $\mathbb{R}^n$ in the standard Euclidean topology.
I shall really appreciate any comment or suggestion. Thank you very much for your effort in advance.