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Let $V$ be a real vector space. It is well known that, given a totally ordered basis of $V$ (say $(b_i)_{i\in I}$ where $I,<$ is totally ordered), $V$ is totally ordered by the lexicographic ordering, the (strict) positive cone being given by the (finite) sums of the type $$\alpha\, b_{i_{Max}}+\sum_{i<i_{Max}}\beta_i\,b_i$$ with $\alpha>0$ (the other coefficients being of arbitrary sign).

One can ask oneself whether the converse is true.

The answer is NO in infinite dimension as shows the counterexample (*) and YES in finite dimension because, if $V$ is of dimension $n$, then our maximal cone is contained in a (unique) closed half space (call $H$ its hyperplan), all elements of the associated open half-space are strictly positive. Take $b_n$ in this open half-space, it is the maximal element of a lexicographic basis whose remainder is constructed within the totally ordered $H$.

Counterexample (*) Take $V=\mathbb{R}[[X]]$ the ring of real series and, for each non-zero $T=\sum_{n\geq m}a_n\,X^n\in \mathbb{R}[[X]]$, let us note $$ val(T)=min\{n\in \mathbb{N}|a_n\not=0\}\ ;\ c(T)=a_{val(T)} $$ then we say that a series $T$ is strictly positive iff $c(T)> 0$. These series form a maximal cone (blunt, i.e. without zero) $C_{\mathbb{R}[[X]]}$. Now, in a ``lexicographic'' basis $(b_i)_{i\in I}$ (in the sense of the question, i.e. the basis being totally ordered by the cone, the lexicographic order induced by the basis is the original - total - ordering), one has the following property that every greater vector strictly dominates all the ray generated by an inferior vector, i.e. $$ b_i\prec_C b_j \Longrightarrow (\forall \beta \in \mathbb{R})(\beta\, b_i\prec_C b_j)\qquad (*) $$ Now, if it existed such a basis in $\mathbb{R}[[X]]$, call it $(T_i)_{i\in I}$, by property $(*)$, we would have all valuations different (i.e. the mapping $i\mapsto val(T_i)$ would be into) and then $I$ would be denumerable which is impossible because $dim(\mathbb{R}[[X]])$ is at least $card(\mathbb{R})$ (consider the series $(\frac{1}{X-a})_{a\not=0}$ which are all linearly independant). Moreover, this example, adapted to the ring of Laurent series $\mathbb{R}[[X,X^{-1}]$ shows a positive cone without supporting hyperplane.

The preceding maximal cone (that with Laurent series) has neither interior point nor supporting hyperplane. Are these two properties equivalent for infinite dimensional spaces ?

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A few hours after having posted the question, a counterexample arose (this is the magic of MO).

It can now be found in the question.

Note However the converse is true in finite dimensions. Let $V$ a real vector space of dimension $n$, then our maximal cone is contained in a (unique) closed half space (call $H$ its hyperplan), all elements of the associated open half-space are strictly positive. Take $b_n$ in this open half-space, it is the maximal element of a lexicographic basis whose remainder is constructed within the totally ordered $H$.

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Sorry again for this auto-reply. The answer is YES as it can be shown that, $C$ being a maximal blunt convex cone, TFAE

i) $C$ admits an internal point (i.e. interior point for the finest locally convex topology on $V$)

ii) $C$ admits a supporting hyperplane

iii) there exists a cofinal half-ray for $\prec_C$.

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