# completeness of cones in an ordered normed space

Let $(X,\|\cdot\|,\le)$ be a normed, ordered vector space over $\mathbb{R}$ and let $X^+=\lbrace x\in X:x\ge0\rbrace$ denote the (positive) cone in $X$. with a metric $d$ induced by the norm $\|\cdot\|$.

Is there an example of a (positive) cone $X^+$ and normed space $X$ such that $(X^+,d)$ complete and the normed space $X,\|\cdot\|$ is not complete?

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Yes, though I'm not sure if you'll like this example. Take any incomplete normed space and equip it with the trivial order, where $x \le y$ iff $x=y$ (i.e. any distinct elements are incomparable). This satisfies the axioms of an ordered vector space, and $X^+ = \{0\}$ which is of course complete.
For a somewhat natural example of this, let $X$ be the vector space of sequences of real numbers which are eventually zero and which sum to zero. That is, its elements are of the form $(x_1, \dots, x_n, 0,0, \dots)$ with $\sum_{i=1}^n x_i = 0$. This is clearly a vector space, and when equipped with the natural pointwise ordering (i.e. $x \ge y$ if $x_i \ge y_i$ for all $i$) the only positive element is 0, so the ordering is trivial. $X$ has countable dimension (a Hamel basis is given by sequences of the form $(0,\dots, 0,1,-1,0,\dots)$) and so by the Baire category theorem $X$ is incomplete in any norm. (You could use an $\ell^p$ norm, for instance.)