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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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2 Answers 2

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.)

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Nate has already answered this perfectly well, but perhaps it is worth noting the question becomes a little more difficult if we require $X^+$ to be large enough to generate $X$, i.e. $X=X^+-X^+$. For an example in this case, let $$X^+=\{(x_n)\in c_0:\forall n(0\leq nx_{2n}\leq x_{2n+1})\},$$ which is complete w.r.t. the supremum norm. Then $X=X^+-X^+$ contains all eventually $0$ sequences. Thus if $X$ were complete w.r.t. the supremum norm it would contain $c_0(=$all sequences converging to $0)$ and, in particular, the sequence $(1/n)$. But if $(1/n)=(x_n)-(y_n)$, for some $(x_n),(y_n)\in X^+$, then $x_{2n+1}\geq nx_{2n}\geq n/(2n)=1/2$, for all $n$, so $(x_n)\notin c_0$, a contradiction.

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