Let $(\mathbb{X}, || \cdot||)$ be a real Banach space.

We define a subset $P$ of $\mathbb{X}$ by $P := \{ x \in \mathbb{X} : x\geq 0\}$.

In general, $P$ is termed a positive cone of $\mathbb{X}$. However, $\mathbb{X}$ is quite abstract and could be any real Banach space, so that I am wondering that whether $P$ always be a closed subset of $\mathbb{X}$ (as primitive)? If so, how to show that?

In fact, there are many different definitions in textbooks for "cone". One is defined as "A subset $C$ of $\mathbb{X}$ is called a cone iff (i) $C$ is nonempty and nontrival ($C \neq \{0\}$); (ii) $C$ is closed and convex; (iii) $\lambda C \subset C$ for any nonnegative real number $\lambda$; (iv) $C \cap (-C) = \{ 0\}$." On the other hand, other textbook defines it as " A subset $C$ of $\mathbb{X}$ is called a cone iff (iii) and (iv) are satisfied."

Therefore, it makes me confused about whether a positive cone $P := \{ x \in \mathbb{X} : x\geq 0\}$ is always a closed subset.

I thought a positive cone $P$ is always closed. Because by the construction of $P$, for any convergent sequence $\{x_n \} \subset P$ such that $ x_n \rightarrow x \in \mathbb{X}$ (i.e., $||x_n - x|| \rightarrow 0$ as $n \rightarrow \infty$), we have $x_n \geq 0$ for all $n$. And, it is clear intuitively that the limit $x$ of this sequence should also be nonnegative. But how to prove it rigorously? Thanks for helping in advance!

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    $\begingroup$ What does $x\ge 0$ mean if $x$ is a point in an arbitrary Banach space? $\endgroup$ – Jack Lee Dec 19 '16 at 23:55
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    $\begingroup$ To paraphrase the previous comment, you won't be able to prove what you're looking for until you've specified the connection between the norm and the order structure. $\endgroup$ – Theoretical Economist Dec 20 '16 at 0:09
  • $\begingroup$ A cone is not necessarily closed, for example in $\ell^2$ consider the sequences with elements $≥0$ and only finitely many non-zero. This is a cone, but not closed. $\endgroup$ – s.harp Dec 20 '16 at 0:11
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    $\begingroup$ What order do you have on $\mathbb{X}$? $\endgroup$ – copper.hat Dec 20 '16 at 0:32
  • $\begingroup$ Thanks @TheoreticalEconomist. What if $\mathbb{X}$ be a Banach space regarding real-valued functions? so that $P$ is a subset of nonnegative functions from $\mathbb{X}$. The order structure is generated by this positive cone and the norm is a complete norm on $\mathbb{X}$. Under this situation, could we prove it now? or to this end, I thought we probably need to make some additional assumption. $\endgroup$ – Paradiesvogel Dec 20 '16 at 0:35

Although your question has been answered in the comments, let me collect a few remarks which might be of interest to you:

First of all, a cone as defined by properties (iii) and (iv) is a purely algebraic object. This means that in order to define a cone $P$, you only need an underlying vector space structure $X$. Here $X$ need not be a normed space. So if your space $X$ has additionally a norm on it, or more generally a linear topology, then there is no reason why $P$ should be closed with respect to it. You need to know how the norm interplays with your order, so that you can make deductions of this kind.

It's also true that the definition of a cone varies from text to text. The most broad definition is that a cone is a set $P$ which satisfies (iii). If it additionally satisfies (iv) then it is called a pointed cone. If it satisfies $P+P\subseteq P$, then it is (called) a convex cone. Some texts might want to study only a specific class of cones, for instance only closed convex cones, so they sometimes adopt the definition accordingly.

Now let me mention a few instances which ensure that $P$ is closed. The property that you wrote in the comment section is obviously equivalent to $P$ being closed, but it isn't very useful because one can use the same method to show immediately that $P$ is closed. I mean, although it is correct, it doesn't provide us with a different way to determine whether $P$ is closed.

Recall that an ordered vector space $(X, P)$ is called Archimedean if for every $x, y\in X$ such that $ny\leq x$ for every $n \in \mathbb{N}$, then $y\leq 0$. This is a purely algebraic property and strangely it has a lot to say about $P$ being closed or not! The following is true:

Proposition: Let $(X, P)$ be an ordered vector space and $\|\cdot\|$ be a norm on $X$. If $P$ has non empty interior, then the following are equivalent:

(a) $P$ is closed,

(b) $P$ is Archimedean.

Notice here a mild assumption about the order and norm interplay: The cone $P$ has non empty interior with respect to the norm. Other than that, the proposition connects a topological property with a purely algebraic one. So it provides an alternative (algebraic) way to test whether a given cone is closed or not.

The proposition above still holds if instead of a norm you have a $T_2$ linear topology.

A simple example of a non closed cone is $\mathbb{R}^2$ equipped with the lexicographical order. The cone $P$ is $P=\{(x, y): x>0 \text{ or } x=0 \text{ and } y\geq 0\}$ and it is obviously not closed in the usual topology. Notice that $P$ has non empty interior and that the lexicographical order is not Archimedean, all in accordance with the previous Proposition.

The last example is that of normed vector lattices and of Banach lattices, which are probably the most important classes of ordered spaces:

Definition: A normed vector lattice $X$ is an ordered space $X$, equipped with a norm such that:

1) For every $z, w\in X$, $\sup\{z,w\}$ exists.

2) If $x, y\in X$ with $|x|\leq |y|$, then $\|x\|\leq \|y\|$. Here by the absolute value we mean $|x|=\sup\{x,-x\}$.

Condition 2) tells us that "small" elements with respect to the order, also have "small" norm. One can easily show that in normed vector lattices the cone $P$ is closed.

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    $\begingroup$ Thanks so much @tree detective! That's really helpful. May I ask one more question that how to show the cone $L^{p}_{+} (\mu)$ of $L^{p}(\mu)$ space is Archimedean? I mean, if we attempt to show that, in this case is it enough and correct to just simply replace a "pointwise" ordering with an "almost everywhere" ordering? Many thanks again. $\endgroup$ – Paradiesvogel Dec 21 '16 at 23:02
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    $\begingroup$ Moreover, I realized that the cone $L^{p}_{+}(\mu)$ does not contain interior point, so that if we attempt to prove $L^{p}_{+}(\mu)$ is a closed cone in $L_p$ space, the proposition you stated above can not be applied. Therefore, how to appropriately use that proposition to show $L^{p}_{+}(\mu)$ is closed? If it does not work, how to prove $L^{p}_{+}(\mu)$ is a closed subset of $L^{p}(\mu)$? $\endgroup$ – Paradiesvogel Dec 22 '16 at 1:04
  • $\begingroup$ You are welcome! Yes, you should use the almost everywhere pointwise ordering because this is how the ordering in these spaces is defined: $f\leq g$ if $f(x)\leq g(x)$ for almost every $x\in \Omega$. The proposition above can't be applied here indeed, for the reason you mentioned. But $L_p$ spaces are Banach lattices (it's pretty obvious they satisfy properties 1) and 2) of the definition of the normed vector lattices). So their cones are closed. (Of course you need to check the proof in the Banach Lattice case, but its rather short). $\endgroup$ – tree detective Dec 22 '16 at 1:17
  • $\begingroup$ The Archimedean property here follows immediately from the definition. It also follows from the Proposition above, as the implication $P$ is closed $\Rightarrow $ $P$ is Archimedean is always correct, regardless of what happens with the interior of $P$. $\endgroup$ – tree detective Dec 22 '16 at 1:19
  • $\begingroup$ Thanks for your reply. I agree with you. Could you please briefly show me how to prove the Proposition you stated above? or please give me some reference of that? $\endgroup$ – Paradiesvogel Dec 22 '16 at 1:47

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