Cone in a metric space We know the definition of a cone in a Real Banach Space.
I want to know if there is any definition for a cone in an abstract metric space.
Have you ever seen such definition anywhere?
 A: There is a natural definition of cone in the context of pointed metric spaces:

A pointed metric space $(X,p)$ is a cone if $(\lambda X,p)$ is isometric to $(X,p)$ for all $\lambda>0$.

Here, $(\lambda X,p)$ denotes the metric space obtained from $(X,p)$ by multiplying the distance function by $\lambda$. Also isometric should be understood in the pointed category, i.e., $(X,x_0)$ and $(Y,y_0)$ are isometric if there exists a distance preserving map $f\colon X\to Y$ such that $f(x_0)=y_0$. This is, for example, the definition you can find in the book by Burago, Burago & Ivanov, Def 8.2.1. It also coincides with the definition in a Banach space setting.
Obviously, you can then say that a subspace $(X,p)$ of your metric space $(\bar X,p)$ is a cone if it is a cone as an abstract metric space...
A: First of all, the notion of $cone$ is a purely algebraic stuff, and
not a metrical one. The $cone$ is naturally defined in the framework
of $linear$ spaces, and not of Banach spaces. One can introduce,
e.g., various "natural'' cones in a Hilbert space, without using
its norm.
However, if $\left(X,\, d\right)$is a complete metric space, and
$\psi:X\rightarrow\left[\,0,\,\infty\right)$is a lower semicontinuous
function, then the partial ordering on $X$ defined by $x\preccurlyeq y$
iff $d\left(x,y\right)\leq\psi\left(y\right)-\psi\left(x\right)$is
very useful in proving the Caristi-Kirk Fixed Point Theorem.
Another metrical variant would be to use the Ralph DeMarr' cone https://projecteuclid.org/euclid.pjm/1103034358,
combined with an Arens-Eells embedding.
A: Maybe this works: given a point $p$ and a subset $A$ of your metric space $(X,d)$ define the cone on $A$ from $p$ to be all points that lie between $p$ and $A$, that is all points $x$ with $d(p,a)=d(p,x)+d(x,a)$ for some $a\in A$.
A: You can isometrically embed any metric space into a Banach space via the Arens-Eells theorem
(original:
https://projecteuclid.org/euclid.pjm/1103043959
simpler proof by E. Michael:
http://www.jstor.org/stable/2034516?origin=JSTOR-pdf
). This embedding is, in some sense, canonical. Convex cones are well-defined in Banach spaces, so you could say that a point x is in the convex cone generated by x_1, ... x_n in the original metric space if f(x) is in the cone of
f(x_1), ..., f(x_n) -- where f is the Arens-Eells embedding.
A: EDIT: As pointed out by Pete below, it seems I misunderstood the question, so what I write below is not relevant.  Apologies!

This is not the general answer, but in Riemannian geometry there is a notion of cone.  If $(M,g)$ is a Riemannian manifold, then its metric cone is $\mathbb{R}^+ \times M$, with $\mathbb{R}^+$ the positive real half-line parametrised by $r>0$, with metric
$$dr^2 + r^2 g$$
The best example is of course $(M,g)$ the unit sphere in $\mathbb{R}^n$ and its cone is then $\mathbb{R}^n \setminus \lbrace 0\rbrace$.  In this case (and in this case only) the metric extends smoothly to the origin, but in general the apex of the cone is singular.
This is used as a local model for conical (!) singularities and there is a nice interplay between the geometry of $M$ and that of its cone.  The most dramatic use of the cone I know is that it turns the problem of determining which Riemannian spin manifolds admit real Killing spinors into a holonomy problem, namely the determination of which metric cones admit parallel spinors.
Some of this generalises to the pseudo-riemannian setting; although this is perhaps not as useful as in the Riemannian setting as the holonomy classification in indefinite signatures (except for Lorentzian) is still lacking.
A: cone is a special type of set which should satisfy some properties.
If P is cone, then it should be closed non empty having at least 2 elements.
All linear combinations by positive constants should also lie in P.  If x lies in p, then its additive inverse may not lie in p
