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Does a metric space have an origin? That is, does it have $(0,0)$.

Does a vector space have an origin?

It seems whatever you can do in a metric space can also be done in a vector space. Is this true?

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They are entirely different concepts. As to an origin, a general metric space does not have anything that behaves like the ordinary number zero does. A vector space does have a unique "zero-like" object, that is, a vector, that we often call $0$, such that $0+v=v+0=v$ for any vector $v$ in the space. Although the concepts of vector space and metric space are entirely different, some familiar spaces, uch as $\mathbb{R^3}$, are simultaneously vector spaces and metric spaces, and there is interaction between the vector structure and the metric structure. – André Nicolas Feb 29 '12 at 19:58
The answers below tell the tale, but I am interested to know what you mean by the penultimate sentence in your question. What sort of things have you seen done in metric spaces that can (apparently) also be done in vector spaces? – arjafi Feb 29 '12 at 20:17

No, a metric space does not have any particular distinguished point called "the origin". A vector space does: it is defined by the property $0 + x = x$ for every $x$.

In general, in a metric space you don't have the operations of addition and scalar multiplication that you have in a vector space. On the other hand, in general a vector space does not have a notion of "distance".

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Normed linear vector spaces may have distance functions. For example $d(\vec x, \vec y)=||\vec x - \vec y||$. – user137035 Aug 25 '15 at 12:21

Vector spaces necessarily have a vector called the "zero vector"; in the special case of the vector space $k^n$ (where $k$ is a field), this vector is often called "the origin", since $k^n$ also can be seen as a geometric object (the $n$-dimensional affine space). But vector spaces don't necessarily have something we call "the origin": the collection of all polynomials with real coefficients is a real vector space, but we don't normally refer to the zero polynomial as "the origin", even though it is the zero vector of this vector space.

Metric spaces are sets with a metric defined on them. For example, the collection of all complex numbers with complex norm $1$, and with metric given by the usual distance between them as complex numbers, is a metric space. Any nonempty subset of the real numbers, with the usual distance function, is a metric space; and any nonempty set $X$, with distance defined by $d(x,y) = 0$ if $x=y$ and $d(x,y)=1$ if $x\neq y$, is a metric space. There need not be anything that we can reasonably call "the origin."

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A metric space is a set with a distance. That's all you know. That means the set may not have an algebraic structure. for example, {chair, apple} is a set. define d(apple, apple) = d(chair, chair) = 0 and d(chair, apple) = d(apple, chair) = 1. That's a metric space, and it doesnt look like a vectorial space at all.

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I think the OP is confusing a vector space with a normed vector space,which indeed shares many properties of general metric spaces.And for a very simple reason: A norm induces a metric on the underlying set on which the map is defined.This is fairly simple to prove from the definitions and the questioner should try and do it. A general vector space does NOT necessarily have these properties,of course.It is not even necessary that a general vector space admits any notion of distance whatsoever.

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Every vector space (over a subfield of $\mathbb{C}$) admits many different norms, and thus can be made into a metric space in multiple ways. – Adam Smith Mar 1 '12 at 18:45
You're kidding,right? Yes,there are many different norms that can be constructed on a vector space,leading to many possible metrics and their respective topologies.How does that make what I just posted incorrect and deserving of a -1? It does NOT follow from the definition of a vector space in and of itself that that's true. My point was the definition of a vector space in and of itself-without a norm-does NOT admit any natural notion of distance and this was the point of confusion for the questioner.The notion of a norm is independent and needs to be considered separately. – Mathemagician1234 Mar 1 '12 at 19:43
You made the following false statement : "It is not even necessary that a general vector space admits any notion of distance whatsoever." The word natural does not appear there. – Adam Smith Mar 1 '12 at 20:49
@Adam The definition of a vector space does NOT admit IN AND OF ITSELF any notion of distance at all. There is nothing in either the axiomatic (10 axioms;additive and multiplicative identities and inverses,addition,scalar multiplication,etc.) or algebraic (an Abelian group over a field closed under scalar multiplication and the distributive law) .THERE IS ABSOLUTELY NOTHING IN EITHER OF THOSE DEFINITIONS THAT IMPLIES A NOTION OF DISTANCE,EITHER DIRECTLY OR INDIRECTLY. Any distance notion must be imposed by further axioms.Algebraicists in the room-am I right or wrong here? – Mathemagician1234 Mar 3 '12 at 5:15

A metric space is a set with a notion of distance defined between points of that set. This notion of distance is a function known as the metric (which must satisfy a set of axioms pertaining to distance). This metric takes in any two points and maps them onto a real number which characterises the distance between those two points.

A vector space is a set containing objects called vectors which interact in some pre-defined way determined by the axioms. Vectors are measured relative to some reference frame and thus have a notion of magnitude and direction from some origin.

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