Should a vector be formally defined to contain a reference to the whole vector space structure?

Question

The definition of vector I usually see is "an element of a vector space". To me, this make little sense from a completely formal standpoint, and I'll explain why.

To speak of an "element of an algebraic structure" is, from my understanding, to speak of an element of its underlying set. So, for example, if we consider the group $\mathbf Z=(\mathbb Z,+)$ (where $\mathbb Z$ is the set of integers), we may use the notation $x\in\mathbf Z$ equivalently to $x\in\mathbb Z$.

So, if I consider the vector space $(\mathbb R^2,\mathbb R,+,\cdot)$, with

$$\begin{align}+\colon\mathbb R^2\times\mathbb R^2\to\mathbb R^2\qquad&\text{s.t.}\quad(x_1,y_1)+(x_2,y_2)=(x_1+x_2,y_1+y_2)\\ \cdot\colon\mathbb R\times\mathbb R^2\to\mathbb R^2\qquad&\text{s.t.}\quad \lambda\cdot(x,y)=(\lambda\cdot x,\lambda \cdot y),\end{align}\tag1$$

I can speak of any element of $\mathbb R^2$ as a "vector". So, $(0,0)$ would be a vector. But, for example, the structure $(\mathbb R^2,+)$ is not a vector space, it's a group. So, we say that in this case $(0,0)$ is not a vector, it's an element of an abelian group.

My problem is the following: $(0,0)$ is always the same object, its value does not depend on the structure defined upon the set which $(0,0)$ is a member of. As I said before, $(0,0)$ being an element of a vector space is the same as saying it's an element of its underlying set, which is $\mathbb R^2$. So, $(0,0)$ being in a group or a vector space or whatever other structure doesn't change it as a mathematical object.

To me, this is a problem. How can we define vector as an element of a vector space, when we can define operations on that same set which don't make a vector space? Defining $+$ as in $(1)$ is not the only way possible, and $x$ being an element of a vector space doesn't add any additional structure to the element itself, taken individually.

A more sensible definition would be to define a vector in a similar way to algebraic structure: a vector is an ordered pair $(\mathbf v,V)$, where $V$ is a vector space such that $\mathbf v\in V$.

Answer 1

Your suggestion to define a vector as an ordered pair $(v, V)$ of a vector space $V$ and an element of (the underlying set of) $V$ works fine. In practice it will almost always be clear from context what $V$ is and so nobody has felt the need to make it an explicit part of the notation, in the same way that it is extremely common to write $V$ itself for a vector space even though this strictly speaking only refers to the underlying set and not to the operations on it.

Answer 2

I would like to argue that the slogan "a vector is an element of a vector space" is a little misleading, for reasons that seem similar to what you mention in the post. If we take the definition "a vector is an element of a vector space" seriously, then every $v$ is a vector, since we can define a vector space structure on $\{v\}$ by letting $v$ play the role of the zero element. Hence, this is quite a useless notion.

I think a more honest account would be as follows. There is a precise definition of a vector space over a field $k$, and if $V$ is a vector space over $k$ which is clear from context, then we often adopt the linguistic convention of calling the elements of (the underlying set of) $V$ "vectors". This linguistic convention is not absolute: I have never heard anyone call the elements of $\mathbb R$ "vectors", even though $\mathbb R$ is a vector space over $\mathbb R$.

Here is an analogy. In analysis class, we learn that $\mathbb R$ can be characterised up to unique isomorphism as a complete ordered field. Later in the course, the skeptics are pleased to learn that we don't need to take the existence of a complete ordered field on faith: given a copy of the rational numbers $\mathbb Q$, one can construct a complete ordered field out of Dedekind cuts, or equivalence classes of Cauchy sequences of rationals. While these constructions technically lead to different fields, this doesn't have any significance whatsoever. Any reasonable theorem of analysis is true for one complete ordered field if and only if it is true for all complete ordered fields. However, there is a slightly puzzling artifact of this phenomenon. Suppose we insist that all we care about $\mathbb R$ is that it is a complete ordered field, and refuse to say exactly what the underlying set of $\mathbb R$ is (this largely mirrors actual mathematical practice). Given a mathematical object $x$, it is a little awkward to ask "is $x$ a real number?", at least if we interpret this question literally. It is surely the case that $x$ is a member of some complete ordered field, since we can choose a set $X$ with cardinality continuum and such that $x\in X$, and then give $X$ the structure of a complete ordered field via transport of structure. This analysis suggests, to me at least, that when we write "let $x$ be a real number", what we actually mean is that we are fixing a complete ordered field for our discussion, and then letting $x$ be a member of that field. Similarly, when we write "let $v$ be a vector", there should be a specific vector space $V$ that has been fixed in advance, and by a "vector" we simply mean "element of $V$". To interpret words like "vector" or "real number" in any other way would seem rather perverse, and would violate the modern structural spirit of mathematics.

Answer 3

Your analysis is correct (even though I would not draw the same conclusion as you that it is a problem), but it is not unique to the definition of a vector. What is a ring element? What is a group element? What is a basis element? What is a homology class? It always depends on the context. You could formalize this with the suggestion in Qiaochu's great answer, by adding the context to the tuple, but there are other ways as well. And of course, the context does not only consist of the underlying set (of a vector space, in case of a vector), but everything we are currently talking about.

Maybe it is more clear if we add some conditions to the elements we are talking about. Take, for example, the notion of an invertible element of a ring $R$. It refers to an element $x \in R$ for which there exists an element $y \in R$ with $xy = 1$ and $yx = 1$. As we know, $x \in R$ actually means $x \in |R|$ where $|R|$ denotes the underlying set of $R$. We usually sweep forgetful functors such as $|-| : \mathbf{Ring} \to \mathbf{Set}$ under the rug, but they are present. Here, your objection, that the notion only depends on the underlying set, is not valid, since $xy = 1$ and $yx = 1$ use the multiplication. (This actually proves that the notion only depends on the underlying monoid. And in fact, most of the multiplicative theory of rings carries over to monoids.) You cannot possibly talk of an invertible element inside of a set with no further context.

But I want to emphasize that the fact that we have a condition on the elements that uses the algebraic structure should not be the only reason to have the ring in the context.

To illustrate this, given a ring $R$, for $n \in \mathbb{N}$ let us call a divisor sequence of length $n$ to be a sequence of elements $(x_1,\dotsc,x_n)$ in $R$ such that $x_i \mid x_{i+1}$. Clearly, this notion does not only depend on the underlying set. Now we may define (I do not claim that this is reasonable, I just do this to make my point): a ring element is a divisor sequence of length $1$. Since we have already agreed that the notion of a divisor sequence does not only depend on the underlying set, it is now clear that the notion of a ring element does not only depend on the underlying set.

Here is another suggestion (which is of course equivalent to the one in Qiaochu's answer). A vector is a linear map of the form $K \to V$, where $V$ is a $K$-vector space. Recall that domain and codomain belong to the data of a map (in particular, of a linear map), so we can recover the ambient vector space from a vector, as well as the base field. Similarly, we may define a ring element to be a ring homomorphism of the form $\mathbb{Z}[T] \to R$, etc. This notion has the advantage that it can also be written down in more general contexts. For example, if we are working in a topos, and we are internalizing the notion of a vector space (or ring object), we will not have an underlying set (!), but we can still use the definition above, namely that of a generalized element.