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It seems like any countable, non-dense subset of the reals is a vector by any other name. Don't get me wrong, a set is a set, and a vector is a vector - it just seems like there isn't much to distinguish the two aside from name and context.

Consider, for example, the set of natural numbers $\mathbb{N}=\{1,2,\ldots\}$. In what way is the vector $\textbf{v}\in\left\{\mathbb{N}^\infty\mid v_n=n\right\}, \textbf{v}=(1,2,\ldots)$ not the same as $\mathbb{N}$? Both have the same number of elements, satisfy total order, contain the natural numbers, and it is possible to substitute one for the other in a great many cases with little to no effort (ex: $\sum_{n\in\mathbb{N}}f(v_n)=\sum_{n\in\mathbb{N}}f(n)$, $a_n=g(v_n)=g(n)=\left(a_n\right)_{n\in\mathbb{N}}$, etc.)

I suppose that the representation of a vector, in writing, is understood to imply its order (whereas the order of the terms in a set do not change the intended meaning) - but I feel like there should be something more substantial than this. After all, the written representation of a mathematical object is distinct from the object itself, and I could easily create a notation where the written order of terms in a vector need not correspond to that of the actual vector.

I hope this isn't a silly question.

To clarify: This is not to say that a set and a vector are the same thing. Rather, the information needed to describe a vector is the same information needed to describe a particular type of set (i.e. the ordering between any two members)

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  • $\begingroup$ I don't understand. Do you know what a vector is? $\endgroup$
    – lulu
    Dec 30, 2018 at 1:06
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    $\begingroup$ Reminds me of the old riddle: "What is the difference between Marilyn Monroe and a submarine?" "Marilyn Monroe is a movie actress, and a submarine is a boat that goes under water." $\endgroup$
    – bof
    Dec 30, 2018 at 1:44
  • $\begingroup$ What does "nowhere dense" have to do with it? Why is a countable dense set like the rational numbers any less like a vector than a countable nowhere dense set? $\endgroup$
    – bof
    Jan 1, 2019 at 3:32
  • $\begingroup$ @bof Some orderings of a dense set cannot be encoded by a vector. Take, for example, $\mathbb{Q}\cap[0,1]$, ordered by $i<j\iff a_i<a_j$. The first element of the vector $(a_1,\ldots,a_\infty)$, would obviously be $0$, but since the next element can be arbitrarily close to $0$, there cannot be a unique $a_2$. Thus, there is no unambiguous vector which encodes both the specified, order and all elements of the rationals in $[0,1]$. Granted, it is possible to create a vector to represent any countable set. But only non-dense sets can be represented this way for any order. $\endgroup$
    – R. Burton
    Jan 1, 2019 at 17:59
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    $\begingroup$ How do you represent $\mathbb Z$ with the natural order as a vector? Better yet, how about the nowhere dense countable set consisting of the midpoints of the "deleted intervals" of the Cantor set? This can be described as the set of numbers whose base $3$ representation has a $1$ in it, and all digits after the first $1$ are $1$'s, i.e., $0.\overline1$, $0.0\overline1$, $0.2\overline1$, $0.00\overline1$, $0.02\overline1$, $0.20\overline1$, $0.22\overline1$, $0.000\overline1$, $0.002\overline1$, $\dots$, This is a countable nowhere dense set, and it is order-isomorphic to $\mathbb Q$. $\endgroup$
    – bof
    Jan 2, 2019 at 3:16

1 Answer 1

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Unfortunately, this probably is not a satisfying answer, but the key difference between the set and the vector is that the set has no order to its terms, while the vector does. You say you could change the written representation of the vector and change the order, but that change in the written representation does not alter the fact that the inherent mathematical object which that representation represents—that is, the vector—still has an order in its underlying structure.

I think a better way to explain this is by trying to better understand what $\Bbb{N}^\infty$ is. For me, I like to think of $\Bbb{N}^\infty$ as the set of all functions from $\Bbb{N}\rightarrow \Bbb{N}$. Thus, here, $v$ is the vector such that $v(n)=n$ for all $n\in \Bbb{N}$, which is like the identity function on $\Bbb{N}$. However, even though $v(n)$ can always just be replaced with $n$ and therefore the identity function may seem redundant, that does not change the underlying structure behind $v(n)$, which is that it is an infinite-dimensional vector, and not a set.

Another way to realize the difference is by examining the vector $w(1)=2$, $w(2)=1$ and $w(n)=n$ for all $n > 3$. Clearly, the set of components of this vector are the natural numbers. However, this is also true for $v$, yet $v\neq w$ because $v$ and $w$ have their components ordered differently: $v$ has the order $1$, then $2$, while $w$ has the order $2$, then $1$. On the other hand, $\Bbb{N}$ has no such underlying order to it, so it is fundamentally distinct from vectors.

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    $\begingroup$ Good answer. I would add that a vector is only a vector in virtue of the larger structure of a vector space. $\endgroup$
    – Ben W
    Dec 30, 2018 at 1:13
  • $\begingroup$ For each vector whose entries are (for continuity's sake) natural numbers, would there be a unique function $Sort:\mathbb{N}\to\mathbb{N}$ which assigns a unique position in $v$ to each $n\in \mathbb{N}$? $\endgroup$
    – R. Burton
    Dec 30, 2018 at 1:14
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    $\begingroup$ @R.Burton I'm not sure if I understand your comment, but I think what you're talking about is called the inverse function and is denoted as $v^{-1}$. Obviously, for $v$, $v^{-1}(n)=n$ since the position of each element $n$ in $v$ is simply $n$. However, the inverse function does not exist for all elements of $\Bbb{N}^\infty$. See $\zeta(n)=0$ for all $n\in\Bbb{N}$. $\endgroup$
    – Noble Mushtak
    Dec 30, 2018 at 1:15
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    $\begingroup$ @R.Burton I see what you mean now. Yes, for any subset of $X \subset \Bbb{N}$, there is a corresponding vector $\mathbf{v}$ such that the set of all components in $\mathbf{v}$ is $X$. However, this does not mean $\mathbf{v}$ is the "equivalent" vector or sequence to $X$, because there are multiple vectors for which the set of all components is $X$. $\endgroup$
    – Noble Mushtak
    Jan 1, 2019 at 20:06
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    $\begingroup$ Take, for instance, the vectors $v$ and $w$ in my answer above. $v$ and $w$ are clearly distinct vectors, but from your description of the $\text{Sort}$ function, $\text{Sort}(\Bbb{N})$ could be $v$ or it could be $w$ since, in both $v$ and $w$, the set of all components is $\Bbb{N}$. $\endgroup$
    – Noble Mushtak
    Jan 1, 2019 at 20:06

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