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I found these two posts but they don’t explain in detail why functions can be thought as tuples. The autor wants to show that $\mathbb R^n$ and $\mathbb R^{\{1,2,...,n\} }$ are actually the same thing.

Note: if $S$ is a set then $\mathbb R^S$ is the set of all functions from $S$ to $\mathbb R $.

Intrepreting tuples as functions

help understanding a paragraph from Linear Algebra Done Right

littleO
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Brian de León
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    Often the term "$n$-tuple of real numbers" is left undefined, because its meaning seems to be intuitively clear. But if you insist on giving a precise definition to this term, it is hard to think of a simpler definition than "a function from ${1,2, \ldots, n }$ to $\mathbb R$." – littleO Nov 02 '20 at 22:45
  • Hi @littleO it’s seems that the whole idea it’s not precisely clear to me. I don’t understand why a n-tuple of real numbers is a function from positive integers to R. I’ve read set theory’s books about this and define a general Cartesian product this way too but didn’t help at all. So.... I don’t know what I’m missing here. – Brian de León Nov 03 '20 at 03:20
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    Suppose that $x$ is an ordered $n$-tuple of real numbers. We typically write the $i$th component of $x$ as $x_i$. But there's an alternate notation which is to write the $i$th component of $x$ as $x(i)$. With that notation maybe it seems more natural to define an ordered $n$-tuple of real numbers to be a function $x: {1,\ldots, n} \to \mathbb R$. – littleO Nov 03 '20 at 03:54

2 Answers2

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The spaces $\Bbb R^{\{1,\dots,n\}}$ and $\Bbb R^n$ are isomorphic as vector spaces. In particular, the map $\Phi:\Bbb R^{\{1,\dots,n\}} \to \Bbb R^n$ defined by $$ \Phi(f) = (f(1),f(2),\dots,f(n)) $$ is an isomorphism.

Ben Grossmann
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Imagine a function $f$ from the set $\{1,2,..,n\}$ to the real numbers. To define such a function we must specify one value for each of the inputs $1,2,..,n$. Therefore, we get $f(1), f(2), .., f(n)$ which we can put in a tuple as $(f(1), f(2),..,f(n))$. Therefore, we can map any function to a tuple in $\mathbb R^n$.

Given any tuple $(x_1,.x_n)\in\mathbb R^n$, we can define the function $f$ by $f(i)=x_i$

Therefore, we can view any tuple as a function from $\{1,2,..,n\}$ to $\mathbb R$.

John Douma
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  • For example f(x) = log(|x|) is an element of {f:{1,2,...,n} → R} then f(1)...f(n) is the same element of that set?. f(x) = f(1)....f(n)?. Does that implies the set {f:{1,2,...,n} → R} is a set of sets?. So, f(x) is a function (an element) but also a set? . Why can you put the function evaluated in the set as a tuple anyways? – Brian de León Nov 02 '20 at 20:26
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    You cannot convert any function to a tuple. We are specifically talking about functions from the domain set ${1..n}$. For example, the point in $\mathbb R^3$ defined by $(2,1,0)$ can be thought of as a function from the set ${1,2,3}$ where $f(1)=2$, $f(2)=1$ and $f(3)=0$. – John Douma Nov 02 '20 at 20:29