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It's written in my linear algebra textbook that $\mathbb{R}^S$ is the set of all possible functions $f: S \to \mathbb{R}$, I fail to see how this applies when $S=2=\lbrace {0,1} \rbrace$ as it's not clear to me how a vector like $(72,15)$ is a function

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    $\begingroup$ I think $2=\{0,1\}$, and then $(72,15)$ represents the function where $f(0)=72$ and $f(1)=15$. $\endgroup$ – Cheerful Parsnip Jan 31 at 18:31
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    $\begingroup$ My comment should answer your question. Is there anything unclear about it? $\endgroup$ – Cheerful Parsnip Jan 31 at 18:39
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    $\begingroup$ @CheerfulParsnip If your comment answers the question, you should make it an answer! I’d gladly +1 $\endgroup$ – Santana Afton Jan 31 at 22:50
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You can think of $\mathbb R^n$ as lists of numbers but you can also think of it as functions $f\colon \{1,\ldots,n\}\to \mathbb R$ where the $j$th coordinate can be regarded as $f(j)$. Then, if $|S|=s$ we have that $\mathbb R^S\cong \mathbb R^s$.

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    $\begingroup$ Small edit - changed $\mathbb{R}^2$ to $\mathbb{R}^n$ since you used $n$ afterwards. $\endgroup$ – Sambo Jan 31 at 23:10
  • $\begingroup$ @Sambo, oops. Thanks. $\endgroup$ – Cheerful Parsnip Jan 31 at 23:14

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