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I'm working through Axler's linear algebra textbook and in it he says that $F^n$, the set of n-tuples with elements in $\mathbb{R}$ or $\mathbb{C}$, as well as $F^\infty$, are specific cases of $F^S$ - the set of functions mapping from $S$ to $F$. He states that we can think of $F^n$ as $F^{\{1,2,...,n\}}$ and analogously for $F^\infty$.

I am having trouble understanding how the set of all points (vectors) in n-dimensions (i.e., $F^n$) is equivalent to the set of all functions mapping the first $n$ natural numbers to $F$.

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Consider the bijection $(\eta_1, \ldots, \eta_n) \mapsto (k \mapsto \eta_k)$ where $k$ runs through $1, \ldots, n.$ In other words, given a vector $\eta = (\eta_1, \ldots, \eta_n) \in \mathbf{K}^n$ consider the function $\varphi_\eta:\{1, \ldots, n\} \to \mathbf{K}$ given by $\varphi_\eta(k)=\eta_k.$ The bijection is $\eta \mapsto \varphi_\eta;$ the inverse is given by $\psi \mapsto (\psi(1), \ldots, \psi(k)).$

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  • $\begingroup$ I think I understand now. Is there an explanation related to cardinality as to why the set of size-$n$ vectors form a one-to-one correspondence with the set of functions {$f:\{1,2,..,n\} \to$ K} ? $\endgroup$
    – Brian
    Feb 27, 2019 at 19:26
  • $\begingroup$ Uh, the number of functions $f:\{1, \ldots, n\} \to \mathbf{K}$ is infinity (consider constant functions. $\endgroup$
    – William M.
    Feb 27, 2019 at 20:05
  • $\begingroup$ Yeah, I meant since both sets are denumerable, how do you justify stating they form a bijection? $\endgroup$
    – Brian
    Feb 27, 2019 at 20:28
  • $\begingroup$ I gave you the inverse. $\endgroup$
    – William M.
    Feb 27, 2019 at 20:45

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