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I'm self-teaching my way through linear algebra done right by Axler, I'm using the 3rd edition. However, I'm a bit confused by the wording in Example 1.22:

$ \mathbf{F}^{\infty}$ is defined to be the set of all sequences of elements of $\mathbf{F}$: $$\mathbf{F}^{\infty} = \{ (x_1,x_2,...):x_{j}\in \mathbf{F} \text{ for } j = 1,2,...\}$$


Following from definition in the first part of chapter 1, I know that $\mathbf{F}^{n}$ denotes any list of length n with elements of $\mathbf{F}$.

  • so the notation here would suggest $\mathbf{F}^{\infty}$ is any list of infinite length

  • but the words suggest that it's something like any possible list of arbitrary length

  • or the set of all possible lists with members of F and arbitrary length $$\{ (x_1),\;(x_2),\;(x_1,x_2),\;(x_3),\;(x_1,x_3),\;(x_2,x_3),\; (x_1,x_2,x_3), ...\}$$

  • or the set of all lists of infinite length and members of $\mathbf{F}$

or something else?

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    $\begingroup$ $F^\infty$ is a set. The elements of this set are infinite lists, where each element of the list is an element of $F$. I think you’re confused because you think a “sequence” can be a finite list. That’s not true. A finite list of length $n$ is called an $n$-tuple, not a sequence. A sequence, by contrast, is always infinite in length. $\endgroup$ Jan 14, 2022 at 2:28

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The elements of $\mathbb{F}^\infty$ are sequences (lists of infinite length), where every element in the sequence belongs to the field $\mathbb{F}$. For example, $(1, 1/2, 1/3, 1/4, \ldots ) \in \mathbb{R}^\infty$, but $(1,2) \not\in \mathbb{R}^\infty$.

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    $\begingroup$ I think it was the infinite length aspect of sequences that I had not caught on to, thank you for clearing that up for me. $\endgroup$ Jan 14, 2022 at 2:46
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    $\begingroup$ @FlynnSullivan There are certainly people who talk about “finite sequences” (meaning tuples of finite length), so this might have caused your confusion. But generally, a sequence (without further specification) is of infinite length. $\endgroup$ Jan 14, 2022 at 7:14

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