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In his linear algebra book, Sheldon Axler defines the set of all sequences of elements of $F$ as:

$$F^\infty = \{(x_1, x_2, \ldots): x_j \in F\text{ for } j = 1, 2, \ldots\}.$$

He also says:

Sometimes we will use the word list without specifying its length. Remember, however, that by definition each list has a finite length that is a nonnegative integer, so that an object that looks like $(x_1,x_2, \ldots)$, which might be said to have infinite length, is not a list.

I feel like there's a contradiction here. If we went by what's said in the quote we could(?) conclude that the elements of $F^\infty$ are finite in length.

Please, elaborate on this.

Clojure
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  • I think this is just distinguishing between lists as being finite and sequences as being (countably) infinite. – Ian Aug 05 '14 at 19:39
  • He defines vectors spaces like $F^n$ in terms of lists, so I am not sure he's making that distinction. – Clojure Aug 05 '14 at 19:43

1 Answers1

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There is no contradiction here because of the following. The quoted portion says that an object of the form $$ (x_1,x_2,\dots) $$ is not a list, by the definition of a list (a list must have finite length). The definition of $F^\infty$ is given as $$ F^\infty = \{ (x_1,x_2,\dots) : x_j \in F \text{ for } j = 1,2,\dots \}. $$

The quoted part does not say that $$ (x_1,x_2,\dots) $$ represents a list of length $n$ for some arbitrary length $n$. Also, the definition of $F^\infty$ is not $$ F^\infty = \{ (x_1,x_2,\dots,x_n) : x_j \in F \text{ for } j = 1,\dots,n, \text{ and $n$ arbitrary positive integer}\}. $$

Perhaps you were confused with one or the other aspect of the definitions as in above. Basically, although $F^n$ is defined as the set of lists of length $n$, $F^\infty$ is not defined in terms of lists and is instead defined in terms of objects of the form $$ (x_1,x_2,\dots). $$