I have a question regarding this problem:
Let $\Bbb R^\Bbb{N}$ be the vector space of all infinite real sequences. Show that even though its infinite subset $X:=\{(1,0,0,\ldots), (0,1,0,0,\ldots), (0,0,1,0,\ldots)\}$ is linearly independent, it does not generate $\Bbb R^\Bbb{N}$, thus it is not its basis. What is the vector space generated by $X$?
As i understand it, $X$ would generate the space of all finite real sequences that have been zero-filled to infinity (or, equivalently, all infinite real sequences that have zeroes almost everywhere). This begs two questions:
- Take the vector $(1,1,1,\ldots)$. Is it linearly independent from $X$, even though it seems to be a (infinite) linear combination of vectors from $X$? How do our usual linear-algebraic intuitions hold up against infinitely-dimensioned spaces?
- How would a basis of $\Bbb R^\Bbb{N}$ look like?
You are correct, the space generated by the vectors your specify is denoted $c_{00}$, and is the space of all sequences which are eventually zero.
The space of all real sequences $\mathbb R^{\mathbb N}$ is a stranger beast. It is uncountably-infinite dimensional (whereas $c_{00}$ is countably-infinite dimensional). It does have a basis (using the axiom of choice), but one could never write such a basis down-it contains uncountably many vectors.
The vector $\mathbf x$ which has a $1$ in every position is in $\mathbb R^{\mathbb N}$, but is the countable sum of the elements you specified. However, if we give $c_{00}$ the norm given by $\lVert(y_n)\rVert=\sup_{n\in\mathbb N}\lvert{y_n\rvert}$, where $(y_n)$ is a sequence in $c_{00}$ then $\mathbf x$ is not even in the closure of $c_{00}$ under this norm (in the space of all sequences where this norm is finite), though clearly the norm of $\mathbf x$ is $1$.
The standard notion of a basis requires that every element be the finite linear combination of basis vectors. There are generalisations to this, for example a Schauder basis, which allows for some kind of norm convergence too (see above).