# Linear independence and ∞-dimensioned vector spaces

I have a question regarding this problem:

Let ℝ be the vector space of all infinite real sequences. Show that even though its infinite subset X≔{(1,0,0,…), (0,1,0,0,…), (0,0,1,0,…)} is linearly independent, it does not generate ℝ, 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,…). 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 ℝ 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 $1$.