Question about following vector space I was confused about the following: Let V be vector space of all real sequences and B its basis. Every vector from V should be able to be represented as a linear combination of finite many vectors from B. But what about, for example, element $a_n = (1, 1, \dots )$ (sequence that is constant $1$)? If the basis is something like this $(1, 0, 0, \dots ), (0, 1, 0, \dots ),\dots $ then there is no way to represent $a_n$ like finite linear combination. Is there something wrong with the definition that every vector from V can be represented as a finite linear combination of vectors from the basis or maybe I have not found the right basis? (This basis work for sequences that have finite many non-zero elements, but what about those that don't?)
 A: In general, if $V$ is an $F$-vector space, we say that $B$ is a basis for $V$ if and only if each $v\in V$ can be uniquely expressed as a linear combination of of vectors in $B$, that is, can be expressed in the form $$v=a_1 u_1+a_2 u_2+\dots+a_n u_n$$ for unique scalars $a_1, a_2,\dots, a_n\in F$. In particular, there is nothing wrong with the definition that your professor presented.
This particular vector space, the $\mathbb{R}$-space of all real sequences, is an example of an infinite-dimensional vector space with dimension $\mathfrak{c}$; in other words, every basis for the space of real sequences must necessarily have as many elements as real numbers. The argument to show the existence of a basis for any vector is based on Zorn's Lemma, which is equivalent to the Axiom of Choice. This is an indicator that, most of the time, we are not going to be able to have an explicit basis on our hands.
As a last comment, the collection that you mention formed by the vectors $(1,0,\dots)$, $(0,1,\dots)$, $\dots$, is a basis for a subspace of the space of real sequences which is linearly isomorphic to the space of real polynomials in one variable $\mathbb{R}[x]$. The detail here is that any basis for $\mathbb{R}[x]$ is countably infinite, while any basis for the space of real sequences is not countably infinite; consequently, $(1,0,\dots)$, $(0,1,\dots)$, $\dots$, can't be a basis for the space of real sequences.
