Understanding Span, Basis, and Dimension I am a bit confused with span, basis, and dimension (when dealing with vector spaces). 
My teacher told us that a span is a finite linear combination. And I know that a basis is a spanning, linearly independent subset, and the dimension is basically the cardinality of a basis. 
MY question is, how can a dimension be infinite dimensional? Isn't a basis supposed to be finite because a span is finite?
I know I must be misinterpreting something, can anyone help clarify?
Thank You. 
 A: You have to be careful with your wording. Saying "a span is finite" doesn't really mean anything. The span of a collection of vectors is the set of all finite linear combinations of those vectors. 
Consider the vector space of all real polynomials $\mathcal{P}(\mathbb{R})$. It has a basis $\{x^n \mid n \in \mathbb{N}\cup\{0\}\}$ which has infinite cardinality, so $\mathcal{P}(\mathbb{R})$ is infinite dimensional. Any finite linear combination of these polynomials gives you an element of $\mathcal{P}(\mathbb{R})$. If you were to take a finite collection of the polynomials, say $\{x^{n_i} \mid i =1, \dots, k\}$, then their span would not contain any polynomial of degree more than $\max_i n_i$, so you would not obtain $\mathcal{P}(\mathbb{R})$.
A: Some more care is needed in your formulations; probably you did not report precisely what your teacher told.
Linear combinations of a (possibly infinite) family of vectors are obtained by multiplying a finite number among them by a nonzero scalar and adding up the results. The restriction to a finite number is essential and inevitable, because in algebra it makes no sense to add up infinitely many nonzero vectors (in analysis such infinite sums may be considered, but this requires additional notions of convergence and limits). One can formally admit infinite sums of vectors provided that only finitely many among them are nonzero (and the others do not affect the value of the sum); in this sense one can say a linear combination of an infinite family of vectors is obtained by associating scalars to every vector, but only for finitely many of them the scalar is nonzero, and adding everything up.
The span of a family of vectors now is the set of all linear combinations of them. Even though any single linear combination involves only finitely many vectors of the family, all the members may contribute to the span, since different linear combinations may involve different vectors in the family. The typical example of a vector space which needs an infinite family of vectors to obtain a basis (it is infinite dimensional) is the $\Bbb R$-vector space $\Bbb R[X]$ of polynomials in$~X$ with real coefficients. The usual basis used for it is formed of the monomials $X^0=1,X^1=X,X^2,X^3,\ldots$, an infinite family. Every individual polynomial (linear combination) can use only finitely many of them (with associated scalars). However none of the monomials is redundant; one needs all of them to be able to produce all polynomials as linear combinations (in other words any proper subset would fail to be a spanning set for $\Bbb R[X]$).
A: Michael gives a good example; I will try to add my own thoughts.
First, it is dangerous to say without qualification that the dimension of a vector space is the cardinality of its basis. For finite-dimensional vector spaces this is all right, but in general proving that every vector space even has a basis is nontrivial (this requires the axiom of choice). Even proving that any two bases of the same vector space have the same cardinality is nontrivial (this requires a weaker form of the axiom of choice).
Second, a point that many students get hung up on is that although a basis may have infinitely many elements, its span is still the set of finite linear combinations of basis vectors. Consider the following example:
Let $\mathbb{R}_c^\infty$ be the set of all sequences $(a_1,~a_2,\ldots)$ of real numbers whose terms are eventually zero, i.e. there exists $N\in\mathbb{N}$ such that $a_n=0$ for $n\geq N$. Define scalar multiplication by $$\lambda(a_1,~a_2,\ldots) = (\lambda a_1,\lambda a_2,\ldots)$$ and vector addition by $$(a_1,~a_2,\ldots)+(b_1,~b_2,\ldots) = (a_1+b_1,a_2+b_2,\ldots).$$
You can check that this is a vector space over the real numbers. It has a basis, consisting of the "standard basis vectors" $e_i = (0, 0, \ldots,0,1,0,\ldots)$, where the $1$ is in the $i$-th position. Indeed, finite linear combinations of the $e_i$-s generate $\mathbb{R}_c^\infty$, but if you allow for infinite sums then you can get $\sum e_i = (1,1,1,\ldots)\notin \mathbb{R}_c^\infty$.
