Dimension of set How do you calculate the dimension of the set $\{(0,0)\}$?
You cannot form a bases, since it has to include $(0,0)$ which makes it linearly dependent. Many have said the dimension to be zero but you can't actually have a concept of dimension here, can you?
 A: Well, if you insist on an explanation, then here goes:
The definition of span is not actually restricted to finite sets. It just so happens that you only deal with finite dimensional vector spaces in your course; and all finite dimensional vector spaces have finite bases.
The real definition is for an arbitrary subset of the vector space, finite or not. 
Here it is. Let V be a vector space over any other field (algebraic system of a set with two binary operations on the set, conveniently denoted by + and $\bullet$, so that all 3 together have the same algebraic structure as $\mathbb R$. Note that $\bullet$ is not the scalar multiplication of V. It is the field multiplication analog of $\mathbb R$'s multiplication).
The span of an arbitrary subset A of V denoted by span(A) is defined to be {$e_V$}, if A is empty, where $e_V$ denotes the additive identity of V; and if A is not empty, then span(A)={$a_1x_1+a_2x_2+...+a_nx_n$: where n is in $\mathbb N$, $a_1,a_2,...a_n$ are all in the field, and $x_1,x_2,...x_n$ are all in V}. Thus, span(A)="The set of all linear combinations of a finite number n, for any natural number n, of vectors in V with scalars in F.
Linear independence is then defined so that a subset A of V is linearly independent if and only if every finite linear combination $a_1x_1+a_2x_2+...+a_nx_n$ with n in $\mathbb N$, $a_1,a_2,....a_n$ in F and $x_1,x_2,....x_n$ in V that gives $a_1x_1+a_2x_2+...+a_nx_n=e_V$ has $a_1,a_2,...a_n$=0 (please understand here that 0 is not the number zero, but instead, we are using the same symbol to denote the additive identity of the field F, since it plays the same role, structurally, as zero does in $\mathbb R$. Note that this allows for the existence of an infinite linearly independent subset of V and thus it is possible for some vector spaces V, and certain subspaces of those vector spaces, to be infinite dimensional.
Linear dependence is defined so that the subset A of V is linearly dependent if and only if there exists some finite linear combination $a_1x_1+a_2x_2+...+a_nx_n$ with n in $\mathbb N$, $a_1,a_2,....a_n$ in F and $x_1,x_2,....x_n$ in V that gives $a_1x_1+a_2x_2+...+a_nx_n=e_V$ and at least one $a_i$ in $a_1,a_2,...a_n\neq$ 0.
Now, you'll need to know some propositional logic here: To prove the empty set is linearly independent, let there exist a_1,a_2....a_n in F and x_1,x_2...x_n in the empty set so that finite linear combination $a_1x_1+a_2x_2+...+a_nx_n$ with n in $\mathbb N$, $a_1,a_2,....a_n$ in F and $x_1,x_2,....x_n$ in V that gives $a_1x_1+a_2x_2+...+a_nx_n=e_V$. The statement is obviously not true, so any predicate of it evaluates to false, thus $a_1,a_2,...a_n$=0 in F. 
This satisfies the definition of linear independence. Thus, since the empty set $\varnothing$ is linearly independent, and span($\varnothing$)={e_V}, then $\varnothing$ spans {e_V} and is thus a basis for {$e_V$}.
Now, since, the empty set contains zero (the number) of vectors, dim {e_V}=0 (the number). It turns out that your case is just a special case of this. There you have it. 
Remark:
It seems natural to define the span of the empty set as being the empty set (since how can the empty set span anything if here's nothing inside it to span anything with?); or to leave it undefined altogether, but there are two reasons for not doing so:
The first is that we want one of the deepest applications ever of the axiom of choice, an axiom of set theory strongly related to linear algebra to extend to all subspaces of a (possibly infinite dimensional) vector space. This result is that every vector space has a basis. It actually requires acceptance of the axiom of choice to prove this, which is not without controversy. Additionally,  an algebraist's acceptance of it in this context turns out to be in the form of Zorn's lemma, which turns out to be a logically equivalent formulation of the axiom in terms of maximal elements of partially ordered sets.
The second is that, we want a result, that says "the span of a subset A of V is the smallest subspace of V containing A," to extend to the empty set.It does; and it turns out that the span of A, empty or not, is the intersection of all the subspaces of V containing A. So, there is a reason for all of this weirdness.
Notice now, for instance, that since {$e_v$} is a subspace of V, the intersection of all subspaces containing {$e_V$} is {$e_V$}, so span($\varnothing$)={$e_V} is this intersection.
WARNING: Do NOT try to learn or use this. This is not what your class is about. It turns out that all of this works for your class, because your class is a special case of this theory; but this is entirely to advanced for you to use. By the time you learn it, and even learn what a field is, let alone the rest of it, the class will be over, and you will be stuck, 5 chapters behind, unable able to consistently figure out how to work any of your homework problems based on this theory. They are, however, workable from this theory though. This is just too abstract for a new student recently beginning their study of vector spaces. I wrote this for you because I enjoy talking about it, and you asked me to; not because you need it any time soon. You just need to take an intro proofs course, abstract algebra; and probably a course on introductory real analysis too before you have the mathematical maturity for arbitrary vector spaces over a field. This was mainly just for your entertainment. 
Adam. V. Nease
A: Dimension is a number of vectors in basis for the set. Since given set has zero elements in basis (zero vector cannot be a member of basis) its dimension is zero.
A: If this is your first course in linear algebra or your second course?
If this is your first course, the short answer is that it's zero (0). If this is your second course, I'll prove it and explain why. I just don't want to confuse you. 
