Must Basis of an Euclidean Space Be Ordered Does the basis of an Euclidean space have to be ordered by definition? Or can be left unordered?
I was also wondering about what is the morphism (i.e. the mapping that can preserve all the structures) on Euclidean spaces? Is it Euclidean transformation (rigid transformation), consisting of rotation, translation and reflection? Or the reflection is not part of it, because the basis of a Euclidean space must be ordered.
Thanks and regards!
 A: Bases must be ordered, otherwise you couldn't speak of coordinates as ordered tuples. And this could make things pretty messy. For instance, if the standard basis for $\mathbb{R}^2$ could be either $e_1, e_2$ or $e_2, e_1$ simultaneously, then it would be difficult to speak of "the point of coordinates $(2,1)$".
Reflexions send bases to bases, because they are isomorphisms. But this doesn't mean they necessarily send a particular basis to the same particular basis. So they can change the order of the vectors of the basis. They don't preserve orientation, but this is an extra piece of structure, not included in the definition of Euclidean space.
A: On bases
Bases do not have to be ordered. I respectfully present the following as a counterpoint to Agustí Roig's response.
First of all, a basis is simply a set of vectors, by definition; and sets are unordered. If you wish to enumerate the vectors in a basis, you will of course do so in some order, but this order is arbitrary.
This arbitrary order of enumeration is the reason why we describe vectors as tuples. The order of the coefficients in a tuple matters only in as much as it must be in agreement with the arbitrary order which was selected for the basis vectors. To wit: let x, y be two arbitrary linearly independent vectors. The tuple [ 7  5 ] with respect to the (enumeration of the) basis v1 = x, v2 = y represents the same vector as the tuple [ 5  7 ] with respect to the (different enumeration of the same) basis v2 = x, v1 = y. The tuple is just a representation of a vector, relative to an arbitrarily chosen order for the basis.
Even more foundationally: "tuples" can be regarded as functions from the integers (e.g. the indices '1', '2', etc.) to the reals, complex numbers, or whichever set you draw your coefficients from. So [ 7  5 ] can be thought of as 'really being' the function mapping

'1' $\mapsto$ 7, 
'2' $\mapsto$ 5.

Our enumeration of the basis has a similar role: choosing the enumeration v1 = x, v2 = y is equivalent to defining a mapping

'1' $\mapsto$ x, 
'2' $\mapsto$ y.

Why do I have '1' and '2' in quotes? Because they're just labels; any other labels would do just as well. For instance, I could replace '1' and '2' with the vectors x and y themselves. Then we could define vectors by coefficient functions such as

x $\mapsto$ 7, 
y $\mapsto$ 5;

that is, the coefficient 7 is associated to the vector x, and the coefficient 5 is associated to y, to represent the vector 7x + 5y. This is what we really mean anyway; never mind any ordering of the vectors in your basis. With this, the arbitrary ordering of the basis disappears, leaving nothing but what it is we really mean by it all.
So: when we order our bases, its only to make it easier to present things — it is not part of the definition, or part of the structure of the objects we really care about.
On morphisms
A morphism on a Euclidean space ought to preserve all of the structures of Euclidean geometry. In particular, it should map circles to other circles, to preserve the structures guaranteed by the third postulate; so it must be an isothety (a rigid transformation up to scaling). This includes, but is not restricted to, the rigid transformations.
Can you make any argument why the morphisms should consist only of rigid transformations (excluding isothetic maps such as v $\mapsto$ 2v)?
A: This began as a reply to Jyotirmoy's comment on Agustí's answer, but it became way too long. 
As I stated in my comment on Agustí's answer, some textbooks define a basis as a set, and I think this is wrong. Obviously, some people disagree (just read some of the other answers and comments). I'll try to explain why a basis should never be thought of as a set.
Do the columns of the following matrix form a basis for $\mathbb{R}^2$? $$\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 1\end{bmatrix}$$ The answer is "yes" if a basis is a set, but obviously the right answer is "no." This makes it clear that whatever sort of collection to which we apply concepts like "linear [in]dependence" or "basis", it must allow repeated vectors; otherwise, a lot of basic theorems from linear algebra would require extra clauses to deal with matrices with repeated rows or columns. This rules out sets as our collections of choice. 
To find out what our collection of choice should be, let's revisit the definition of a basic concept like linear dependence (in somewhat imprecise "collection-agnostic" language (whose meaning should nevertheless be clear)): 

The vectors $v_1,\ldots,v_n$ are linearly dependent if there are scalars $c_1,\ldots,c_n$, not all zero, such that $c_1v_1+\cdots+c_nv_n=0$.

It's obvious here that the natural object to apply such a concept to is an indexed tuple of vectors (indexed by $\{1,\ldots,n\}$ above, but we could formulate the concept equally well for an arbitrary finite index set): we begin with an indexed tuple of vectors and then consider a tuple of scalars indexed by the same set, and then we apply a function to these two tuples that is now completely specified (taking the linear combination). We could have started with a completely unindexed, unordered collection of vectors with repetition allowed (i. e., a "multiset"), but then to consider a linear combination, we would have had to choose some indexing of the multiset first and then to consider some scalars indexed by our arbitrary index set in order to form a linear combination. The core point is that the natural input to the "linear combination" operation is a tuple of vectors and a tuple of scalars indexed by the same finite set. This is the fundamental operation on a vector space, so to me, it makes sense to formulate everything in terms of indexed tuples of vectors whenever possible.
I talked about indexing lists of vectors in the previous paragraph, but indexing isn't the same as ordering. When I say "order," though, I'm actually being a bit imprecise. It's usually not the order relation itself that matters, but the choice of index set does matter. For example, if you want to write down the coordinate matrix of a vector or the matrix of a linear transformation relative to a basis, then your index set pretty much has to be $\{1,...,n\}$ (or $\{0,...,n-1\}$ if you've ever programmed in a language other than Matlab or Fortran) because whatever you're using to index the elements of the basis, you're also using to index the rows or columns of your matrices. Since one of the main uses of bases is to reduce things to matrices, the sets $\{1,...,n\}$ are very convenient for indexing. In this very common context, all the basic concepts like "basis," "linear independence," "spanning," etc. would apply to finite sequences of vectors. In no context, however, do they naturally apply to finite sets of vectors. 
A: Orientation does not play any role in Euclidean geometry. So it is most natural to define a Euclidean space as a finite-dimensional vector space with an inner product. Lengths, distances and angles are then defined in terms of this inner product. We are interested in transformations which preserve lengths and angles, which are precisely the orthogonal transformations including reflections. There is no need to explicitly choose a basis at any point.
