I'm learning about the Gram-Schmidt process:
I have some subspace basis $A$ with three vectors:
$$A = \{a_1,a_2,a_3\}.$$
Based on it, we will create an orthonormal basis $B$ with three vectors, using the Gram-Schmidt process:
$$B = \{b_1,b_2,b_3\}.$$
Each vector in $B$ is calculated this way:
$$\begin{align}b_1 &= a_1,\\ b_2 &= a_2 - \operatorname{proj}_{b_1}(a_2),\\ b_3 &= a_3 - \operatorname{proj}_{b_1}(a_3) - \operatorname{proj}_{b_2}(a_3).\end{align}$$
Alright. At plain sight, I would summarize this method as follows.
The element at index $n$ of $B$ will be the element at index $n$ of $A$, subtracted by its projection over each of the previous elements of $B$.
As you can observe from the bold text, I seem to imply that there is some order in the terms of $A$ and $B$ (there is a first, a second, and a third, all ordered by an index number). This goes against my understanding of sets - a set doesn't have a particular order (that I'm aware of), so I find it odd how this formula seems to be operating on "previous" elements.
I mean, if we had $A = \{x,y,z\}$, technically, it should be equivalent to $A = \{z,y,x\}$. But wouldn't that affect the process? Would you clarify this, please?
Yes, in fact, bases themselves are usually understood as sequences (in my experience anyway), just as Daniel Fischer describes. The order is very important.
For example, given a linear transformation $\phi$, say, the matrix of $\phi$ with respect to basis $\mathcal{B}=(v_1, v_2, \dots , v_n)$, denoted $M_{\mathcal{B}}^{\mathcal{B}}(\phi)$, is (usually) distinct from the matrix of $\phi$ with respect to $\mathcal{B}'=(v_2, v_1, v_3, v_4, \dots , v_n)$, denoted $M_{\mathcal{B}'}^{\mathcal{B}'}(\phi)$, etc.
So you're right, it will change the outcome of the process if you change the order of the basis elements, since it would be a different basis!