Here's the definition of a sequence as laid out in the text:
Let $m$ be an integer. A sequence $(a_n)_{n=m}^\infty$ of rational numbers is any function from the set $\{n \in \mathbf{Z} : n \geq m\}$ to $\mathbf{Q}$.
I can make sense out of this definition, but I was under the impression that a sequence has an ordering to it. I do not see any order implied on the "outputs" in the definition. Am I missing something?
On
As $\mathbb{Z}$ is ordered, the set $\{ n \in \mathbb{Z} : n \geq m \}$ is ordered. Hence the sequence $(a_n)_{n \geq m}$ is ordered (by the subscript $n$).
For a concrete example, take the set $S = \{2, 3, 4, 5, \ldots \}$ and the function
$$ \begin{align*} f: \{ n \in \mathbb{Z} : n \geq 2 \} &\longrightarrow \mathbb{Q} \\ n &\mapsto n^2 \end{align*} $$
that squares inputs. Then the sequence is
$$ (a_2, a_3, a_4, \ldots) = (4, 9, 16, \ldots), $$
and the ordering of the sequence is inherited from the ordering of the integers.
On
Informally, a sequence is an ordered list of things. The ordering is the order in which the things appear, not an order among the things (the ouptputs of the function in Tao's definition). So $$ 1, 0, 1, 0, \ldots $$ is (the start of) a sequence of integers.
Tao orders the position of the elements of the sequence by labeling them with integers starting at some $m$. (The usual starting points are $m=0$ or $m=1$. Tao may have some use later for the extra generality.)
Yup, if you look only at the "outputs" there is no sequential order. That's why the sequence is not the same thing as its "outputs". The sequence is the mapping, and the domain/index set is a subset of the naturals, which does have the standard order on it. If you want to get poetical about it, "elements of the sequence remember where they came from".
So, for instance, suppose we have a sequence that starts at index $i=1$, and the sequence is $1, 1, 2, 3, 5, ...$, then $a_1$ and $a_2$ are different elements of the sequence, even though they both have the value $1$.
Added in response to questions in the comment section of the answer by davidlowryduda:
An ordered set is a different thing from just a plain set. Even if you write $\{3,2,5,4,\ldots \}$ instead of $\{1,2,3,4,\ldots \}$ you still have the same set. The thing that can be confusing for people learning about formal mathematical structures is that many sets come with "default" structures: the usual "$+$" addition operation on $\mathbb R$, the usual "$\times$" multiplication operation on $\mathbb C$, and the usual "$\lt$" ordering on $\mathbb Z$. And mathematicians will often assume they are there and available without mentioning them.
If we want to make it explicit, we can refer to "the ordered set $(\mathbb Z, \lt)$". And if you wanted to define a second ordering, $\lt'$, on $\mathbb Z$, where $ 3 \lt' 2 \lt' 5 \lt' 4 \lt' \ldots$, you could, and you could define a sequence as a function on the ordered set $(\mathbb Z, \lt')$, but it's hard to see what would be gained by that. Firstly, you don't cover any new ground, and secondly, it's incredibly confusing. People expect to use the standard order on $\mathbb Z$, and it takes a lot of mental effort to use a different order. You had better be doing something pretty amazing that absolutely depends on a non-standard order for them to follow along with one.
(The closest thing I can think of to what you're describing is "re-arranging a sequence". You might want to check that out.)
If we go back to the very beginning, perhaps the definition should have said "... is any function from the ordered set ...", and it also could have specified "with the standard order". Although this does raise an interesting question: why does it matter that the domain of our sequence function has an order defined on it? Do we need it? I don't know what sequence this text presents things in, but, while you can take care of many of the simple details about sequences without referring to an order, the definitions of "limit" and "Cauchy sequence" do depend on the order, and without those two concepts there isn't much use for sequences.