Must a sequence be well-founded? 
Must a sequence be well-founded?

Is $\Bbb Z=(\ldots-1,0,1,2,\ldots)$ a sequence?
Conventionally we think of $\Bbb N=(0,1,2,3,\ldots)$ as a sequence, but what about if it has no starting value?
Obviously we can reorder any countable set $X$ into a well-founded sequence e.g. by an injection $f:X\to \Bbb N$ but what about in its raw form, do we call $\Bbb Z$ a sequence?
 A: It depends on context.
In most contexts, a sequence is understood by default to mean a $\mathbb{N}$-indexed sequence.  So you can talk about e.g. a $\mathbb{Z}$-indexed sequence, but if someone says just “a sequence” without specifying the index set, they should be assumed to be talking about $\mathbb{N}$-indexed sequences.
In some areas of mathematics, the convention is different.  In set theory, for instance, sequences indexed by arbitrary ordinals are very commonly used, and so “a sequence” may be used to mean a sequence indexed by some ordinal, even if that’s not explicitly specified.
More generally: if you do specify the domain, how general can it be — arbitrary total order, well-order, poset, …?  Well, there’s no standard fixed definition of sequence to restrict this; a sequence is just a function, and the circumstances where one calls a function a sequence are just a matter of field-specific convention.  $\mathbb{Z}$-indexed sequences are certainly fairly commonly used; I wouldn’t be surprised to hear sequence used for functions on arbitrary total orders.  I’d be slightly surprised to hear sequence used for posets, and very surprised to hear it used for functions on an arbitrary set with no specified order at all.
A: Calling it a sequence can result in conflict with statements of some theorems. It is better to call it a sequence indexed by integers. 
A: That is not a sequence, but it is a net.
Definition A nonempty set $A$ together with a binary relation $\le $ is said to be a directed set if $\le$ satisfies


*

*$a \le a, \quad\forall a \in A$ 

*$a \le b \text{ and } b \le c \text{ implies } a \le c, \quad \forall a,b,c \in A$

*$\forall a,b \in A \,\exists c \in A \text{ such that } a\le c \text{ and } b \le c$
Definition Let $(A, \le)$ be a directed set and let $X$ be any set. Any function $f : A \to X$ is said to be a net in $X$.
We see that $\mathbb{Z}$ with its standard order $\le$ is a directed set (and so is every totally ordered set) so the identity function $\operatorname{id} :\mathbb{Z}  \to \mathbb{Z}$ is a net in $\mathbb{Z}$.
Sequences are precisely nets with the domain $(\mathbb{N}, \le)$.
Bear in mind that the informal terminology "an $S$-indexed sequence", where $S$ is only a set, may also be used to describe a function $f : S \to X$.
A: Given any set $I$, it makes sense to talk about "$I$-indexed sequences".
We don't need to make any extra assumptions on $I$ for this to make sense. However, some features we might equip $I$ can induce additional structure on $I$-indexed sequences.
For example, choosing a total ordering on $I$ lets us talk about whether one place in the sequence comes before or after another place. Well-orderings are common to consider, because you can do transfinite induction over a well-ordered index set.
A: The terminology of "sequence" is not completely nailed down.
In the strictest sense, "a sequence in $X$" is a function $\mathbb N \to X$. In a looser sense, $\mathbb N$ might be replaced by another upwards unbounded countable linear order, by any countable linear order, or even by any linear order. In the loosest sense it can refer to any function $I \to X$ from some index set $I$.
If you say sequence without further context, I expect the strictest definition; but if you talk about an $I$-sequence, I accept that no matter what kind of index set $I$ is.
