Is it possible to define a monotonically increasing sequence on $\mathbb Z$ in such a way that the sequence is $\ldots,-3, -2, -1, 0, 1, 2, 3,\ldots$ Sorry if the question is a bit vague. We know that any monotonically increasing sequence $(a_n)$ must have a lower bound $a_1.$ I was wondering why the sequence $\ldots, -3, -2, -1, 0, 1, 2, 3, \ldots$ is not an example of a monotonically increasing sequence with no lower bound. Anyway, if none of it makes sense, then my question is why must every monotonically increasing sequence have a lower bound?

A _sequence on $\mathbb{Z}$_ is defined as a function from $\mathbb{N}$ to $\mathbb{Z}$.

By the Peano axioms, there is no element $n \in \mathbb{N}$ such that $0$ (or $1$, it's just a symbol) is the successor of $n$.

In a **monotonically increasing sequence** you can only find smaller number by "going backwards" on the indexes (since $a_{n+1} \geq a_n \; \forall n \in \mathbb{N}$), but sooner or later you'll reach $0$ _by definition of $\mathbb{N}$ itself_.

Thus the proposition.

Note that on a not monotonically increasing sequence you don't have any restrictions on how small your sequence could get (take for example $a_n = -n$).