Can a sequence of length 1 be considered strictly ascending? I think this is a simple question of conventions: can a number sequence be considered to be strictly ascending if it only has one element, or must there be at least two?
In my case, specifically, I'm actually trying to describe longest strictly-ascending contiguous sub-sequences. If my sequence is "3 5 2 4 6 6 1", then the longest is clearly "2 4 6", length 3.
But what if my sequence is "5 4 3 2 1"? Is the longest strictly-ascending contiguous sub-sequence of length 1 (and there are 5 of them), or are there none?
My guess is: it's none, and there has to be at least two elements. A sequence of length 1 would be both strictly ascending and descending, and that seems bad. But maybe this is one of those "no strong conventions, so state your assumptions clearly" things?
 A: I personally make such decisions based on combining two sequences.

Given sequences $a_1,\dots,a_n$ and $b_1,\dots,b_m$ then the concatenation sequence $a_1,\dots,a_n,b_1,\dots,b_m$ is increasing iff $a_i$ and $b_i$ are increasing and $a_n<b_1.$ (Or $\leq,$ if not wanting strict increasing.)

This result gives us reason to call sequences of length $1$ “increasing,” because without it, you’d have to name exceptions to this statement.
We of course have to look for possible harm in such definitions. Is there a reason not to? The only argument I can think of is a slight discomfort with defining it this way.
This doesn’t answer empty sequences. For that you have the result:

The concatenation of two finite sequences $a,b$ is increasing if the both sequences are increasing, and every element of the $a$ is less than every element of $b.$

This result also implies we should define an empty sequence as increasing, or we have to add special cases.
This same approach is why we define empty sums as $0$ and empty products as $1.$

Another reason to define it this way is to think of $a:\{1,\dots,n\}\to \mathbb R$ to be (strictly) increasing if it is an order (mono)morphism. And the map $a:\{1\}\to\mathbb R$ is always a partial order morphism. This is a “category theory” approach to the definition.
Indeed, concatenation of sequences is really a statement about appending two orders. If $P_1|P_2$ is the concatenation of two orders, then you get an isomorphism $$\{1,\dots,n\}|\{1,\dots,m\}\cong\{1,\dots,n+m\}.$$
In the category of totally ordered finite sets, we might include the empty order, just to have an “initial object” in the category.
