Is there a standard name for this "continuous induction" principle?

I am working on a paper, and I want to prove that some statement $P(x)$ holds for every value of a parameter $x \in [0,\infty)$. I plan to proceed as follows:

1. Show that $P(0)$;

2. Show that if $P(x)$ then there exists $\epsilon > 0$ such that $P(y)$ for all $x \le y \le x+\epsilon$;

3. Show that if $x_n \uparrow x$ and $P(x_n)$ for every $n$, then $P(x)$.

It will follow that $P(x)$ holds for every $x \in [0,\infty)$. Is there a standard concise word or phrase for this principle?

Of course, I could include a proof of a lemma saying that 1,2,3 implies $\forall x P(x)$. But that seems overly pedantic; this fact must be familiar to professional mathematicians, but I want to know what word will remind them of it.

It's sort of a continuous analogue of (transfinite) induction (1 is like the base case, 2 like the successor case, and 3 the case of a limit ordinal). It is also somewhat akin to showing that a property holds for every $x$ in a connected space by showing that the set on which it holds is nonempty and clopen; indeed, if I had both upper and lower limits, that is exactly what it would be. But I can't seem to find a phrase that fits exactly.

As an analogy, suppose I wanted to show that some statement $Q(n)$ holds for every natural number $n$. I might show (1) $Q(0)$ and (2) if $Q(n)$ then $Q(n+1)$. I would then write "By induction, $Q(n)$ for every $n$"; I would not bother to give a proof of the induction principle. Basically I want to know what phrase can play the role of "by induction" for the principle described above.

Solution 1:

Yes, a standard term is “connectedness”. Namely, let $S$ be the set of all $x$ such that $P(x)$ holds. You are just proving, in this order:
- $S$ is not empty;
- $S$ is open [actually, open for a slightly unusual toplogy, but depending on your property $P$ it might be as easy to show that $S$ is open for the usual topology]: if $S$ contains a point $x$ then it contains some ball centered at $x$;
- $S$ is closed: a limit of points belonging to $S$ also belongs to $S$.

Since $[0,+\infty[$ is connected, any closed-and-open subset is either empty or the full set $[0,+\infty[$. Since $S$ is not empty, it is the full set.