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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.

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  • $\begingroup$ I think that is called the method of continuity (at least that's how it is called in the PDE book by Gilbarg and Trudinger). $\endgroup$ – user99914 May 11 '15 at 5:56
  • $\begingroup$ The proof is simple. Let $A \subseteq [0, \infty)$ be the set $\{x \in \Bbb R \mid \lnot P(x)\}$. Then $A$ is bounded below, and if it's non-empty $\inf A = a$ exists. Show that this leads to a contradiction. $\endgroup$ – Arthur May 11 '15 at 6:02
  • $\begingroup$ @Arthur: Yeah, this was my current best plan. "Let $A = \{ x \mid \lnot P(x)\}$ and let $a = \inf A$. (Insert proof of 1) Therefore $a \ge 0$, and if $a = 0$ then $P(a)$. Otherwise, suppose $a > 0$. (Insert proof of 2) Then $P(a)$. But (insert proof of 3) and therefore $\inf A > a$, a contradiction." But it still seems too wordy, like there should be a shorthand phrase for it. $\endgroup$ – Nate Eldredge May 11 '15 at 6:11
  • $\begingroup$ You can also consider $A = \{x: P(y) \text{is true } \forall y\le x\} \subset [0,\infty)$. Then this is nonempty open and closed. $\endgroup$ – user99914 May 11 '15 at 6:20
  • $\begingroup$ Here you go. He calls it "real induction," but it's also been called "continuous induction." He cites lots of sources where it's been used. Most of the paper is proving things using it. $\endgroup$ – Akiva Weinberger May 11 '15 at 10:53
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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.

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  • $\begingroup$ So what would that "unusual topology" be? I thought about something like the lower limit topology but that doesn't work as it isn't connected. $\endgroup$ – Nate Eldredge May 11 '15 at 11:47
  • $\begingroup$ I've seen it called "real induction" (and sometimes "continuous induction") — see the link I posted above. While it is similar to connectedness, it isn't exactly the same thing. (Though, I have to admit, Clark's proof that $[0,1]$ is connected is really short.) $\endgroup$ – Akiva Weinberger May 11 '15 at 12:44

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