# Theorems Implying their Own Generalization [closed]

Are there any examples of theorems which were later found to imply their own generalization?

Here's an example of what I mean: Hypothetically, suppose you proved Fermat's Little Theorem: $a^p \equiv a \pmod{p}$ for $a \in \mathbb{Z}$, $p$ prime. Suppose, subsequently, you were able to prove Euler's Theorem: $a^{\phi(n)} \equiv 1 \pmod{n}$ using Fermat's Little Theorem and perhaps some other results. This may not be possible, I'm just using it as an example.

I'm just looking for an example.

• Most of the time when a theorem implies it's own generalization in a straightforward way, it's called a lemma and the generalization is the theorem. One source would be even lemma that proves something for two (like the intersection of two open sets is open) which leads immediately to a theorem about finite many. Commented Feb 22, 2016 at 2:29
• Anytime you see the phrase "without loss of generality (WLOG)", it is an example of a theorem proving it's generality. Specifically though, I think Jensen's inequality is a nice generalization of the definition of concavity. Commented Feb 22, 2016 at 15:40
• Nice question and nice examples, but note that from a logical point of view this is a softish question: e.g., Cauchy's theorem for arbitrary paths is true, so it is implied by any statement, true or false. So the question is really about methods of proof rather than material implication. (Sorry to be pedantic!) Commented Feb 22, 2016 at 19:49

Cauchy's theorem for triangles can be used to prove Cauchy's theorem for arbitrary closed paths.

Let $f$ be analytic on a simply connected region $R$. Then $\int_T f(z) dz = 0$, for any triangle $T$ in $R$. This is useful in proving the more general fact that $\int_C f(z) dz = 0$, for any closed path $C$ in $R$.

• @M10687: I don't know if your wording "which were later found" was meant to indicate a historical advance from specific to general or not. I gave my favorite nontrivial example, but this is a common phenomenon. For example, statements about simple functions are frequently used (as lemmas) to prove statements about measurable functions in measure theory. Commented Feb 22, 2016 at 15:28
• This example actually illustrates exactly what I meant by the question, I completely forgot about this. Thanks! Commented Feb 23, 2016 at 1:34

It's not quite what you are asking for, but it may be of interest to you to know that there is a large class of combinatorial problems where the general case can be proved by proving a few specific instances. A simple example is to note that $$\begin{array}{rcl} 1 &=& \frac{1(1 + 1)}{2}\\ 1 + 2 &=& \frac{2(2+1)}{2}\\ 1 + 2 + 3 &=& \frac{3(3+1)}{2} \end{array}$$

and conclude that $$1 + 2 + \ldots + n = \frac{n(n+1)}{2}$$

which is justified because (by considering second differences), $1 + 2 + \ldots + n$ is quadratic in $n$ so that it suffices to verify $3$ values of a proposed quadratic solution. This kind of idea is exploited with great success in the beautiful book A=B.

• how do you justify that $1+2+\cdots + n$ is a polynomial in $n$? Commented Feb 24, 2016 at 22:20
• @IttayWeiss: by considering second differences. See (en.wikipedia.org/wiki/Finite_difference), although the facts that you need aren't very prominent there - so I'd appreciate a pointer to a better online reference to this very well-known material if anyone reading this has one. If $f(n) = 1 + 2 + \ldots + n$, then $\Delta f(n) = f(n) - f(n-1) = n$ and $\Delta^2f(n) = \Delta f(n) -\Delta f(n-1) = 1$ is constant function of $n$. This implies that $f(n)$ is a quadratic function of $n$. Commented Feb 24, 2016 at 23:10

Rolle's theorem in elementary analysis implies its generalisation, namely the mean value theorem.

Here's a simple example of two theorems, usually encountered (in the USA) in high school Algebra 2:

Thm 1. (The Remainder Theorem) Let $$p(x)$$ be any polynomial, and $$a \in \mathbb{R}$$ a constant. Then there exists some polynomial $$q(x)$$ such that $$p(x)=(x-a)q(x) + p(a)$$.

Thm 2. (The Factor Theorem) Let $$p(x)$$ be any polynomial, and $$a \in \mathbb{R}$$ a constant. Then $$(x-a)$$ is a factor of $$p(x)$$ if and only if $$p(a)=0$$.

At first glance, it appears that Thm. 1 is the general case, and Thm. 2 a corollary or special case of Thm 1. And this is not wrong. But in fact you can also run the argument in the other direction: it is fairly straightforward to prove Thm. 1 by applying Theorem 2 to the polynomial $$\hat{p}(x)=p(x)-p(a)$$.