Theorems Implying their Own Generalization

Question

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.

Answer 1

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 $R$.

Answer 2

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.

Answer 3

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

Answer 4

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 $\hat{p}(x)=p(x)-p(a)$.