Looking for a theorem that can be reduced to a particular case

Question

I am looking for a theorem that can be reduced to a particular case, i.e., such that there is a particular case for which if the theorem holds, then this easily implies its general validity.

About the context: I am writing an article about the Four Color Theorem, and in this article I say that there are theorems like the ones mentioned above. The objective is to introduce the argument that if every triangulated planar graph has a proper coloring with $n$ colors, the same will be true for every planar graph (triangulated or not). My reviewer said it would be interesting to quote a simple example of such a theorem.

I don't care much if the problem is open or not, just that it is simple and, preferably, recognizable.

More specifically, the theorem I'm looking for is something like "$P$ is true for the set $X$" and such that there is a set $Y\subset X$ such that if $P$ is true for the set $Y$, then $P$ is true for the set $X$.

Answer 1

It's quite a common trick in Real/Functional Analysis to show a result holds for functions of norm 1, then show the general case by normalizing (dividing through by the norm) and reducing to the special case.

e.g. Hölder's inequality https://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality#Proof_of_H%C3%B6lder's_inequality

Answer 2

The Pythagorean theorem is a particular case of the Cosine rule, when $ \theta = 90^\circ$.

However, the Pythagorean theorem implies the Cosine rule.

Note that in any given geometry, these 2 statements are equivalent.

Answer 3

Two functions are equal if they coincide at all points. Now, we have the following result: For two continuous functions to coincide, it is sufficent that they coincide on a DENSE subset. For example, two continuous function of one variable coincide on an interval $(a,b)$ of $\mathbb{R}$, if they coincide on each rational number in that interval.

Answer 4

The Jacobian conjecture can be reduced to the degree $3$ case. You asked for a theorem though; examples of that sort are given in this MO thread.

Answer 5

The diagonal argument which shows that $\mathbb{Q}$ is countable and furthermore that $\mathbb{N}^2$ is countable also shows that for all $n$ $\mathbb{N}^n$ is countable. You do this by induction using as a base case $\mathbb{N}$ itself and for the induction step you have $\mathbb{N}^n$ is in bijection with $\mathbb{N}$ thus you order the elements of $\mathbb{N}^n$ and reapply the same diagonal argument on $\mathbb{N}^n\times \mathbb{N}$