If n is the biggest number that might not have a property, then n + 1 is the smallest number that must have the property. Do I need to prove it?

Question

Everything is in the question. I'm not talking about strictly a number, but rather an object and a property, let's say, $B$. Every object $O$ has a value assigned to it so that the function mapping objects to values is bijective. For simplicity's sake, let's say they are just numbers. If $n$ is the biggest number that might not have a property, the $n+1$ is the smallest number that must have that property. It's easy to see that's true, trivial I would say, but if I'm writing a paper using this fact, do I have to prove it? If yes, how do I do that?

P.S If a number has this property, then every number bigger than the number, has to have this property too.

Answer 1

The claim is false.

Let $P(n)$ be the property

$n$ is an odd number or $n > 10$ clearly the biggest number for which $P$ does not hold is $10$: it is an even number and it is not strctly greater than $10$, hence $P(10)$ is clearly false, and for every $k> 10$ clearly $P(k)$ holds.

Nevertheless the smallest number $n$ for which $P(n)$ holds it is not $11=10+1$, but it is $1$, because it is odd.

The problem is due in the fact that in your hypothesis

$n$ is the biggest that does not have the property $P$

you are requiring that $n$ is the biggest element of the set $$\{k \in \mathbb N\colon P(k) \text{ does not hold}\}$$ but there is no reason in general for suspecting that if a property $P$ does not hold for a given $n$ then it should not hold also forall $k < n$.

So there could be a $k < n$ for which $P(n)$, i.e. which does not belong to the set above, and the smallest element for which $P$ holds should be smaller (not necessarily strictly) of such $k$.

Hope this helps.