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I apologize for the lack of proper terminology; I have zero experience in this field.

What I mean by "proof by maximality": One way to show that a set $A$ has a certain property $p$ is to assume there is a largest proper subset $X \subsetneq A$ which verifies $p$ and then show that there is another subset of $A$ larger than $X$, $Y\supsetneq X$, which also verifies $p$, a contradiction. Then one concludes that $A$ verifies $p$.

Now, these kinds of proofs seem strikingly similar to proofs by induction, where one assumes that a proposition holds for $n$ (i.e. $n$ is the largest number that verifies it) and then proves that it must hold for $n+1$ as well and therefore, by the principle of induction, it is true for all the elements of $\mathbb{N}.$ The "base case" is similar as well, since in proofs by "maximality" one first has to show that there exists a subset verifying the property $p$.

I was wondering if there is a link between these two proof techniques and, if at all, how one can be turned into the other. I am having trouble trying to formalize this. Where can I learn more about it?

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    $\begingroup$ Yes; see e.g.Enderton, Mathematical Logic, Ch.1.4 Induction and Recursion. $\endgroup$ – Mauro ALLEGRANZA Mar 20 at 16:38
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    $\begingroup$ See Structural induction : "a proof method used in mathematical logic that is a generalization of mathematical induction used to prove that some proposition $P(x)$ holds for all elements of some sort of recursively defined structure, such as formulas, lists, or trees." $\endgroup$ – Mauro ALLEGRANZA Mar 20 at 16:44
  • $\begingroup$ Thanks @MauroALLEGRANZA, very helpful. Will look into those. $\endgroup$ – The Footprint Mar 20 at 16:46
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    $\begingroup$ The principle you describe is invalid unless $A$ is finite. The structural induction principles referred to by Mauro assume some restriction on the subsets $X \subseteq A$. E.g., your principle is valid for all finite subsets of the natural numbers but not valid for all subsets of the natural numbers. Being finite is a property that satisfies the conditions of your principle - there is no maximal finite set of natural numbers. But not every set of natural numbers is finite. $\endgroup$ – Rob Arthan Mar 20 at 20:41
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Well, not familiar with your argument. But the induction axiom goes in the direction of maximality.

Let ${\Bbb N}_0$ be the set of natural numbers. Let $N$ be a subset of ${\Bbb N}_0$ such that $0\in N$ and with each $n\in N$ we have $n+1\in N$. Then $N = {\Bbb N}_0$.

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  • $\begingroup$ I don't see how a proof by maximality can be turned into one by induction, though. $\endgroup$ – The Footprint Mar 20 at 16:36

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