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First question:

The set of natural numbers $N$, with no addition, no multiplication, just as we have it from the Peano Axioms, still has some (algebraic) structure, because it is not a mere set, because one can say 4 is the successor of 3, 5 is the successor of 4, e.t.c. (on a set there are no relations between the elements so i assume something structural is different here). This can be formulated by the successor function $S: N\to N$ (here i define $S$ after i have constructed the naturals). Is this correct? Can we say $N$ has algebraic structure ?

Second question:

Consider $(N,0,+)$. Given $A\subset N$, $A= \{1,2,3,5\} $, can i say that A has some structure? Addition is not closed on A but the elements of $A$ are related to each other, in a way i cannot express mathematically, but still 3 is the successor of 2 and 5 is the successor of the successor of 3. Is my thinking correct, and if yes how do we form this mathematically?

This question can be generalized to whether we can establish some structure on lets say $S^2\subset R^3 $, where $S^2=\{(x,y,z)|x^2+y^2+z^2=1 \}$, without making $R^3 $ a topological space and inheriting a topology on $S^2$.

Thanks

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    $\begingroup$ Before you define addition and multiplication, $(\Bbb N, S)$ has an order, and that's it. You can construct $\leq$ from $S$, and you can recover $S$ from $\leq$, so they are equivalent. $\endgroup$
    – Arthur
    Commented Feb 6, 2018 at 8:22

2 Answers 2

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Strictly speaking, no, $\mathbb{N}$ as given from the Peano axioms does not have a defined structure - in particular, it has no well-defined algebraic structure.

The notion of successor as you have stated it is derived not from an algebraic property, but rather an order relation on the natural numbers: the relation $<$, "less than". In fact, $\mathbb{N}$ is a well-ordered set under this order relation.

A well-ordered set is a set $S$ with some order relation $<_S$ such that any subset $M$ of $S$ contains an element $m$ such that for any $n \in M$, $m <_S n$. This element is called minimum.

This property of $\mathbb{N}$ expresses the notion of successor, since in any subset composed of two elements we have a well defined inferior and a superior element. Note that $\mathbb{Q}$ does not have this property.

This is however not an algebraic structure, but an order structure.

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  • $\begingroup$ Ok, but what about $A\subset N$? $\endgroup$
    – kot
    Commented Feb 6, 2018 at 8:34
  • $\begingroup$ As a subset of $\mathbb{N}$, it shares the same order structure. Any subset of a well-ordered set is well-ordered. The way you express it mathematically is the one I reported in the block quote. $\endgroup$ Commented Feb 6, 2018 at 8:38
  • $\begingroup$ Ok, thanks Niki $\endgroup$
    – kot
    Commented Feb 6, 2018 at 8:42
  • $\begingroup$ So is it wrong to say $(N,+)$ is a commutative monoid? Since $N$ has an order structure by construction must it be an ordered commutative monoid? $\endgroup$
    – kot
    Commented Jul 17, 2018 at 12:51
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Answers:

  1. Yes, in a language with only the successor function $S$ you can express that the set of natural numbers $\mathbb{N}$ is a total discrete order with a least element. Maybe this is the "algebraic structure" of $\mathbb{N}$ you refer to.

  2. Since $A$ is finite, you can easily find an intensional way (using first-order logic formulas using only $+$, $0$ and $S$ as non-logical symbol) to describe all and only the elements of $A$.

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  • $\begingroup$ don't know logic at all, but thanks that was helpful. $\endgroup$
    – kot
    Commented Feb 6, 2018 at 8:37

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