0

My textbook (Amann and Escher, Analysis I) gives a theorem which says that the operations of addition and multiplication (and a partial order $\leq$) exist and are uniquely defined by a whole host of characterizations. My question here is in particular about the claim of uniqueness with respect to addition.

The claim with respect to uniqueness is that addition is the unique operation on $\mathbb{N}$ which obeys

Addition is associative, commutative and has the identity element 0.

We note that $0$ has already been defined as the privileged 0 element in $\mathbb{N}$ (i.e. the middle element in the triple $(\mathbb{N},0,\nu)$, where $\nu$ is the successor function).

The way the text seems to prove uniqueness is as follows. It first shows that if two operations $\circledast, \circledcirc$ obey the following three conditions, then they are the same (incidentally, doesn't C2 imply C3 given the $\nu(0) =: 1$ identification?):

(C1) $0 \circledast 0 = 0$

(C2) $n \circledast 1 = \nu(n)$

(C3) $n \circledast \nu(m) = \nu(n \circledast m)$

We then show that $+$ obeys the (Ci), so that any other operation obeying the (Ci) is the same as +. But the authors just leave it here! Apparently the suggestion is that the (Ci) are sufficient and necessary for any operation obeying the characterization of $+$ in the theorem. Sufficiency was shown, but I am not sure to show necessity. Why should commutativity, associativity, and 0 as the identity for a given operation imply the (Ci)?

I here attach the proof in its full gory detail in case my question is not clear (I apologize for the length, so please feel free to ignore -- hopefully my question in bold above is clear!) enter image description here enter image description here enter image description here enter image description here

EE18
  • 1,211
  • You have misread the theorem, the claim is not that (i) suffices and that isn't what's proven (and it's obviously false). The theorem is that all of those conditions collectively characterize addition, multiplication, and the partial order simultaneously, so you need to use every condition in which addition is mentioned, namely (i), (iii), (iv), (vii), (viii). I don't understand why the authors state this very complicated theorem and then prove a restricted theorem with a different hypothesis (5.1), that seems sloppy and confusing to me. – Qiaochu Yuan Sep 17 '23 at 20:44
  • That's well taken @QiaochuYuan May I ask though why then they would give us parts (a) and (b) of the proof? If, to your point, (i) does not suffice to get us (5.2), then what was the point of having (a) and (b) there at all? That was my reason for posting -- I figured there had to be some way to actually use (a) and (b), because otherwise they're just there doing nothing, no? – EE18 Sep 17 '23 at 22:31
  • I don't know. Maybe there's some way to use the conditions stated in the theorem to prove (5.1) but it's not immediately obvious to me. I don't really understand the authors' choices here. – Qiaochu Yuan Sep 17 '23 at 22:46
  • Fair enough :( this is disappointing as these authors have been tremendous and precise thus far. Thanks for your help here! @QiaochuYuan – EE18 Sep 17 '23 at 22:48
  • See this post and see this one regarding Landau's proof that IMO is the ref [Lan30] above. – Mauro ALLEGRANZA Sep 20 '23 at 11:26
  • Your suspicion is correct -- it is Landau! I actually bought it yesterday for this very reason. Thank you also for your very helpful links! @MauroALLEGRANZA – EE18 Sep 20 '23 at 13:00

0 Answers0