This is a follow-up question to Why does the definition of addition require proofs? In Landau's Foundations of Analysis, his definition of addition on the natural numbers seems a bit strange to me -- this '$+$' operator seems to pop out of thin air. I'm wondering if this "definition" is typical of textbooks on real analysis these days. How else is it being defined?
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On the contrary, it doesn't "pop out of thin air" at all; I think you are misunderstanding the argument.
In Landau's exposition the successor operation (denoted by the apostrophe or 'prime' symbol) is built into the Peano axioms; this operation, together with the use of the Induction Axiom, is then used (in Theorem 4) to prove that there exists a unique function $f$ on $\mathbb{N} \times \mathbb{N}$ with the properties that
- $f(x,1)=x'$
- $f(x,y')=f(x,y)'$
Once the proof of this theorem is complete, you can then introduce the notational convention that $x+y$ simply means $f(x,y)$.
What may be making Landau's exposition hard to follow is that, unlike what I wrote above, Landau does not bother messing around with writing $f(x,y)$; he introduces the notation $x+y$ at the same time that he states and proves the theorem. (That's why in the textbook it says "Theorem 4, and at the same time Definition 1".) That is a rhetorical technique that can be hard to decode if you're not used to it.
To the question of whether this is "typical of textbooks on real analysis these days": I think in general real analysis textbooks take one of two approaches. Some texts begin by defining (in approximately this order) group, field, ordered field, and complete ordered field (perhaps interposing Archimedian ordered field in between the third and fourth items of that list), then prove that any two complete ordered fields are isomorphic. This theorem then entitles them to say something like "From now on we will use the symbol $\mathbb{R}$ to refer to a complete ordered field; because any two complete ordered fields are isomorphic, we don't need to concern ourselves with what the real numbers 'are really'. All we need to know is that they are a complete ordered field, and ensure that everything else we say about them from now on can be derived from the field, order, and completeness axioms."
Of course it can be objected that:
- Proving uniqueness does not prove existence; for the latter, we would need to construct a real ordered field out of more basic elements.
- It seems perverse to develop an entire theory of the real numbers without ever saying what a real number is.
In response to these objections one might rebut: - Sure, but which set of basic elements? You can construct the reals by taking Dedkind cuts on the rationals, but then you have to ask "What are the rationals?" You can construct the rationals as equivalence classes of integers, but then you have to ask "What are the integers?" You can construct the integers as equivalence classes of natural numbers, but then you have to ask "What are the natural numbers?" You can construct the natural numbers out of ZF set theory, but then you have to ask "What are sets?" No matter how "deep" you go, eventually you have to stop and just stipulate some axioms about undefined terms. So why not just start at the level of "complete ordered field"?
Landau's "Preface for the Teacher" deals directly with these issues. In his context (1929 Germany) it's clear that the mainstream approach was to just start with the axioms of a complete ordered field and build up from there. In such an approach, addition does not get defined at all. It (and "natural number") are undefined terms; one just specifies what axioms it satisfies and moves on. Landau objected to this approach, feeling that the foundations of analysis should be pushed deeper back.
mweiss
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By starting with the $+$ operator right away, uniqueness is guaranteed by the notation. If $a=b$ and $c=d$ than we must have $a+c=b+d$. Even if you use the $f$ notation, you have $f(a,c)=f(b,d)$. Technically, I think you have to start with a subset of $N^3$ and establish functionality. – Dan Christensen May 08 '14 at 21:36
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No, I think you are still misunderstanding. Notation doesn't guarantee uniqueness, and one can define things that don't exist. Landau is proving simultaneously both that (a) a function exists that meets certain recursively-specifiied criteria, and (b) that function is unique. – mweiss May 08 '14 at 21:49
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The function notation (whether prefix $f$ or infix $+$) guarantees a unique output from any given input. Landau doesn't actually prove the existence of any function here. As far as I can tell (not sure), he assumes the existence of a certain binary operator $+$ on $N,$ and shows that, if such a function exists, then it must be unique. Missing is a formal proof of existence of that function using the axioms of set theory (for Cartesian products and subsets) as well as Peano's axioms, and formal proof of the functionality of the set of ordered triples so constructed. – Dan Christensen May 09 '14 at 03:39
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I think we are reading completely different texts. My interpretation of Landau is 100% the opposite of yours. He is not assuming the existence of a binary operator, he is proving it in exactly the way that you say he is not: From a direct application of the Peano axioms. – mweiss May 09 '14 at 03:48
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I am referring to the proof of "Theorem 4" (on page 4 and 5) at http://www.lib.uni-bonn.de/PhiMSAMP/Data/Book/PhiMSAMP-bk_Borovik.pdf I'm finding the proof impossible to follow. Which part do you think proves the existence of the add function? – Dan Christensen May 09 '14 at 04:28
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Some time ago, I was able to formally prove the existence of the required add function from first principles. It wasn't easy. (728 lines in the DC Proof format). I just don't see that Landau's proof has covered any of the ground there. http://dcproof.com/ConstructAdd0.htm – Dan Christensen May 09 '14 at 04:53
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The existence proof is in Part B, where he says "Now we will show for each $x$ it is actually possible to define $x + y$ for all $y$ in such a way that..." What Landau actually does (in different notation) is prove the existence not of a single binary function $add$, but rather, for each $x$ he proves the existence of a unary function $add_x$ that satisfies the conditions $add_x(1)=x'$ and $add_x(y')=(add_x(y))'$. – mweiss May 09 '14 at 13:32
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In the base case for Part B, Landau seems to be assuming that $1+1=1'.$ Have I missed something? Where did that come from? – Dan Christensen May 09 '14 at 16:21
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I'm not alone in my suspicions. See "Notes and Comments on the Addition Theorem by Leonard Blackburn" at http://www.math.umn.edu/~jodeit/course/LBonADDed.pdf – Dan Christensen May 09 '14 at 16:34
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I'm not sure what about this argument you are finding difficult. I think there are some unstated assumptions you and I are making about what it means to "prove something exists". In particular Landau is starting with PA but is not using ZFC axioms explicitly. This relates to what I said in my answer about choosing a starting point. Many (most?) real analysis authors start with the axioms of a complete ordered field; Landau pushes that back to Peano arithmetic. You seem to want him to go further back to ZFC. Those are all reasonable choices, just different places to base the theory. – mweiss May 09 '14 at 16:40
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No, he's not assuming anything. Landau needs to prove that for all x a function $add_x$ exists with the property that $add_x(1)=x'$ and $add_x(y')=add_x(y)'$. He does this by induction on $x$. For the base case, he needs to show that a function $add_1$ exists. So he defines the function as follows: For any $y$, set $add_1(y)=y'$. This exists and is unambiguous because the successor function is built into the Peano axioms. It is a straightforward matter to verify that $add_1$ has the necessary properties: $add_1(1)=1'$ and $add_1(y')=y''=(add_1(y))'$. – mweiss May 09 '14 at 16:46
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The paper you linked to is a critique of a different proof from a different source. Moreover it acknowledges that even in that other proof the base case (which seems to be more-or-less analogous to Landau's, except that in that other version the naturals begin with 0 rather than with 1) is correct. – mweiss May 09 '14 at 16:48
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I too am starting from PA. If you are not using some recursion theorem, I would think that, to prove the existence of a recursive function on $N$, you have to start with a set of ordered n-tuples (triples in this case) then prove it is function and that this function has the required properties. You can't just assume, for example, that $1+1=1'.$ I think Landau assumes this in the line, $x+1=1'=x'$ where $x=1.$ – Dan Christensen May 09 '14 at 17:10
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He is not assuming $1+1=1'$, he is defining $1+1=1'$. – mweiss May 09 '14 at 17:25
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He is at a point in a proof where $1 + $ ___ does not yet mean anything. He has to specify what it means. So he says "From now on $1 + y$ will mean $y'$." The latter exists because the successor function is baked into PA. Having defined what $1 + y$ means, it is now trivial to check that $1 + 1=1'$. Likewise one can check that $1 + y' = (y')' = (1+y)'$. – mweiss May 09 '14 at 17:28
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I think perhaps you are bothered that he uses the notation $1+y=y'$, rather than say something like $(1,y,y') \in Add$. That difference is purely cosmetic. – mweiss May 09 '14 at 17:30
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I really find it strange that we are having such difficulty communicating with each other, and if anybody else wants to jump in and try to figure out what is underlying our different readings of Landau's proof I would appreciate it. But I don't think there's much point in carrying this discussion out any further, we are just saying the same things over and over. – mweiss May 09 '14 at 17:31