I hope I am not going to drive people buggy by what most might consider obvious but I must ask.
Peano’s axioms define the basic properties of the set of natural numbers but does not attempt to enumerate them because they are the “given” and can be expressed many different ways, slash marks, Arabic numbers, Roman numerals etc.
But I take it that:
given x e N, x = 501, I am allowed to write,
x’ = 501’ = 502 because we are familiar with the system of Arabic numbers and know what symbol follows 501.
Theorem 3 establishes that every natural number except 1 has a precursor.
If Peano’s axioms allow 501’ -> 502. It does it also allow writing 502 -> 501’?
If not, does Theorem 3 allow writing 502 -> 501’.
Hmmm, going out on a limb here but in thinking about it I guess I am pondering whether the successor function, though injective and surjective on N-{1} is also bijective. I am torn because I was thinking that any injective function that is also surjective must also be bijective but that would mean the the successor function would automatically define a precursor function, I think ... and that does not seem right, otherwise I would have run across it in the literature.
Does Theorem 3 establish that the successor function is a bijection?
My question is prompted by considering how to prove Theorem #4 which establish addition as an operation.
Well, once again I feel like a curious fly brain again caught in a spider web, ah well ...
What I mean by 501' -> 502, and am trying to say, is that I have the right to write that 502 is the unique successor of 501, by virtue of Peano's axioms and of ones knowledge of Arabic numerals.
What I unsure of is whether Peano axioms allow me the right to claim that 501 is the unique predecessor of 502.
The successor function in not surjective on N but I was asking about N -{1}, doesn't each element in that set have an n, element of N mapped to it? In that light isn't the successor function onto as well as injective and hence bijective? But if that is the case then why is theorem 3 needed to prove that each n except 1 has a predecessor unless, perhaps, it takes theorem 3 to prove that the successor function IS injective (?) .
Yes, there is a function $p:\Bbb N\setminus\{1\} \to \Bbb N$ such that:
$s \circ p = \text{id}_{\Bbb N\setminus\{1\}}$ and
$p \circ s = \text{id}_{\Bbb N}$
because $s$ is a bijection, simply define $p(k)$ to be the UNIQUE $n$ such that $s(n) = k$.
(Yes: injective + surjective = bijective).
If you strip away everything but the well-ordered property and infinitude of the natural numbers, then the set $\Bbb N$ differs from $\Bbb N\setminus\{1\}$ "in name(s) only", we have the bijection:
$x \leftrightarrow x'$
OK, after looking up a couple of definitions I can follow this.
It seems strange that a set can have a proper subset that is equivalent but I see that this works out because N is infinite. In effect you are just eliminating a name of a particular set and bumping the name of each following set back one, since N is infinite this Ponzi scheme doesn't fall apart.
Yes, they do.
The statement that the successor $s$ is bijective is not formulated precisely enough. A function is a triple consisting of a domain, a codomain and an actual rule mapping inputs to outputs. It is not clear what you consider as the domain and codomain of $s$. If the domain and codomain are the same (in particular, $\Bbb N\setminus\{1\}$), then the successor is not a surjection. It is a bijection, though, if the domain is $\Bbb N$ and the codomain is $\Bbb N\setminus\{1\}$.
Yes, it takes theorem 3 to derive from Peano axioms the fact that $s:\Bbb N\to\Bbb N\setminus\{1\}$ is a surjection (not an injection: that is directly stated in one of the axioms).
Gottcha, thanks for the point by point response, that is really helpful.
Thanks to all. The internet age, so fudging awesome, its like you can take a seat in the math faculty lounge if you like .... and it's free !!!
When I was in the service I used to take correspondence courses which I loved because you could work at your own rate ... but the correspondence lag time was geological compared to this, a dialogue took weeks.
Thanks for the knowledge and the patience.
And, of course, you know what ' means. That's what is important here, not "Arabic numbers" "what symbol follows 501". We could also write DI'= DII.
The Peano's axioms, as you list them here, says nothing about the symbol "->". Where did you see that. It certainly is true that 501'= 502 and 502= 501'. That is true because "equals" is symmetric.Theorem 3 establishes that every natural number except 1 has a precursor.
If Peano’s axioms allow 501’ -> 502. It does it also allow writing 502 -> 501’?
And "Theorem 3" does not mention the symbols "-->" either.If not, does Theorem 3 allow writing 502 -> 501’.
The definition of "bijection" is a function that is both injective and surjective.Hmmm, going out on a limb here but in thinking about it I guess I am pondering whether the successor function, though injective and surjective on N-{1} is also bijective. I am torn because I was thinking that any injective function that is also surjective must also be bijective but that would mean the the successor function would automatically define a precursor function, I think ... and that does not seem right, otherwise I would have run across it in the literature.
Does Theorem 3 establish that the successor function is a bijection?
My question is prompted by considering how to prove Theorem #4 which establish addition as an operation.
Well, once again I feel like a curious fly brain again caught in a spider web, ah well ...
Thank you HallsofIvy, but to tell you the truth I don't why using "->" is wrong.
Peano's axiom 2 says, at least the version I am using:
"For each x there is exactly one natural number, called the successor of x, which will be denoted by x'."
No mention of "->" or "="
According to Wikipedia's definition of math symbols:
"a -> b" means "if a then b",
both emakarov and you, Hallsof Ivy, have each said that the symbol "->" is at best ambiguous, or undefined, or unclear, certainly untenable in this instance, but doesn't "->" say exactly what is meant by axiom 2,
if x then x' ?
and by extension
if 501 then 502
Respectfully, Ray
Edit, maybe I am writing too soon,
axiom #4 says:
if x' = y', then x = y
but isn't that derivative of axiom #2 which already says that for x its successor x' is unique, what more does #4 add?
Wait, this is what I am looking for I think, the power to claim that a natural number has a predecessor, the right to say for a given natural number c, not 1, I can write b' as its predecessor.
Well, nuts, I need to wait and think more before I begin a post.
Sorry, I really need to beat this to death.
Saying there is a UNIQUE natural number x' for every x, is somewhat different than saying that if x' = y', x = y.
For example (so you can appreciate the distinction), there is a UNIQUE integer $a^2$, for every integer $a$.
On the other hand, if $a^2 = b^2$, it might not be that $a = b$, it might be that $a = -b$. Squaring is not injective (on the integers).
So saying the successor of a natural number x (that is, x') is unique, means we have a FUNCTION. Contrast that with another sort of successorship (in people):
S(x) = a son of x. This is not a function, because x may have more than one son, and we don't know which is intended.
Not all functions are injective. So axiom #4 truly adds something new.
People often confuse the two statements for a subset $f \subseteq A \times B$ of pairs $(a,b)$ with $a \in A,b \in B$:
$[((a,b) \in f) \wedge ((a,b') \in f)] \implies (b=b')$ and:
$[(a,b) \in f) \wedge ((a',b) \in f)] \implies (a=a')$
because they LOOK similar. The first says $f$ is a function, the two taken together says $f$ is an injective function ("one-to-one").
In other words, there is a subtle and IMPORTANT difference between:
"For every x, there is a unique y with x' = y", and:
"For every y, there is a unique x with x' = y".
In a phrase "If $P$, then $Q$", $P$ and $Q$ have to be propositions, i.e., something than can be either true or false. Some other synonyms for "proposition" are "statement" and "claim". In contrast, $x$ and $x'$ appearing in Peano axioms are expressions (or terms) that have numbers (instead of truth values "true" and "false") as their values. Thus, $x$ cannot be true or false, though it may have the value of 5. Therefore, it does not makes sense to write $x\to x'$. You can say that there is a way to turn $x$ into $x'$ or that for each $x$ there exists a unique $y$ denoted by $x'$ with a certain property, but these are not implications. You can give them some notations, for example, $x\mapsto x'$, but you need to provide a definition of such notations.
Deveno ... well, after reading your explanation I don't know why I was confused, essentially axiom #2 says the "successor relation" passes the "vertical test" and is therefore a "function". Axiom #4 says the "successor function" passes the "horizontal test" and is therefore an "injection", together they say the successor function is a one-to-one function (but not yet a one-to-one correspondence, surjective, by those two axioms alone).
I think I need to put this subject aside for a while and come back in a week or so and look at everything again and digest all of the information that I have been given. Thank you.
PS. I did google Hilberts Hotel, my favorite was the cartoon at
The Infinite Hotel Paradox - Jeff Dekofsky | TED-Ed
where they beyond the beyond and discussed real numbers a very finite amount.