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Let $ A $ and $ B $ be arbitrary sets. Is there a function $ f:A times Ato B $ such that all functions from $A$ to $B$ are representable by $f$?
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$mathbfdef:$ let $A, B$ be sets, and $ f : A times A rightarrow B$ be any
function. A function$ g : A rightarrow B $ is representable by $f$ iff there is $ a in A$ such
that for all $ x in A, g(x) = f(x, a)$.
$mathbfTheorem:$ [Cantor’s Little Theorem] Let $ A, B $ be sets and $f : A × A → B$
be any function such that all functions $g : A → B$ are representable by $f$. Then
every function $Æ : B → B$ has a fixed point.
$mathbfproof:$ Suppose that $f : A × A → B$ is such a function that all functions
$g : A → B$ are representable by $f$. Let $Æ : B → B$ be any function. Define
a function $È : A → A × A$ , called Cantor’s diagonalization, by $È(x) = (x, x)$.
Let $h = Æ ◦ f ◦ È$;Since the function $h : A → B$ is representable by $f$ , we have an $a ∈ A$
such that for all $x ∈ A, h(x) = f(x, a)$. In particular, $h(a) = f(a, a)$. But
$h(a) = Æ(f(È(a))) = Æ(f(a, a))$. Writing $f(a, a) = b$, we have $Æ(b) = b$. Thus
$Æ$ has a fixed point, namely,$b$.
$mathbfQuestion:$ It says Let $ A, B $ be sets and $f : A × A → B$
be any function such that $mathbfall$ functions $g : A → B$ are representable by $f$.This assumption seems logically wrong to me because the cardinality of the set of all functions from $A$ to $B$ is bigger than the cardinality of $A$,so how can there exists a function $f : A × A → B$
such that $mathbfall$ functions $g : A → B$ are representable by it?In the proof the assumption is applied to the function "h" (defined in the proof ).
link of the paper containing the theorem:https://mat.iitm.ac.in/home/asingh/public_html/papers/cantor.pdf
elementary-set-theory logic proof-explanation fixedpoints
asked Jul 15 at 10:25
bumba
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$mathbfdef:$ let $A, B$ be sets, and $ f : A times A rightarrow B$ be any
function. A function$ g : A rightarrow B $ is representable by $f$ iff there is $ a in A$ such
that for all $ x in A, g(x) = f(x, a)$.
$mathbfTheorem:$ [Cantor’s Little Theorem] Let $ A, B $ be sets and $f : A × A → B$
be any function such that all functions $g : A → B$ are representable by $f$. Then
every function $Æ : B → B$ has a fixed point.
$mathbfproof:$ Suppose that $f : A × A → B$ is such a function that all functions
$g : A → B$ are representable by $f$. Let $Æ : B → B$ be any function. Define
a function $È : A → A × A$ , called Cantor’s diagonalization, by $È(x) = (x, x)$.
Let $h = Æ ◦ f ◦ È$;Since the function $h : A → B$ is representable by $f$ , we have an $a ∈ A$
such that for all $x ∈ A, h(x) = f(x, a)$. In particular, $h(a) = f(a, a)$. But
$h(a) = Æ(f(È(a))) = Æ(f(a, a))$. Writing $f(a, a) = b$, we have $Æ(b) = b$. Thus
$Æ$ has a fixed point, namely,$b$.
$mathbfQuestion:$ It says Let $ A, B $ be sets and $f : A × A → B$
be any function such that $mathbfall$ functions $g : A → B$ are representable by $f$.This assumption seems logically wrong to me because the cardinality of the set of all functions from $A$ to $B$ is bigger than the cardinality of $A$,so how can there exists a function $f : A × A → B$
such that $mathbfall$ functions $g : A → B$ are representable by it?In the proof the assumption is applied to the function "h" (defined in the proof ).
link of the paper containing the theorem:https://mat.iitm.ac.in/home/asingh/public_html/papers/cantor.pdf
elementary-set-theory logic proof-explanation fixedpoints
asked Jul 15 at 10:25
bumba
246
1
And yet you seem fine with the idea that there are sets $B$ such that any function $Bto B$ has a fixed point? I would find that as counterintuitive, if not more. I can only think of one such set (or two, really, because of vacuous truths), and in that case your problem isn't really a problem.
– Arthur
Jul 15 at 10:31
It seems to be saying: "if there is such a function f, then ...."
– Bram28
Jul 15 at 10:32
1
@Arthur I can think of lots of sets $B$ such that every function $f:Bto B$ has a fixed point, but they are all isomorphic, they are all one-element sets. I don't believe there are any other examples. Do you have one?
– bof
Jul 15 at 10:46
1
@bumba But the consequence is important. If I said "If pigs can fly, then chickens have teeth", that's a true statement (with current gene editing technology, at least). But your objections here amount to "but pigs can't fly, how can that statement be true?" It is true exactly because the consequence isn't (except for a few trivial cases where the assumption is also true).
– Arthur
Jul 15 at 10:54
1
@Arthur Nope. If there were no functions $f:emptysettoemptyset$ then it would be true "vacuously" that every function $f:emptysettoemptyset$ had a fixed point. But in fact there is one function $f:emptysettoemptyset$ and it has no fixed points.
– bof
Jul 15 at 11:16
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$mathbfdef:$ let $A, B$ be sets, and $ f : A times A rightarrow B$ be any
function. A function$ g : A rightarrow B $ is representable by $f$ iff there is $ a in A$ such
that for all $ x in A, g(x) = f(x, a)$.
$mathbfTheorem:$ [Cantor’s Little Theorem] Let $ A, B $ be sets and $f : A × A → B$
be any function such that all functions $g : A → B$ are representable by $f$. Then
every function $Æ : B → B$ has a fixed point.
$mathbfproof:$ Suppose that $f : A × A → B$ is such a function that all functions
$g : A → B$ are representable by $f$. Let $Æ : B → B$ be any function. Define
a function $È : A → A × A$ , called Cantor’s diagonalization, by $È(x) = (x, x)$.
Let $h = Æ ◦ f ◦ È$;Since the function $h : A → B$ is representable by $f$ , we have an $a ∈ A$
such that for all $x ∈ A, h(x) = f(x, a)$. In particular, $h(a) = f(a, a)$. But
$h(a) = Æ(f(È(a))) = Æ(f(a, a))$. Writing $f(a, a) = b$, we have $Æ(b) = b$. Thus
$Æ$ has a fixed point, namely,$b$.
$mathbfQuestion:$ It says Let $ A, B $ be sets and $f : A × A → B$
be any function such that $mathbfall$ functions $g : A → B$ are representable by $f$.This assumption seems logically wrong to me because the cardinality of the set of all functions from $A$ to $B$ is bigger than the cardinality of $A$,so how can there exists a function $f : A × A → B$
such that $mathbfall$ functions $g : A → B$ are representable by it?In the proof the assumption is applied to the function "h" (defined in the proof ).
link of the paper containing the theorem:https://mat.iitm.ac.in/home/asingh/public_html/papers/cantor.pdf
elementary-set-theory logic proof-explanation fixedpoints
asked Jul 15 at 10:25
bumba
246
$mathbfdef:$ let $A, B$ be sets, and $ f : A times A rightarrow B$ be any
function. A function$ g : A rightarrow B $ is representable by $f$ iff there is $ a in A$ such
that for all $ x in A, g(x) = f(x, a)$.
$mathbfTheorem:$ [Cantor’s Little Theorem] Let $ A, B $ be sets and $f : A × A → B$
be any function such that all functions $g : A → B$ are representable by $f$. Then
every function $Æ : B → B$ has a fixed point.
$mathbfproof:$ Suppose that $f : A × A → B$ is such a function that all functions
$g : A → B$ are representable by $f$. Let $Æ : B → B$ be any function. Define
a function $È : A → A × A$ , called Cantor’s diagonalization, by $È(x) = (x, x)$.
Let $h = Æ ◦ f ◦ È$;Since the function $h : A → B$ is representable by $f$ , we have an $a ∈ A$
such that for all $x ∈ A, h(x) = f(x, a)$. In particular, $h(a) = f(a, a)$. But
$h(a) = Æ(f(È(a))) = Æ(f(a, a))$. Writing $f(a, a) = b$, we have $Æ(b) = b$. Thus
$Æ$ has a fixed point, namely,$b$.
$mathbfQuestion:$ It says Let $ A, B $ be sets and $f : A × A → B$
be any function such that $mathbfall$ functions $g : A → B$ are representable by $f$.This assumption seems logically wrong to me because the cardinality of the set of all functions from $A$ to $B$ is bigger than the cardinality of $A$,so how can there exists a function $f : A × A → B$
such that $mathbfall$ functions $g : A → B$ are representable by it?In the proof the assumption is applied to the function "h" (defined in the proof ).
link of the paper containing the theorem:https://mat.iitm.ac.in/home/asingh/public_html/papers/cantor.pdf
elementary-set-theory logic proof-explanation fixedpoints
asked Jul 15 at 10:25
bumba
246
edited Jul 15 at 11:59
asked Jul 15 at 10:25
bumba
246
asked Jul 15 at 10:25
bumba
246
asked Jul 15 at 10:25
bumba
246
246
1
And yet you seem fine with the idea that there are sets $B$ such that any function $Bto B$ has a fixed point? I would find that as counterintuitive, if not more. I can only think of one such set (or two, really, because of vacuous truths), and in that case your problem isn't really a problem.
– Arthur
Jul 15 at 10:31
It seems to be saying: "if there is such a function f, then ...."
– Bram28
Jul 15 at 10:32
1
@Arthur I can think of lots of sets $B$ such that every function $f:Bto B$ has a fixed point, but they are all isomorphic, they are all one-element sets. I don't believe there are any other examples. Do you have one?
– bof
Jul 15 at 10:46
1
@bumba But the consequence is important. If I said "If pigs can fly, then chickens have teeth", that's a true statement (with current gene editing technology, at least). But your objections here amount to "but pigs can't fly, how can that statement be true?" It is true exactly because the consequence isn't (except for a few trivial cases where the assumption is also true).
– Arthur
Jul 15 at 10:54
1
@Arthur Nope. If there were no functions $f:emptysettoemptyset$ then it would be true "vacuously" that every function $f:emptysettoemptyset$ had a fixed point. But in fact there is one function $f:emptysettoemptyset$ and it has no fixed points.
– bof
Jul 15 at 11:16
 |Â
show 7 more comments
1
And yet you seem fine with the idea that there are sets $B$ such that any function $Bto B$ has a fixed point? I would find that as counterintuitive, if not more. I can only think of one such set (or two, really, because of vacuous truths), and in that case your problem isn't really a problem.
– Arthur
Jul 15 at 10:31
It seems to be saying: "if there is such a function f, then ...."
– Bram28
Jul 15 at 10:32
1
@Arthur I can think of lots of sets $B$ such that every function $f:Bto B$ has a fixed point, but they are all isomorphic, they are all one-element sets. I don't believe there are any other examples. Do you have one?
– bof
Jul 15 at 10:46
1
@bumba But the consequence is important. If I said "If pigs can fly, then chickens have teeth", that's a true statement (with current gene editing technology, at least). But your objections here amount to "but pigs can't fly, how can that statement be true?" It is true exactly because the consequence isn't (except for a few trivial cases where the assumption is also true).
– Arthur
Jul 15 at 10:54
1
@Arthur Nope. If there were no functions $f:emptysettoemptyset$ then it would be true "vacuously" that every function $f:emptysettoemptyset$ had a fixed point. But in fact there is one function $f:emptysettoemptyset$ and it has no fixed points.
– bof
Jul 15 at 11:16
1
1
And yet you seem fine with the idea that there are sets $B$ such that any function $Bto B$ has a fixed point? I would find that as counterintuitive, if not more. I can only think of one such set (or two, really, because of vacuous truths), and in that case your problem isn't really a problem.
– Arthur
Jul 15 at 10:31
And yet you seem fine with the idea that there are sets $B$ such that any function $Bto B$ has a fixed point? I would find that as counterintuitive, if not more. I can only think of one such set (or two, really, because of vacuous truths), and in that case your problem isn't really a problem.
– Arthur
Jul 15 at 10:31
It seems to be saying: "if there is such a function f, then ...."
– Bram28
Jul 15 at 10:32
It seems to be saying: "if there is such a function f, then ...."
– Bram28
Jul 15 at 10:32
1
1
@Arthur I can think of lots of sets $B$ such that every function $f:Bto B$ has a fixed point, but they are all isomorphic, they are all one-element sets. I don't believe there are any other examples. Do you have one?
– bof
Jul 15 at 10:46
@Arthur I can think of lots of sets $B$ such that every function $f:Bto B$ has a fixed point, but they are all isomorphic, they are all one-element sets. I don't believe there are any other examples. Do you have one?
– bof
Jul 15 at 10:46
1
1
@bumba But the consequence is important. If I said "If pigs can fly, then chickens have teeth", that's a true statement (with current gene editing technology, at least). But your objections here amount to "but pigs can't fly, how can that statement be true?" It is true exactly because the consequence isn't (except for a few trivial cases where the assumption is also true).
– Arthur
Jul 15 at 10:54
@bumba But the consequence is important. If I said "If pigs can fly, then chickens have teeth", that's a true statement (with current gene editing technology, at least). But your objections here amount to "but pigs can't fly, how can that statement be true?" It is true exactly because the consequence isn't (except for a few trivial cases where the assumption is also true).
– Arthur
Jul 15 at 10:54
1
1
@Arthur Nope. If there were no functions $f:emptysettoemptyset$ then it would be true "vacuously" that every function $f:emptysettoemptyset$ had a fixed point. But in fact there is one function $f:emptysettoemptyset$ and it has no fixed points.
– bof
Jul 15 at 11:16
@Arthur Nope. If there were no functions $f:emptysettoemptyset$ then it would be true "vacuously" that every function $f:emptysettoemptyset$ had a fixed point. But in fact there is one function $f:emptysettoemptyset$ and it has no fixed points.
– bof
Jul 15 at 11:16
 |Â
show 7 more comments
5 Answers
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Have you seen the classic proof that $sqrt 2$ is irrational? In modern terms, it's really more a proof of a statement like this:
If $a, b$ are integers such that $a^2 = 2b^2$, then for any natural number $n$ we have $2^nmid a, b$.
This is an if-then-formulated theorem. If we stop worrying about circularity here for a moment, note that it is well-known that the assumption, i.e. the existence of such integers $a, b$, is (nearly) impossible. Yet the theorem is true. How could this be?
This is exactly because the conclusion, that $a, b$ are both divisible by any power of $2$, is impossible (unless $a = b = 0$, which is an uninteresting triviality).
In your case, you have the statement (slightly paraphrased to make the if-then structure a bit clearer)
If $ A, B $ be sets and $f : A × A → B$ is a function such that all functions $g : A → B$ are representable by $f$, then every function $Æ : B → B$ has a fixed point.
You seem hung up on the fact that the assumption, i.e. the existence of such a function $f$, seems impossible when the fact is that this is (almost) a proof by contradiction, just like the classical proof of the irrationality of $sqrt 2$. That's because the conclusion, that any function $Bto B$ has a fixed point, is impossible (unless $|B|<2$, which is an uninteresting triviality). Your theorem is still true.
answered Jul 15 at 11:09
Arthur
99k793175
PS. That $|B|<2$ is not really an uninteresting triviality. I just formulated it that way to make the parallel to a result you ought to know well that much clearer.
– Arthur
Jul 15 at 11:10
I really like better to prove the irratinoality of $sqrt 2$ in terms of parity of $2$-valuations though.
– Arnaud Mortier
Jul 15 at 11:19
@ArnaudMortier But that's not the classic proof. And it's not so well known that I can rely on it as a reference point.
– Arthur
Jul 15 at 11:19
@bumba there is no need to read the whole paper, it is very clear. What do you find unclear in our answers?
– Arnaud Mortier
Jul 15 at 11:20
@bumba Your question, as stated, doesn't require reading the paper at all. You seem to have problems parsing the statement of the theorem and comprehending how it could possibly be true. I have answered that as best I can. If you have a question about the proof or application of the theorem, which would require people to read the paper, then you will have to make a new question post about that.
– Arthur
Jul 15 at 11:21
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Here is another theorem:
Theorem: If $A$ is a nonempty finite set such that there is an injective map $Atimes Ato A$, then every map $Ato A$ has a fixed point.
You would say, but this assumption doesn't make sense, there are no injective maps $Atimes Ato A$ since $Atimes A$ is bigger than $A$. And you would be almost right: $Atimes A$ is bigger than $A$ except in very exceptional cases, and this theorem is precisely telling you which are these exceptional cases.
answered Jul 15 at 11:11
Arnaud Mortier
19.2k22159
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Your mistake is thinking that the cardinality of the set $B^A$ of all functions from $A$ to $B$ must be bigger than the cardinality of $A$.
For example, if $A$ is non empty and $B$ is empty, then there is no function from $A$ to $B$, so the cardinality of $B^A$ is $0$, which is less than the cardinality of $A$.
Also, if $A$ has at least two elements and $B$ has a single element $b$, then there is a unique function $f colon A to B$ (mapping all elements of $A$ to $b$), and so the cardinality of $B^A$ is $1$, which is less than the cardinality of $A$.
answered Jul 15 at 11:07
Luca Bressan
3,86621036
in the proof the assumption is applied to the function "h" (defined in the proof ) which is not one of your trivial type
– bumba
Jul 15 at 11:56
What makes you think that it's not a "trivial" function as you call it?
– Luca Bressan
Jul 15 at 12:01
sorry it seems that i am wrong
– bumba
Jul 15 at 12:33
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First of all it is my personal opinion that a false hypothesis does not have to be a "wrong hypothesis", for me a wrong hypothesis is one that does not allow you to conclude your thesis.
With that said, as others have already noted the hypothesis
all functions $g colon A to B$ are represented by $f$
is not at all logically false, indeed it is verified whenever $B$ has only one element, and that is also the only case of course, since for every $B$ with al least two elements you can find a function with no fixed point (hence violating the thesis of Cantor's Little theorem).
So this theorem may seem pretty useless since it applies to so few families of sets, nevertheless if you replace sets and functions with objects and morphisms of a cartesian closed category the same proof provides you with a criterion for fixed points.
This can applied to the category of dcpos and continuous functions between them, in this way you get Kleene fixed point theorem.
There are other applications of this theorem in [this paper] from Lawvere who I believe was the first person to notice the importance of this theorem for cartesian closed categories.
I hope this helps.
answered Jul 16 at 9:59
Giorgio Mossa
13.3k11748
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If $B$ is empty, then there is a unique function $phi : B to B$ and it has no fixed point. If $B$ has $1$ element, then there is a unique function $phi: B to B$ and it has a fixed point. If $B$ has $2$ elements (or more), then Cantor's theorem as it is usually stated tells us that for any set $A$, there is no surjective function from $A$ to the set of functions $A to B$ and hence no function $A times A to B$ that represents every function from $A$ to $B$ in the sense of your question. Hence the hypothesis of your "Theorem" is false if $B$ has more than $1$ element.
So your statement is correct for non-empty sets $A$ and $B$ and is very closely related to Cantor's theorem as normally stated but it is extremely misleading: the existence of a fixed point for every function $phi : B to B$ just means that $B$ has exactly one $1$ element.
answered Jul 15 at 21:44
Rob Arthan
27.1k42863
1
If $B$ is empty, then not every function $Bto B$ has a fixed point -- indeed, the empty function does not.
– Henning Makholm
Jul 15 at 22:34
@HenningMakholm: thanks for picking up my error, which I believe I have now fixed.
– Rob Arthan
Jul 17 at 23:16
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5 Answers
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5 Answers
5
active
oldest
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active
oldest
votes
active
oldest
votes
up vote
3
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Have you seen the classic proof that $sqrt 2$ is irrational? In modern terms, it's really more a proof of a statement like this:
If $a, b$ are integers such that $a^2 = 2b^2$, then for any natural number $n$ we have $2^nmid a, b$.
This is an if-then-formulated theorem. If we stop worrying about circularity here for a moment, note that it is well-known that the assumption, i.e. the existence of such integers $a, b$, is (nearly) impossible. Yet the theorem is true. How could this be?
This is exactly because the conclusion, that $a, b$ are both divisible by any power of $2$, is impossible (unless $a = b = 0$, which is an uninteresting triviality).
In your case, you have the statement (slightly paraphrased to make the if-then structure a bit clearer)
If $ A, B $ be sets and $f : A × A → B$ is a function such that all functions $g : A → B$ are representable by $f$, then every function $Æ : B → B$ has a fixed point.
You seem hung up on the fact that the assumption, i.e. the existence of such a function $f$, seems impossible when the fact is that this is (almost) a proof by contradiction, just like the classical proof of the irrationality of $sqrt 2$. That's because the conclusion, that any function $Bto B$ has a fixed point, is impossible (unless $|B|<2$, which is an uninteresting triviality). Your theorem is still true.
answered Jul 15 at 11:09
Arthur
99k793175
PS. That $|B|<2$ is not really an uninteresting triviality. I just formulated it that way to make the parallel to a result you ought to know well that much clearer.
– Arthur
Jul 15 at 11:10
I really like better to prove the irratinoality of $sqrt 2$ in terms of parity of $2$-valuations though.
– Arnaud Mortier
Jul 15 at 11:19
@ArnaudMortier But that's not the classic proof. And it's not so well known that I can rely on it as a reference point.
– Arthur
Jul 15 at 11:19
@bumba there is no need to read the whole paper, it is very clear. What do you find unclear in our answers?
– Arnaud Mortier
Jul 15 at 11:20
@bumba Your question, as stated, doesn't require reading the paper at all. You seem to have problems parsing the statement of the theorem and comprehending how it could possibly be true. I have answered that as best I can. If you have a question about the proof or application of the theorem, which would require people to read the paper, then you will have to make a new question post about that.
– Arthur
Jul 15 at 11:21
add a comment |Â
up vote
3
down vote
Have you seen the classic proof that $sqrt 2$ is irrational? In modern terms, it's really more a proof of a statement like this:
If $a, b$ are integers such that $a^2 = 2b^2$, then for any natural number $n$ we have $2^nmid a, b$.
This is an if-then-formulated theorem. If we stop worrying about circularity here for a moment, note that it is well-known that the assumption, i.e. the existence of such integers $a, b$, is (nearly) impossible. Yet the theorem is true. How could this be?
This is exactly because the conclusion, that $a, b$ are both divisible by any power of $2$, is impossible (unless $a = b = 0$, which is an uninteresting triviality).
In your case, you have the statement (slightly paraphrased to make the if-then structure a bit clearer)
If $ A, B $ be sets and $f : A × A → B$ is a function such that all functions $g : A → B$ are representable by $f$, then every function $Æ : B → B$ has a fixed point.
You seem hung up on the fact that the assumption, i.e. the existence of such a function $f$, seems impossible when the fact is that this is (almost) a proof by contradiction, just like the classical proof of the irrationality of $sqrt 2$. That's because the conclusion, that any function $Bto B$ has a fixed point, is impossible (unless $|B|<2$, which is an uninteresting triviality). Your theorem is still true.
answered Jul 15 at 11:09
Arthur
99k793175
PS. That $|B|<2$ is not really an uninteresting triviality. I just formulated it that way to make the parallel to a result you ought to know well that much clearer.
– Arthur
Jul 15 at 11:10
I really like better to prove the irratinoality of $sqrt 2$ in terms of parity of $2$-valuations though.
– Arnaud Mortier
Jul 15 at 11:19
@ArnaudMortier But that's not the classic proof. And it's not so well known that I can rely on it as a reference point.
– Arthur
Jul 15 at 11:19
@bumba there is no need to read the whole paper, it is very clear. What do you find unclear in our answers?
– Arnaud Mortier
Jul 15 at 11:20
@bumba Your question, as stated, doesn't require reading the paper at all. You seem to have problems parsing the statement of the theorem and comprehending how it could possibly be true. I have answered that as best I can. If you have a question about the proof or application of the theorem, which would require people to read the paper, then you will have to make a new question post about that.
– Arthur
Jul 15 at 11:21
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Have you seen the classic proof that $sqrt 2$ is irrational? In modern terms, it's really more a proof of a statement like this:
If $a, b$ are integers such that $a^2 = 2b^2$, then for any natural number $n$ we have $2^nmid a, b$.
This is an if-then-formulated theorem. If we stop worrying about circularity here for a moment, note that it is well-known that the assumption, i.e. the existence of such integers $a, b$, is (nearly) impossible. Yet the theorem is true. How could this be?
This is exactly because the conclusion, that $a, b$ are both divisible by any power of $2$, is impossible (unless $a = b = 0$, which is an uninteresting triviality).
In your case, you have the statement (slightly paraphrased to make the if-then structure a bit clearer)
If $ A, B $ be sets and $f : A × A → B$ is a function such that all functions $g : A → B$ are representable by $f$, then every function $Æ : B → B$ has a fixed point.
You seem hung up on the fact that the assumption, i.e. the existence of such a function $f$, seems impossible when the fact is that this is (almost) a proof by contradiction, just like the classical proof of the irrationality of $sqrt 2$. That's because the conclusion, that any function $Bto B$ has a fixed point, is impossible (unless $|B|<2$, which is an uninteresting triviality). Your theorem is still true.
answered Jul 15 at 11:09
Arthur
99k793175
Have you seen the classic proof that $sqrt 2$ is irrational? In modern terms, it's really more a proof of a statement like this:
If $a, b$ are integers such that $a^2 = 2b^2$, then for any natural number $n$ we have $2^nmid a, b$.
This is an if-then-formulated theorem. If we stop worrying about circularity here for a moment, note that it is well-known that the assumption, i.e. the existence of such integers $a, b$, is (nearly) impossible. Yet the theorem is true. How could this be?
This is exactly because the conclusion, that $a, b$ are both divisible by any power of $2$, is impossible (unless $a = b = 0$, which is an uninteresting triviality).
In your case, you have the statement (slightly paraphrased to make the if-then structure a bit clearer)
If $ A, B $ be sets and $f : A × A → B$ is a function such that all functions $g : A → B$ are representable by $f$, then every function $Æ : B → B$ has a fixed point.
You seem hung up on the fact that the assumption, i.e. the existence of such a function $f$, seems impossible when the fact is that this is (almost) a proof by contradiction, just like the classical proof of the irrationality of $sqrt 2$. That's because the conclusion, that any function $Bto B$ has a fixed point, is impossible (unless $|B|<2$, which is an uninteresting triviality). Your theorem is still true.
answered Jul 15 at 11:09
Arthur
99k793175
edited Jul 15 at 11:14
answered Jul 15 at 11:09
Arthur
99k793175
answered Jul 15 at 11:09
Arthur
99k793175
answered Jul 15 at 11:09
Arthur
99k793175
99k793175
PS. That $|B|<2$ is not really an uninteresting triviality. I just formulated it that way to make the parallel to a result you ought to know well that much clearer.
– Arthur
Jul 15 at 11:10
I really like better to prove the irratinoality of $sqrt 2$ in terms of parity of $2$-valuations though.
– Arnaud Mortier
Jul 15 at 11:19
@ArnaudMortier But that's not the classic proof. And it's not so well known that I can rely on it as a reference point.
– Arthur
Jul 15 at 11:19
@bumba there is no need to read the whole paper, it is very clear. What do you find unclear in our answers?
– Arnaud Mortier
Jul 15 at 11:20
@bumba Your question, as stated, doesn't require reading the paper at all. You seem to have problems parsing the statement of the theorem and comprehending how it could possibly be true. I have answered that as best I can. If you have a question about the proof or application of the theorem, which would require people to read the paper, then you will have to make a new question post about that.
– Arthur
Jul 15 at 11:21
add a comment |Â
PS. That $|B|<2$ is not really an uninteresting triviality. I just formulated it that way to make the parallel to a result you ought to know well that much clearer.
– Arthur
Jul 15 at 11:10
I really like better to prove the irratinoality of $sqrt 2$ in terms of parity of $2$-valuations though.
– Arnaud Mortier
Jul 15 at 11:19
@ArnaudMortier But that's not the classic proof. And it's not so well known that I can rely on it as a reference point.
– Arthur
Jul 15 at 11:19
@bumba there is no need to read the whole paper, it is very clear. What do you find unclear in our answers?
– Arnaud Mortier
Jul 15 at 11:20
@bumba Your question, as stated, doesn't require reading the paper at all. You seem to have problems parsing the statement of the theorem and comprehending how it could possibly be true. I have answered that as best I can. If you have a question about the proof or application of the theorem, which would require people to read the paper, then you will have to make a new question post about that.
– Arthur
Jul 15 at 11:21
PS. That $|B|<2$ is not really an uninteresting triviality. I just formulated it that way to make the parallel to a result you ought to know well that much clearer.
– Arthur
Jul 15 at 11:10
PS. That $|B|<2$ is not really an uninteresting triviality. I just formulated it that way to make the parallel to a result you ought to know well that much clearer.
– Arthur
Jul 15 at 11:10
I really like better to prove the irratinoality of $sqrt 2$ in terms of parity of $2$-valuations though.
– Arnaud Mortier
Jul 15 at 11:19
I really like better to prove the irratinoality of $sqrt 2$ in terms of parity of $2$-valuations though.
– Arnaud Mortier
Jul 15 at 11:19
@ArnaudMortier But that's not the classic proof. And it's not so well known that I can rely on it as a reference point.
– Arthur
Jul 15 at 11:19
@ArnaudMortier But that's not the classic proof. And it's not so well known that I can rely on it as a reference point.
– Arthur
Jul 15 at 11:19
@bumba there is no need to read the whole paper, it is very clear. What do you find unclear in our answers?
– Arnaud Mortier
Jul 15 at 11:20
@bumba there is no need to read the whole paper, it is very clear. What do you find unclear in our answers?
– Arnaud Mortier
Jul 15 at 11:20
@bumba Your question, as stated, doesn't require reading the paper at all. You seem to have problems parsing the statement of the theorem and comprehending how it could possibly be true. I have answered that as best I can. If you have a question about the proof or application of the theorem, which would require people to read the paper, then you will have to make a new question post about that.
– Arthur
Jul 15 at 11:21
@bumba Your question, as stated, doesn't require reading the paper at all. You seem to have problems parsing the statement of the theorem and comprehending how it could possibly be true. I have answered that as best I can. If you have a question about the proof or application of the theorem, which would require people to read the paper, then you will have to make a new question post about that.
– Arthur
Jul 15 at 11:21
add a comment |Â
up vote
1
down vote
Here is another theorem:
Theorem: If $A$ is a nonempty finite set such that there is an injective map $Atimes Ato A$, then every map $Ato A$ has a fixed point.
You would say, but this assumption doesn't make sense, there are no injective maps $Atimes Ato A$ since $Atimes A$ is bigger than $A$. And you would be almost right: $Atimes A$ is bigger than $A$ except in very exceptional cases, and this theorem is precisely telling you which are these exceptional cases.
answered Jul 15 at 11:11
Arnaud Mortier
19.2k22159
add a comment |Â
up vote
1
down vote
Here is another theorem:
Theorem: If $A$ is a nonempty finite set such that there is an injective map $Atimes Ato A$, then every map $Ato A$ has a fixed point.
You would say, but this assumption doesn't make sense, there are no injective maps $Atimes Ato A$ since $Atimes A$ is bigger than $A$. And you would be almost right: $Atimes A$ is bigger than $A$ except in very exceptional cases, and this theorem is precisely telling you which are these exceptional cases.
answered Jul 15 at 11:11
Arnaud Mortier
19.2k22159
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Here is another theorem:
Theorem: If $A$ is a nonempty finite set such that there is an injective map $Atimes Ato A$, then every map $Ato A$ has a fixed point.
You would say, but this assumption doesn't make sense, there are no injective maps $Atimes Ato A$ since $Atimes A$ is bigger than $A$. And you would be almost right: $Atimes A$ is bigger than $A$ except in very exceptional cases, and this theorem is precisely telling you which are these exceptional cases.
answered Jul 15 at 11:11
Arnaud Mortier
19.2k22159
Here is another theorem:
Theorem: If $A$ is a nonempty finite set such that there is an injective map $Atimes Ato A$, then every map $Ato A$ has a fixed point.
You would say, but this assumption doesn't make sense, there are no injective maps $Atimes Ato A$ since $Atimes A$ is bigger than $A$. And you would be almost right: $Atimes A$ is bigger than $A$ except in very exceptional cases, and this theorem is precisely telling you which are these exceptional cases.
answered Jul 15 at 11:11
Arnaud Mortier
19.2k22159
edited Jul 15 at 11:35
answered Jul 15 at 11:11
Arnaud Mortier
19.2k22159
answered Jul 15 at 11:11
Arnaud Mortier
19.2k22159
answered Jul 15 at 11:11
Arnaud Mortier
19.2k22159
19.2k22159
add a comment |Â
add a comment |Â
up vote
0
down vote
Your mistake is thinking that the cardinality of the set $B^A$ of all functions from $A$ to $B$ must be bigger than the cardinality of $A$.
For example, if $A$ is non empty and $B$ is empty, then there is no function from $A$ to $B$, so the cardinality of $B^A$ is $0$, which is less than the cardinality of $A$.
Also, if $A$ has at least two elements and $B$ has a single element $b$, then there is a unique function $f colon A to B$ (mapping all elements of $A$ to $b$), and so the cardinality of $B^A$ is $1$, which is less than the cardinality of $A$.
answered Jul 15 at 11:07
Luca Bressan
3,86621036
in the proof the assumption is applied to the function "h" (defined in the proof ) which is not one of your trivial type
– bumba
Jul 15 at 11:56
What makes you think that it's not a "trivial" function as you call it?
– Luca Bressan
Jul 15 at 12:01
sorry it seems that i am wrong
– bumba
Jul 15 at 12:33
add a comment |Â
up vote
0
down vote
Your mistake is thinking that the cardinality of the set $B^A$ of all functions from $A$ to $B$ must be bigger than the cardinality of $A$.
For example, if $A$ is non empty and $B$ is empty, then there is no function from $A$ to $B$, so the cardinality of $B^A$ is $0$, which is less than the cardinality of $A$.
Also, if $A$ has at least two elements and $B$ has a single element $b$, then there is a unique function $f colon A to B$ (mapping all elements of $A$ to $b$), and so the cardinality of $B^A$ is $1$, which is less than the cardinality of $A$.
answered Jul 15 at 11:07
Luca Bressan
3,86621036
in the proof the assumption is applied to the function "h" (defined in the proof ) which is not one of your trivial type
– bumba
Jul 15 at 11:56
What makes you think that it's not a "trivial" function as you call it?
– Luca Bressan
Jul 15 at 12:01
sorry it seems that i am wrong
– bumba
Jul 15 at 12:33
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Your mistake is thinking that the cardinality of the set $B^A$ of all functions from $A$ to $B$ must be bigger than the cardinality of $A$.
For example, if $A$ is non empty and $B$ is empty, then there is no function from $A$ to $B$, so the cardinality of $B^A$ is $0$, which is less than the cardinality of $A$.
Also, if $A$ has at least two elements and $B$ has a single element $b$, then there is a unique function $f colon A to B$ (mapping all elements of $A$ to $b$), and so the cardinality of $B^A$ is $1$, which is less than the cardinality of $A$.
answered Jul 15 at 11:07
Luca Bressan
3,86621036
Your mistake is thinking that the cardinality of the set $B^A$ of all functions from $A$ to $B$ must be bigger than the cardinality of $A$.
For example, if $A$ is non empty and $B$ is empty, then there is no function from $A$ to $B$, so the cardinality of $B^A$ is $0$, which is less than the cardinality of $A$.
Also, if $A$ has at least two elements and $B$ has a single element $b$, then there is a unique function $f colon A to B$ (mapping all elements of $A$ to $b$), and so the cardinality of $B^A$ is $1$, which is less than the cardinality of $A$.
answered Jul 15 at 11:07
Luca Bressan
3,86621036
answered Jul 15 at 11:07
Luca Bressan
3,86621036
answered Jul 15 at 11:07
Luca Bressan
3,86621036
answered Jul 15 at 11:07
Luca Bressan
3,86621036
3,86621036
in the proof the assumption is applied to the function "h" (defined in the proof ) which is not one of your trivial type
– bumba
Jul 15 at 11:56
What makes you think that it's not a "trivial" function as you call it?
– Luca Bressan
Jul 15 at 12:01
sorry it seems that i am wrong
– bumba
Jul 15 at 12:33
add a comment |Â
in the proof the assumption is applied to the function "h" (defined in the proof ) which is not one of your trivial type
– bumba
Jul 15 at 11:56
What makes you think that it's not a "trivial" function as you call it?
– Luca Bressan
Jul 15 at 12:01
sorry it seems that i am wrong
– bumba
Jul 15 at 12:33
in the proof the assumption is applied to the function "h" (defined in the proof ) which is not one of your trivial type
– bumba
Jul 15 at 11:56
in the proof the assumption is applied to the function "h" (defined in the proof ) which is not one of your trivial type
– bumba
Jul 15 at 11:56
What makes you think that it's not a "trivial" function as you call it?
– Luca Bressan
Jul 15 at 12:01
What makes you think that it's not a "trivial" function as you call it?
– Luca Bressan
Jul 15 at 12:01
sorry it seems that i am wrong
– bumba
Jul 15 at 12:33
sorry it seems that i am wrong
– bumba
Jul 15 at 12:33
add a comment |Â
up vote
0
down vote
First of all it is my personal opinion that a false hypothesis does not have to be a "wrong hypothesis", for me a wrong hypothesis is one that does not allow you to conclude your thesis.
With that said, as others have already noted the hypothesis
all functions $g colon A to B$ are represented by $f$
is not at all logically false, indeed it is verified whenever $B$ has only one element, and that is also the only case of course, since for every $B$ with al least two elements you can find a function with no fixed point (hence violating the thesis of Cantor's Little theorem).
So this theorem may seem pretty useless since it applies to so few families of sets, nevertheless if you replace sets and functions with objects and morphisms of a cartesian closed category the same proof provides you with a criterion for fixed points.
This can applied to the category of dcpos and continuous functions between them, in this way you get Kleene fixed point theorem.
There are other applications of this theorem in [this paper] from Lawvere who I believe was the first person to notice the importance of this theorem for cartesian closed categories.
I hope this helps.
answered Jul 16 at 9:59
Giorgio Mossa
13.3k11748
add a comment |Â
up vote
0
down vote
First of all it is my personal opinion that a false hypothesis does not have to be a "wrong hypothesis", for me a wrong hypothesis is one that does not allow you to conclude your thesis.
With that said, as others have already noted the hypothesis
all functions $g colon A to B$ are represented by $f$
is not at all logically false, indeed it is verified whenever $B$ has only one element, and that is also the only case of course, since for every $B$ with al least two elements you can find a function with no fixed point (hence violating the thesis of Cantor's Little theorem).
So this theorem may seem pretty useless since it applies to so few families of sets, nevertheless if you replace sets and functions with objects and morphisms of a cartesian closed category the same proof provides you with a criterion for fixed points.
This can applied to the category of dcpos and continuous functions between them, in this way you get Kleene fixed point theorem.
There are other applications of this theorem in [this paper] from Lawvere who I believe was the first person to notice the importance of this theorem for cartesian closed categories.
I hope this helps.
answered Jul 16 at 9:59
Giorgio Mossa
13.3k11748
add a comment |Â
up vote
0
down vote
up vote
0
down vote
First of all it is my personal opinion that a false hypothesis does not have to be a "wrong hypothesis", for me a wrong hypothesis is one that does not allow you to conclude your thesis.
With that said, as others have already noted the hypothesis
all functions $g colon A to B$ are represented by $f$
is not at all logically false, indeed it is verified whenever $B$ has only one element, and that is also the only case of course, since for every $B$ with al least two elements you can find a function with no fixed point (hence violating the thesis of Cantor's Little theorem).
So this theorem may seem pretty useless since it applies to so few families of sets, nevertheless if you replace sets and functions with objects and morphisms of a cartesian closed category the same proof provides you with a criterion for fixed points.
This can applied to the category of dcpos and continuous functions between them, in this way you get Kleene fixed point theorem.
There are other applications of this theorem in [this paper] from Lawvere who I believe was the first person to notice the importance of this theorem for cartesian closed categories.
I hope this helps.
answered Jul 16 at 9:59
Giorgio Mossa
13.3k11748
First of all it is my personal opinion that a false hypothesis does not have to be a "wrong hypothesis", for me a wrong hypothesis is one that does not allow you to conclude your thesis.
With that said, as others have already noted the hypothesis
all functions $g colon A to B$ are represented by $f$
is not at all logically false, indeed it is verified whenever $B$ has only one element, and that is also the only case of course, since for every $B$ with al least two elements you can find a function with no fixed point (hence violating the thesis of Cantor's Little theorem).
So this theorem may seem pretty useless since it applies to so few families of sets, nevertheless if you replace sets and functions with objects and morphisms of a cartesian closed category the same proof provides you with a criterion for fixed points.
This can applied to the category of dcpos and continuous functions between them, in this way you get Kleene fixed point theorem.
There are other applications of this theorem in [this paper] from Lawvere who I believe was the first person to notice the importance of this theorem for cartesian closed categories.
I hope this helps.
answered Jul 16 at 9:59
Giorgio Mossa
13.3k11748
answered Jul 16 at 9:59
Giorgio Mossa
13.3k11748
answered Jul 16 at 9:59
Giorgio Mossa
13.3k11748
answered Jul 16 at 9:59
Giorgio Mossa
13.3k11748
13.3k11748
add a comment |Â
add a comment |Â
up vote
0
down vote
If $B$ is empty, then there is a unique function $phi : B to B$ and it has no fixed point. If $B$ has $1$ element, then there is a unique function $phi: B to B$ and it has a fixed point. If $B$ has $2$ elements (or more), then Cantor's theorem as it is usually stated tells us that for any set $A$, there is no surjective function from $A$ to the set of functions $A to B$ and hence no function $A times A to B$ that represents every function from $A$ to $B$ in the sense of your question. Hence the hypothesis of your "Theorem" is false if $B$ has more than $1$ element.
So your statement is correct for non-empty sets $A$ and $B$ and is very closely related to Cantor's theorem as normally stated but it is extremely misleading: the existence of a fixed point for every function $phi : B to B$ just means that $B$ has exactly one $1$ element.
answered Jul 15 at 21:44
Rob Arthan
27.1k42863
1
If $B$ is empty, then not every function $Bto B$ has a fixed point -- indeed, the empty function does not.
– Henning Makholm
Jul 15 at 22:34
@HenningMakholm: thanks for picking up my error, which I believe I have now fixed.
– Rob Arthan
Jul 17 at 23:16
add a comment |Â
up vote
0
down vote
If $B$ is empty, then there is a unique function $phi : B to B$ and it has no fixed point. If $B$ has $1$ element, then there is a unique function $phi: B to B$ and it has a fixed point. If $B$ has $2$ elements (or more), then Cantor's theorem as it is usually stated tells us that for any set $A$, there is no surjective function from $A$ to the set of functions $A to B$ and hence no function $A times A to B$ that represents every function from $A$ to $B$ in the sense of your question. Hence the hypothesis of your "Theorem" is false if $B$ has more than $1$ element.
So your statement is correct for non-empty sets $A$ and $B$ and is very closely related to Cantor's theorem as normally stated but it is extremely misleading: the existence of a fixed point for every function $phi : B to B$ just means that $B$ has exactly one $1$ element.
answered Jul 15 at 21:44
Rob Arthan
27.1k42863
1
If $B$ is empty, then not every function $Bto B$ has a fixed point -- indeed, the empty function does not.
– Henning Makholm
Jul 15 at 22:34
@HenningMakholm: thanks for picking up my error, which I believe I have now fixed.
– Rob Arthan
Jul 17 at 23:16
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If $B$ is empty, then there is a unique function $phi : B to B$ and it has no fixed point. If $B$ has $1$ element, then there is a unique function $phi: B to B$ and it has a fixed point. If $B$ has $2$ elements (or more), then Cantor's theorem as it is usually stated tells us that for any set $A$, there is no surjective function from $A$ to the set of functions $A to B$ and hence no function $A times A to B$ that represents every function from $A$ to $B$ in the sense of your question. Hence the hypothesis of your "Theorem" is false if $B$ has more than $1$ element.
So your statement is correct for non-empty sets $A$ and $B$ and is very closely related to Cantor's theorem as normally stated but it is extremely misleading: the existence of a fixed point for every function $phi : B to B$ just means that $B$ has exactly one $1$ element.
answered Jul 15 at 21:44
Rob Arthan
27.1k42863
If $B$ is empty, then there is a unique function $phi : B to B$ and it has no fixed point. If $B$ has $1$ element, then there is a unique function $phi: B to B$ and it has a fixed point. If $B$ has $2$ elements (or more), then Cantor's theorem as it is usually stated tells us that for any set $A$, there is no surjective function from $A$ to the set of functions $A to B$ and hence no function $A times A to B$ that represents every function from $A$ to $B$ in the sense of your question. Hence the hypothesis of your "Theorem" is false if $B$ has more than $1$ element.
So your statement is correct for non-empty sets $A$ and $B$ and is very closely related to Cantor's theorem as normally stated but it is extremely misleading: the existence of a fixed point for every function $phi : B to B$ just means that $B$ has exactly one $1$ element.
answered Jul 15 at 21:44
Rob Arthan
27.1k42863
edited Jul 17 at 23:28
answered Jul 15 at 21:44
Rob Arthan
27.1k42863
answered Jul 15 at 21:44
Rob Arthan
27.1k42863
answered Jul 15 at 21:44
Rob Arthan
27.1k42863
27.1k42863
1
If $B$ is empty, then not every function $Bto B$ has a fixed point -- indeed, the empty function does not.
– Henning Makholm
Jul 15 at 22:34
@HenningMakholm: thanks for picking up my error, which I believe I have now fixed.
– Rob Arthan
Jul 17 at 23:16
add a comment |Â
1
If $B$ is empty, then not every function $Bto B$ has a fixed point -- indeed, the empty function does not.
– Henning Makholm
Jul 15 at 22:34
@HenningMakholm: thanks for picking up my error, which I believe I have now fixed.
– Rob Arthan
Jul 17 at 23:16
1
1
If $B$ is empty, then not every function $Bto B$ has a fixed point -- indeed, the empty function does not.
– Henning Makholm
Jul 15 at 22:34
If $B$ is empty, then not every function $Bto B$ has a fixed point -- indeed, the empty function does not.
– Henning Makholm
Jul 15 at 22:34
@HenningMakholm: thanks for picking up my error, which I believe I have now fixed.
– Rob Arthan
Jul 17 at 23:16
@HenningMakholm: thanks for picking up my error, which I believe I have now fixed.
– Rob Arthan
Jul 17 at 23:16
add a comment |Â
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1
And yet you seem fine with the idea that there are sets $B$ such that any function $Bto B$ has a fixed point? I would find that as counterintuitive, if not more. I can only think of one such set (or two, really, because of vacuous truths), and in that case your problem isn't really a problem.
– Arthur
Jul 15 at 10:31
It seems to be saying: "if there is such a function f, then ...."
– Bram28
Jul 15 at 10:32
1
@Arthur I can think of lots of sets $B$ such that every function $f:Bto B$ has a fixed point, but they are all isomorphic, they are all one-element sets. I don't believe there are any other examples. Do you have one?
– bof
Jul 15 at 10:46
1
@bumba But the consequence is important. If I said "If pigs can fly, then chickens have teeth", that's a true statement (with current gene editing technology, at least). But your objections here amount to "but pigs can't fly, how can that statement be true?" It is true exactly because the consequence isn't (except for a few trivial cases where the assumption is also true).
– Arthur
Jul 15 at 10:54
1
@Arthur Nope. If there were no functions $f:emptysettoemptyset$ then it would be true "vacuously" that every function $f:emptysettoemptyset$ had a fixed point. But in fact there is one function $f:emptysettoemptyset$ and it has no fixed points.
– bof
Jul 15 at 11:16