Comments on: Proof of negation and proof by contradiction http://math.andrej.com/2010/03/29/proof-of-negation-and-proof-by-contradiction/ Mathematics for computers Mon, 30 Jan 2012 07:14:48 +0000 hourly 1 http://wordpress.org/?v=3.2-beta2-18055 By: Robert Furber http://math.andrej.com/2010/03/29/proof-of-negation-and-proof-by-contradiction/comment-page-1/#comment-14300 Robert Furber Sat, 30 Oct 2010 03:42:33 +0000 http://math.andrej.com/?p=453#comment-14300 You are quite right, of course. I got confused because I didn't realise you had assumed something that was Bishop-false. This must be why Bishop's definition of compact isn't any of the usual ones, it's totally bounded and complete, and applies only to metric spaces. You are quite right, of course. I got confused because I didn’t realise you had assumed something that was Bishop-false. This must be why Bishop’s definition of compact isn’t any of the usual ones, it’s totally bounded and complete, and applies only to metric spaces.

]]> By: Andrej Bauer http://math.andrej.com/2010/03/29/proof-of-negation-and-proof-by-contradiction/comment-page-1/#comment-14276 Andrej Bauer Wed, 27 Oct 2010 16:47:21 +0000 http://math.andrej.com/?p=453#comment-14276 @Robert: I took the liberty to edit your comment to insert LaTeX and also to fix the proof a bit (I think HTML ate a part of it). Please let me know if I did injustice to your proof. (Also, I think it goes without saying that $f$ is assumed to be continuous in Bishop sense, i.e., uniformly continuous on every closed interval, otherwise how do you define the integral?) You have shown, I think, that if every Bishop-continuous map on $[0,1]$ attains its maximum then the Intermediate Value Theorem holds. But my claim (specialized to $X = [0,1]$) was: if every sequence in $[0,1]$ has a convergent subsequence, then every continuous map on $[0,1]$ attains its maximum. I am very sorry, but how does what you have shown invalidate what I claimed? If we put both things together we only get: "If every sequence in $[0,1]$ has a convergent subsequence then the Intermediate Value Theorem holds". This is not in contradiction with possible failure of the Intermediate Value Theorem in Bishop-style mathematics. In fact, it is easy to show Bishop-style that the claim actually holds. For suppose every sequence in $[0,1]$ has a convergent subsequence and $f : [0,1] \to \mathbb{R}$ is continuous and $f(0) < 0 < f(1)$. We can find a sequence $(x_n)_n$ such that $|f(x_n)| \leq 2^{-n}$. Take a convergent subsequence of $(x_n)$, and its limit is where $f$ has a zero. It is an exercise to show that if every sequence of $[0,1]$ has a convergent subsequence then every continuous map on $[0,1]$ attains its maximum. @Robert: I took the liberty to edit your comment to insert LaTeX and also to fix the proof a bit (I think HTML ate a part of it). Please let me know if I did injustice to your proof. (Also, I think it goes without saying that $f$ is assumed to be continuous in Bishop sense, i.e., uniformly continuous on every closed interval, otherwise how do you define the integral?)

You have shown, I think, that if every Bishop-continuous map on $[0,1]$ attains its maximum then the Intermediate Value Theorem holds.

But my claim (specialized to $X = [0,1]$) was: if every sequence in $[0,1]$ has a convergent subsequence, then every continuous map on $[0,1]$ attains its maximum.

I am very sorry, but how does what you have shown invalidate what I claimed? If we put both things together we only get: “If every sequence in $[0,1]$ has a convergent subsequence then the Intermediate Value Theorem holds”. This is not in contradiction with possible failure of the Intermediate Value Theorem in Bishop-style mathematics. In fact, it is easy to show Bishop-style that the claim actually holds. For suppose every sequence in $[0,1]$ has a convergent subsequence and $f : [0,1] \to \mathbb{R}$ is continuous and $f(0) < 0 < f(1)$. We can find a sequence $(x_n)_n$ such that $|f(x_n)| \leq 2^{-n}$. Take a convergent subsequence of $(x_n)$, and its limit is where $f$ has a zero.

It is an exercise to show that if every sequence of $[0,1]$ has a convergent subsequence then every continuous map on $[0,1]$ attains its maximum.

]]> By: Robert Furber http://math.andrej.com/2010/03/29/proof-of-negation-and-proof-by-contradiction/comment-page-1/#comment-14264 Robert Furber Tue, 26 Oct 2010 22:21:55 +0000 http://math.andrej.com/?p=453#comment-14264 <cite> I am pretty sure with a bit more work we could show that f attains its supremum, and in fact this must have been proved by someone constructively. </cite> Unfortunately not. Recall that it is not possible to prove the intermediate value theorem constructively (see page 5 of Bishop's book, the introduction section, send me a mail if you don't have the book and want me to just scan that part for you). However, a method to prove your statement can be adapted to prove the intermediate value theorem: Suppose that $f$ is a function $[0,1] \to \mathbb{R}$, and $f(0) < 0 < f(1)$. Let $F(x) = \int_0^x -f(y) dy$. Use your procedure to find the $x$ in $[0,1]$ where $F$ attains its maximum. By the fundamental theorem of calculus $F$ is differentiable and $F' = -f$, and since $x$ is where $F$ has a maximum, $F'(x) = 0$, so $-f(x) = 0$, so $f(x) = 0$, and the intermediate value has been found. [End proof] Therefore no such method can exist (a proof of negation). Bishop proves the necessary calculus theorems in his book. Interestingly, Bishop shows how to compute the least upper bound of a continuous function on an interval, but doesn't mention that it can't be shown to take that value.
I am pretty sure with a bit more work we could show that f attains its supremum, and in fact this must have been proved by someone constructively.

Unfortunately not. Recall that it is not possible to prove the intermediate value theorem constructively (see page 5 of Bishop’s book, the introduction section, send me a mail if you don’t have the book and want me to just scan that part for you). However, a method to prove your statement can be adapted to prove the intermediate value theorem:

Suppose that $f$ is a function $[0,1] \to \mathbb{R}$, and $f(0) < 0 < f(1)$. Let $F(x) = \int_0^x -f(y) dy$. Use your procedure to find the $x$ in $[0,1]$ where $F$ attains its maximum. By the fundamental theorem of calculus $F$ is differentiable and $F' = -f$, and since $x$ is where $F$ has a maximum, $F'(x) = 0$, so $-f(x) = 0$, so $f(x) = 0$, and the intermediate value has been found. [End proof]

Therefore no such method can exist (a proof of negation).

Bishop proves the necessary calculus theorems in his book. Interestingly, Bishop shows how to compute the least upper bound of a continuous function on an interval, but doesn’t mention that it can’t be shown to take that value.

]]> By: Andrej Bauer http://math.andrej.com/2010/03/29/proof-of-negation-and-proof-by-contradiction/comment-page-1/#comment-14246 Andrej Bauer Mon, 25 Oct 2010 04:37:20 +0000 http://math.andrej.com/?p=453#comment-14246 @Zach: you have to be careful there. Intuitionistically $a \Rightarrow b$ implies $\lnot b \Rightarrow \lnot a$, but in general the other direction does not hold (and in fact, if it does then we get classical logic). However, if I remember correctly, as soon as either $a$ or $b$ is $\lnot\lnot$-stable (equivalent to its double negation) then we do get $(a \Rightarrow b) \iff (\lnot b \Rightarrow \lnot a)$. In the case of $\lnot \phi$, which is just an abbreviation for $\phi \Rightarrow \bot$, we do have $\lnot\lnot$-stability of $\bot$, so indeed we could prove the contrapositive if we wished. But the contrapositive is $\lnot \bot \Rightarrow \lnot \phi$, which is just $\lnot \phi$ again, so we end up where we started. @Zach: you have to be careful there. Intuitionistically $a \Rightarrow b$ implies $\lnot b \Rightarrow \lnot a$, but in general the other direction does not hold (and in fact, if it does then we get classical logic). However, if I remember correctly, as soon as either $a$ or $b$ is $\lnot\lnot$-stable (equivalent to its double negation) then we do get $(a \Rightarrow b) \iff (\lnot b \Rightarrow \lnot a)$. In the case of $\lnot \phi$, which is just an abbreviation for $\phi \Rightarrow \bot$, we do have $\lnot\lnot$-stability of $\bot$, so indeed we could prove the contrapositive if we wished. But the contrapositive is $\lnot \bot \Rightarrow \lnot \phi$, which is just $\lnot \phi$ again, so we end up where we started.

]]> By: Zach Norwood http://math.andrej.com/2010/03/29/proof-of-negation-and-proof-by-contradiction/comment-page-1/#comment-14174 Zach Norwood Wed, 20 Oct 2010 02:57:21 +0000 http://math.andrej.com/?p=453#comment-14174 @Andrej: Is a proof by negation valid for the intuitionist because it can easily be turned into a proof by contraposition? (For example, if, in trying to prove $\lnot\phi$, I prove $\phi\Rightarrow b$, where $\lnot b$ is known to be true, then I've established the contrapositive $\lnot b\Rightarrow\lnot\phi$, which establishes $\lnot\phi$.) In fact, is the formalization of such a proof actually a proof of the contrapositive, at least as the intuitionist requires the proof to be formalized? Playing the same game with proofs by contradiction (as opposed to proofs by negation) requires excluded middle, of course (as you say). @Andrej: Is a proof by negation valid for the intuitionist because it can easily be turned into a proof by contraposition? (For example, if, in trying to prove $\lnot\phi$, I prove $\phi\Rightarrow b$, where $\lnot b$ is known to be true, then I’ve established the contrapositive $\lnot b\Rightarrow\lnot\phi$, which establishes $\lnot\phi$.) In fact, is the formalization of such a proof actually a proof of the contrapositive, at least as the intuitionist requires the proof to be formalized? Playing the same game with proofs by contradiction (as opposed to proofs by negation) requires excluded middle, of course (as you say).

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