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1. • CommentRowNumber1.
   • CommentAuthorTobyBartels
   • CommentTimeJun 8th 2012
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   Author: TobyBartels
   Format: MarkdownItexTo anybody who can think constructively (Mike at least): Given a real number $a$, and let $X$ be the Kuratowski-finite set $\{y, z\}$ such that $y = z$ iff $a \ne 0$ (in the strong sense of apartness). Consider the ring of polynomials with real coefficients and variables from the set $X$. Are the polynomials $a y$ and $a z$ equal? (The polynomial functions from $\mathbb{R}^X$ to $\mathbb{R}$ defined by them are certainly equal, but I'm asking about polynomials as formal expressions. For example, the polynomials $x$ and $x^2$ are not equal, even in the ring of one-variable polynomials with coefficients from $\mathbb{F}_2$.)

   To anybody who can think constructively (Mike at least):

   Given a real number $a$, and let $X$ be the Kuratowski-finite set $\{y, z\}$ such that $y = z$ iff $a \ne 0$ (in the strong sense of apartness). Consider the ring of polynomials with real coefficients and variables from the set $X$. Are the polynomials $a y$ and $a z$ equal?

   (The polynomial functions from $\mathbb{R}^X$ to $\mathbb{R}$ defined by them are certainly equal, but I’m asking about polynomials as formal expressions. For example, the polynomials $x$ and $x^2$ are not equal, even in the ring of one-variable polynomials with coefficients from $\mathbb{F}_2$.)

2. • CommentRowNumber2.
   • CommentAuthorMike Shulman
   • CommentTimeJun 8th 2012
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   Author: Mike Shulman
   Format: MarkdownItexI must be missing something; why are the polynomial functions they define equal?

   I must be missing something; why are the polynomial functions they define equal?

3. • CommentRowNumber3.
   • CommentAuthorIngoBlechschmidt
   • CommentTimeJun 9th 2012
   • (edited Jun 9th 2012)
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   Author: IngoBlechschmidt
   Format: MarkdownItexI believe the following argument shows that the polynomial functions are equal: Let $\varphi \in \mathbb{R}^{X}$, we want to show $a \varphi(y) = a \varphi(z)$. For any $n \geq 1$, we have $a \neq 0$ or $\left|a\right| \le \frac{1}{n}$. In both cases the inequality $\left|a \varphi(y) - a \varphi(z)\right| \leq \frac{1}{n} \cdot \left|\varphi(y) - \varphi(z)\right|$ follows. As $\left|\varphi(y) - \varphi(z)\right|$ is fixed, this shows that indeed $\left|a \varphi(y) - a \varphi(z)\right| = 0$. I don't have an answer to the original question.

   I believe the following argument shows that the polynomial functions are equal:

   Let $\varphi \in \mathbb{R}^{X}$, we want to show $a \varphi(y) = a \varphi(z)$. For any $n \geq 1$, we have $a \neq 0$ or $\left|a\right| \le \frac{1}{n}$. In both cases the inequality $\left|a \varphi(y) - a \varphi(z)\right| \leq \frac{1}{n} \cdot \left|\varphi(y) - \varphi(z)\right|$ follows. As $\left|\varphi(y) - \varphi(z)\right|$ is fixed, this shows that indeed $\left|a \varphi(y) - a \varphi(z)\right| = 0$.

   I don’t have an answer to the original question.

4. • CommentRowNumber4.
   • CommentAuthorMike Shulman
   • CommentTimeJun 9th 2012
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   Author: Mike Shulman
   Format: MarkdownItexNice argument! I'm afraid I don't have an answer to the original question either, though. I don't see any reason why they should be equal, but I don't immediately see how to build a counterexample either.

   Nice argument! I’m afraid I don’t have an answer to the original question either, though. I don’t see any reason why they should be equal, but I don’t immediately see how to build a counterexample either.

5. • CommentRowNumber5.
   • CommentAuthorjcmckeown
   • CommentTimeJun 10th 2012
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   Author: jcmckeown
   Format: MarkdownI shouldn't be distracting myself with this, but what does "=" mean, here? Or is that part of the question, too?

   I shouldn't be distracting myself with this, but what does "=" mean, here? Or is that part of the question, too?

6. • CommentRowNumber6.
   • CommentAuthorMike Shulman
   • CommentTimeJun 10th 2012
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   Author: Mike Shulman
   Format: MarkdownItexWell, the real numbers are a set, as are Toby's set $X$ and the polynomial ring, so $=$ means equality in those sets. Are you asking about how those sets are defined? The set $X$, for instance, is defined as the quotient of the two-element set $\{y,z\}$ by the equivalence relation which sets $y\sim z$ iff $a\neq 0$ (and of course $y\sim y$ and $z\sim z$ unconditionally).

   Well, the real numbers are a set, as are Toby’s set $X$ and the polynomial ring, so $=$ means equality in those sets. Are you asking about how those sets are defined? The set $X$, for instance, is defined as the quotient of the two-element set $\{y,z\}$ by the equivalence relation which sets $y\sim z$ iff $a\neq 0$ (and of course $y\sim y$ and $z\sim z$ unconditionally).

7. • CommentRowNumber7.
   • CommentAuthorTobyBartels
   • CommentTimeJun 10th 2012
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   Author: TobyBartels
   Format: MarkdownItexJC is right to ask; the question is what equality in $\mathbb{R}[X]$ should be. Of course, $\mathbb{R}[X]$ is the free commutative $\mathbb{R}$-algebra on the set $X$, but what's an $\mathbb{R}$-algebra? If this is simply a set equipped with some operations and commutative diagrams (including one unary scalar multiplication operation for each real number), then $a y = a z$ in $\mathbb{R}[X]$ iff $a = 0$ or $a \ne 0$ (so if $a y = a z$ for every $a$, then the [[lesser limited principle of omniscience]] holds). However, if an $\mathbb{R}$-algebra is equipped with a [[tight apartness]] that all operations respect and a binary scalar multiplication operation $\mathbb{R} \times B \to B$ that also respects it and the usual apartness on $\mathbb{R}$ (in the sense of being [[strongly extensional function|strongly extensional]]), then $a y = a z$ necessarily. So that answers my question, I guess.

   JC is right to ask; the question is what equality in $\mathbb{R}[X]$ should be. Of course, $\mathbb{R}[X]$ is the free commutative $\mathbb{R}$-algebra on the set $X$, but what’s an $\mathbb{R}$-algebra? If this is simply a set equipped with some operations and commutative diagrams (including one unary scalar multiplication operation for each real number), then $a y = a z$ in $\mathbb{R}[X]$ iff $a = 0$ or $a \ne 0$ (so if $a y = a z$ for every $a$, then the lesser limited principle of omniscience holds). However, if an $\mathbb{R}$-algebra is equipped with a tight apartness that all operations respect and a binary scalar multiplication operation $\mathbb{R} \times B \to B$ that also respects it and the usual apartness on $\mathbb{R}$ (in the sense of being strongly extensional), then $a y = a z$ necessarily. So that answers my question, I guess.

8. • CommentRowNumber8.
   • CommentAuthorMike Shulman
   • CommentTimeJun 11th 2012
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   Author: Mike Shulman
   Format: MarkdownItexHmm, I was about to say that I thought Ingo's argument would work for polynomials too. Namely, define the norm of a polynomial to be the supremum of the absolute values of its coefficients. Then since either ${|a|}\lt \frac{1}{n}$ or $a\neq 0$, in both cases ${\Vert a y - a z \Vert }\lt \frac{2}{n}$, and this holds for all $n$, so ${\Vert a y - a z \Vert} = 0$ and thus $a y = a z$. What definition of $\mathbb{R}[X]$ is that using?

   Hmm, I was about to say that I thought Ingo’s argument would work for polynomials too. Namely, define the norm of a polynomial to be the supremum of the absolute values of its coefficients. Then since either ${|a|}\lt \frac{1}{n}$ or $a\neq 0$, in both cases ${\Vert a y - a z \Vert }\lt \frac{2}{n}$, and this holds for all $n$, so ${\Vert a y - a z \Vert} = 0$ and thus $a y = a z$. What definition of $\mathbb{R}[X]$ is that using?

9. • CommentRowNumber9.
   • CommentAuthorTobyBartels
   • CommentTimeJun 11th 2012
   • (edited Jun 11th 2012)
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   Author: TobyBartels
   Format: MarkdownItexHow do you know that this supremum exists? (It always exists as a [[lower real number]], but the [[comparison]] law is invalid for those, so we must prove that the supremum is located.) In particular, the norm of $a y - a z$ is $0$ if $y = z$, but this norm is $a$ if $y \ne z$. With the second definition of $\mathbb{R}[X]$ (with apartness), I we can prove that $a y - a z$ must have a located norm (although we can't prove this for an arbitrary polynomial in $\mathbb{R}[X]$), so the argument will still go through; but not with the first definition.

   How do you know that this supremum exists? (It always exists as a lower real number, but the comparison law is invalid for those, so we must prove that the supremum is located.) In particular, the norm of $a y - a z$ is $0$ if $y = z$, but this norm is $a$ if $y \ne z$.

   With the second definition of $\mathbb{R}[X]$ (with apartness), I we can prove that $a y - a z$ must have a located norm (although we can’t prove this for an arbitrary polynomial in $\mathbb{R}[X]$), so the argument will still go through; but not with the first definition.

10. • CommentRowNumber10.
    • CommentAuthorMike Shulman
    • CommentTimeJun 11th 2012
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    Author: Mike Shulman
    Format: MarkdownItexAh, right. (You can probably tell that I don't work with constructive real numbers much.)

    Ah, right. (You can probably tell that I don’t work with constructive real numbers much.)