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Keith Kearnes
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This is a long comment rather than a complete answer. But before writing it let me insert that I don't agree that universal algebra is the study of varieties. It is the study of algebraic structures. (In my universe, universal algebra is synonymous with algebra.)

 

Taylor showed that there is no algorithm to determine if a finitely presented variety belongs to $({\mathcal V}_{\mathbb R}]$. He doesdid this by interpreting the undecidable satisfiability problem for Diophantine equations into this problem.

This is a long comment rather than a complete answer. But before writing it let me insert that I don't agree that universal algebra is the study of varieties. It is the study of algebraic structures. (In my universe, universal algebra is synonymous with algebra.)

Taylor showed that there is no algorithm to determine if a finitely presented variety belongs to $({\mathcal V}_{\mathbb R}]$. He does this by interpreting the undecidable satisfiability problem for Diophantine equations into this problem.

This is a long comment rather than a complete answer. But before writing it let me insert that I don't agree that universal algebra is the study of varieties. (In my universe, universal algebra is synonymous with algebra.)

 

Taylor showed that there is no algorithm to determine if a finitely presented variety belongs to $({\mathcal V}_{\mathbb R}]$. He did this by interpreting the undecidable satisfiability problem for Diophantine equations into this problem.

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Keith Kearnes
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Regarding Q2.1: There are continuumly many continuous, associative, binary operations on $\mathbb R$. There can't be more, and it is not hard to construct this many. For a trivial construction of many continuous associative operations, defineconsider constant operations $x*_ry = r$ for different $r\in \mathbb R$. Less trivially, select an interval $[a,b]$ and define $f(x) = \min(\max(x,a),b)$. The operation $x*y = \max(f(x),f(y))$ is continuous, associative, depends on both variables, and has range $[a,b]$. You get a lot of these by varying $a$ and $b$. From some given continuous, associative, binary operations you can often generate more by conjugating by continuous permutations of $\mathbb R$. ($x*_{\pi}y:=\pi^{-1}(\pi(x)*\pi(y))$.)

Regarding Q2.1: There are continuumly many continuous, associative, binary operations on $\mathbb R$. There can't be more, and it is not hard to construct this many. For a trivial construction of many continuous associative operations, define $x*_ry = r$ for different $r\in \mathbb R$. Less trivially, select an interval $[a,b]$ and define $f(x) = \min(\max(x,a),b)$. The operation $x*y = \max(f(x),f(y))$ is continuous, associative, depends on both variables, and has range $[a,b]$. You get a lot of these by varying $a$ and $b$. From some given continuous, associative, binary operations you can often generate more by conjugating by continuous permutations of $\mathbb R$. ($x*_{\pi}y:=\pi^{-1}(\pi(x)*\pi(y))$.)

Regarding Q2.1: There are continuumly many continuous, associative, binary operations on $\mathbb R$. There can't be more, and it is not hard to construct this many. For a trivial construction of many continuous associative operations, consider constant operations $x*_ry = r$ for different $r\in \mathbb R$. Less trivially, select an interval $[a,b]$ and define $f(x) = \min(\max(x,a),b)$. The operation $x*y = \max(f(x),f(y))$ is continuous, associative, depends on both variables, and has range $[a,b]$. You get a lot of these by varying $a$ and $b$. From some given continuous, associative, binary operations you can often generate more by conjugating by continuous permutations of $\mathbb R$. ($x*_{\pi}y:=\pi^{-1}(\pi(x)*\pi(y))$.)

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Keith Kearnes
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This is a long comment rather than a complete answer. But before writing it let me insert that I don't agree that universal algebra is the study of varieties. It is the study of algebraic structures. (In my universe, universal algebra is synonymous with algebra.)

Regarding Q2.1: There are continuumly many continuous, associative, binary operations on $\mathbb R$. There can't be more, and it is not hard to construct this many. For a trivial construction of many continuous associative operations, define $x*y = r$$x*_ry = r$ for different $r\in \mathbb R$. Less trivially, select an interval $[a,b]$ and define $f(x) = \min(\max(x,a),b)$. The operation $x*y = \max(f(x),f(y))$ is continuous, associative, depends on both variables, and has range $[a,b]$. You get a lot of these by varying $a$ and $b$. From some given continuous, associative, binary operations you can often generate more by conjugating by continuous permutations of $\mathbb R$. ($x*_{\pi}y:=\pi^{-1}(\pi(x)*\pi(y))$.)

This is a long comment rather than a complete answer. But before writing it let me insert that I don't agree that universal algebra is the study of varieties. It is the study of algebraic structures.

Regarding Q2.1: There are continuumly many continuous, associative, binary operations on $\mathbb R$. There can't be more, and it is not hard to construct this many. For a trivial construction of many continuous associative operations, define $x*y = r$ for different $r\in \mathbb R$. Less trivially, select an interval $[a,b]$ and define $f(x) = \min(\max(x,a),b)$. The operation $x*y = \max(f(x),f(y))$ is continuous, associative, depends on both variables, and has range $[a,b]$. You get a lot of these by varying $a$ and $b$. From some given continuous, associative, binary operations you can often generate more by conjugating by continuous permutations of $\mathbb R$. ($x*_{\pi}y:=\pi^{-1}(\pi(x)*\pi(y))$.)

This is a long comment rather than a complete answer. But before writing it let me insert that I don't agree that universal algebra is the study of varieties. It is the study of algebraic structures. (In my universe, universal algebra is synonymous with algebra.)

Regarding Q2.1: There are continuumly many continuous, associative, binary operations on $\mathbb R$. There can't be more, and it is not hard to construct this many. For a trivial construction of many continuous associative operations, define $x*_ry = r$ for different $r\in \mathbb R$. Less trivially, select an interval $[a,b]$ and define $f(x) = \min(\max(x,a),b)$. The operation $x*y = \max(f(x),f(y))$ is continuous, associative, depends on both variables, and has range $[a,b]$. You get a lot of these by varying $a$ and $b$. From some given continuous, associative, binary operations you can often generate more by conjugating by continuous permutations of $\mathbb R$. ($x*_{\pi}y:=\pi^{-1}(\pi(x)*\pi(y))$.)

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Keith Kearnes
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Keith Kearnes
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Keith Kearnes
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Keith Kearnes
  • 14.6k
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  • 86
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