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Since your algebra mixes together the objects $x$ and $y$ with their pair $(x,y)$ in the same algebra, it has the effect of erasing "ordered-pair" as a separate type in this context, and so there is no reason to expect that condition $(*)$ is all the structure one will expect to find. For example, in your algebra you can form iterated terms like $(x,(y,z))$ and inquire whether $x=(x,y)$ is possible, while in a more highly typed context, such an equation wouldn't even be sensible necessarily. For this reason, there are numerous pairing functions that exhibit all kinds of other extra algebraic structure in the algebra you are considering.

For example, many of the usual ordered-pair definitions in set theory have the property that $(x,y)$ has higher rank than $x$ and $y$, and in particular, $x\neq (x,y)\neq y$.

Similarly, for most of the pairing functions, $\emptyset\neq (x,y)$, and so these pairing function are never a bijection of $U\times U$ with $U$.

But there are other pairing functions that do constitute a bijection between $U\times U$ and $U$, and in this sense it would be correct to write $U\times U=U$. This would include some of the usual flat pairing functions one sees in set theory, where actually every set $x$ is $(y,z)$ for some $y$ and $z$, and so the pairing function is a bijection of $U\times U$ with $U$. Indeed, with the flat pairing functions I have in mind, $V_\theta\times V_\theta=V_\theta$ for any infinite ordinal $\theta$, and this includes all the Grothendieck universes you were considering.

One can easily design artificial pairing functions that have other extra properties, such as having fixed-point objects $x$ for which $x=(x,x)$ and hence $x=(x,(x,x))=(((x,x),x),x)$ and so on, or having no such fixed-points $x$. One could also make pairing functions that had various instances of finite cycles $x=(x,y)$, $y=(y,z)$ and $z=(z,x)$ and so on. One can easily arrange crazy stuff, since of course the only requirement that $(*)$ imposes is that $(\cdot,\cdot)$ is injective.

If $(x,y)$ is a pairing function and $\pi:U\to U$ is any injective function, then $(x,y)_\pi:=\pi((x,y))$ is another pariing function. And indeed, all pairing functions arise this way from any given surjective pairing function.

Since your algebra mixes together the objects $x$ and $y$ with their pair $(x,y)$ in the same algebra, it has the effect of erasing "ordered-pair" as a separate type in this context, and so there is no reason to expect that condition $(*)$ is all the structure one will expect to find. For example, in your algebra you can form iterated terms like $(x,(y,z))$ and inquire whether $x=(x,y)$ is possible, while in a more highly typed context, such an equation wouldn't even be sensible necessarily. For this reason, there are numerous pairing functions that exhibit all kinds of other extra algebraic structure in the algebra you are considering.

For example, many of the usual ordered-pair definitions in set theory have the property that $(x,y)$ has higher rank than $x$ and $y$, and in particular, $x\neq (x,y)\neq y$.

Similarly, for most of the pairing functions, $\emptyset\neq (x,y)$, and so these pairing function are never a bijection of $U\times U$ with $U$.

But there are other pairing functions that do constitute a bijection between $U\times U$ and $U$, and in this sense it would be correct to write $U\times U=U$. This would include some of the usual flat pairing functions one sees in set theory, where actually every set $x$ is $(y,z)$ for some $y$ and $z$, and so the pairing function is a bijection of $U\times U$ with $U$. Indeed, with the flat pairing functions I have in mind, $V_\theta\times V_\theta=V_\theta$ for any infinite ordinal $\theta$, and this includes all the Grothendieck universes you were considering.

One can easily design artificial pairing functions that have other extra properties, such as having fixed-point objects $x$ for which $x=(x,x)$ and hence $x=(x,(x,x))=(((x,x),x),x)$ and so on, or having no such fixed-points $x$. One could also make pairing functions that had various instances of finite cycles $x=(x,y)$, $y=(y,z)$ and $z=(z,x)$ and so on. One can easily arrange crazy stuff, since of course the only requirement that $(*)$ imposes is that $(\cdot,\cdot)$ is injective.

If $(x,y)$ is a pairing function and $\pi:U\to U$ is any injective function, then $(x,y)_\pi:=\pi((x,y))$ is another pariing function. And indeed, all pairing functions arise this way from any given surjective pairing function.

Since your algebra mixes together the objects $x$ and $y$ with their pair $(x,y)$ in the same algebra, it has the effect of erasing "ordered-pair" as a separate type in this context, and so there is no reason to expect that condition $(*)$ is all the structure one will expect to find. For example, in your algebra you can form iterated terms like $(x,(y,z))$ and inquire whether $x=(x,y)$ is possible, while in a more highly typed context, such an equation wouldn't even be sensible necessarily. For this reason, there are numerous pairing functions that exhibit all kinds of other extra algebraic structure in the algebra you are considering.

For example, many of the usual ordered-pair definitions in set theory have the property that $(x,y)$ has higher rank than $x$ and $y$, and in particular, $x\neq (x,y)\neq y$.

Similarly, for most of the pairing functions, $\emptyset\neq (x,y)$, and so these pairing function are never a bijection of $U\times U$ with $U$.

But there are other pairing functions that do constitute a bijection between $U\times U$ and $U$, and in this sense it would be correct to write $U\times U=U$. This would include some of the usual flat pairing functions one sees in set theory, where actually every set $x$ is $(y,z)$ for some $y$ and $z$, and so the pairing function is a bijection of $U\times U$ with $U$. Indeed, with the flat pairing functions I have in mind, $V_\theta\times V_\theta=V_\theta$ for any infinite ordinal $\theta$, and this includes all the Grothendieck universes you were considering.

One can easily design artificial pairing functions that have other extra properties, such as having fixed-point objects $x$ for which $x=(x,x)$ and hence $x=(x,(x,x))=(((x,x),x),x)$ and so on, or having no such fixed-points $x$. One could also make pairing functions that had various instances of finite cycles $x=(x,y)$, $y=(y,z)$ and $z=(z,x)$ and so on. One can easily arrange crazy stuff, since of course the only requirement that $(*)$ imposes is that $(\cdot,\cdot)$ is injective.

Since your algebra mixes together the objects $x$ and $y$ with their pair $(x,y)$ in the same algebra, it has the effect of erasing "ordered-pair" as a separate type in this context, and so there is no reason to expect that condition $(*)$ is all the structure one will expect to find. For example, in your algebra you can form iterated terms like $(x,(y,z))$ and inquire whether $x=(x,y)$ is possible, while in a more highly typed context, such an equation wouldn't even be sensible necessarily. For this reason, there are numerous pairing functions that exhibit all kinds of other extra algebraic structure in the algebra you are considering.

For example, many of the usual ordered-pair definitions in set theory have the property that $(x,y)$ has higher rank than $x$ and $y$, and in particular, $x\neq (x,y)\neq y$.

Similarly, for most of the pairing functions, $\emptyset\neq (x,y)$, and so these pairing function are never a bijection of $U\times U$ with $U$.

But there are other pairing functions that do constitute a bijection between $U\times U$ and $U$, and in this sense it would be correct to write $U\times U=U$. This would include some of the usual flat pairing functions one sees in set theory, where actually every set $x$ is $(y,z)$ for some $y$ and $z$, and so the pairing function is a bijection of $U\times U$ with $U$. Indeed, with the flat pairing functions I have in mind, $V_\theta\times V_\theta=V_\theta$ for any infinite ordinal $\theta$, and this includes all the Grothendieck universes you were considering.

One can easily design artificial pairing functions that have other extra properties, such as having fixed-point objects $x$ for which $x=(x,x)$ and hence $x=(x,(x,x))=(((x,x),x),x)$ and so on, or having no such fixed-points $x$. One could also make pairing functions that had various instances of finite cycles $x=(x,y)$, $y=(y,z)$ and $z=(z,x)$ and so on. One can easily arrange crazy stuff, since of course the only requirement that $(*)$ imposes is that $(\cdot,\cdot)$ is injective.

If $(x,y)$ is a pairing function and $\pi:U\to U$ is any injective function, then $(x,y)_\pi:=\pi((x,y))$ is another pariing function. And indeed, all pairing functions arise this way from any given surjective pairing function.

Since your algebra mixes together the objects $x$ and $y$ with their pair $(x,y)$ in the same algebra, it has the effect of erasing "ordered-pair" as a separate type in this context, and so there is no reason to expect that condition $(*)$ is all the structure one will expect to find. For example, in your algebra you can form iterated terms like $(x,(y,z))$ and inquire whether $x=(x,y)$ is possible, while in a more highly typed context, such an equation wouldn't even be sensible necessarily. For this reason, there are numerous pairing functions that exhibit all kinds of other extra algebraic structure in the algebra you are considering.

For example, many of the usual ordered-pair definitions in set theory have the property that $(x,y)$ has higher rank than $x$ and $y$, and in particular, $x\neq (x,y)\neq y$.

Similarly, for most of the pairing functions, $\emptyset\neq (x,y)$, and so these pairing function are never a bijection of $U\times U$ with $U$.

But there are other pairing functions that do constitute a bijection between $U\times U$ and $U$, and in this sense it would be correct to write $U\times U=U$. This would include some of the usual flat pairing functions one sees in set theory, where actually every set $x$ is $(y,z)$ for some $y$ and $z$, and so the pairing function is a bijection of $U\times U$ with $U$. Indeed, with the flat pairing functions I have in mind, $V_\theta\times V_\times=V_\theta$ for any infinite ordinal $\theta$, and this includes all the Grothendieck universes you were considering.

One can easily design artificial pairing functions that have other extra properties, such as having fixed-point objects $x$ for which $x=(x,x)$ and hence $x=(x,(x,x))=(((x,x),x),x)$ and so on, or having no such fixed-points $x$. One could also make pairing functions that had various instances of finite cycles $x=(x,y)$, $y=(y,z)$ and $z=(z,x)$ and so on. One can easily arrange crazy stuff, since of course the only requirement that $(*)$ imposes is that $(\cdot,\cdot)$ is injective.

Since your algebra mixes together the objects $x$ and $y$ with their pair $(x,y)$ in the same algebra, it has the effect of erasing "ordered-pair" as a separate type in this context, and so there is no reason to expect that condition $(*)$ is all the structure one will expect to find. For example, in your algebra you can form iterated terms like $(x,(y,z))$ and inquire whether $x=(x,y)$ is possible, while in a more highly typed context, such an equation wouldn't even be sensible necessarily. For this reason, there are numerous pairing functions that exhibit all kinds of other extra algebraic structure in the algebra you are considering.

For example, many of the usual ordered-pair definitions in set theory have the property that $(x,y)$ has higher rank than $x$ and $y$, and in particular, $x\neq (x,y)\neq y$.

Similarly, for most of the pairing functions, $\emptyset\neq (x,y)$, and so these pairing function are never a bijection of $U\times U$ with $U$.

But there are other pairing functions that do constitute a bijection between $U\times U$ and $U$, and in this sense it would be correct to write $U\times U=U$. This would include some of the usual flat pairing functions one sees in set theory, where actually every set $x$ is $(y,z)$ for some $y$ and $z$, and so the pairing function is a bijection of $U\times U$ with $U$. Indeed, with the flat pairing functions I have in mind, $V_\theta\times V_\theta=V_\theta$ for any infinite ordinal $\theta$, and this includes all the Grothendieck universes you were considering.

One can easily design artificial pairing functions that have other extra properties, such as having fixed-point objects $x$ for which $x=(x,x)$ and hence $x=(x,(x,x))=(((x,x),x),x)$ and so on, or having no such fixed-points $x$. One could also make pairing functions that had various instances of finite cycles $x=(x,y)$, $y=(y,z)$ and $z=(z,x)$ and so on. One can easily arrange crazy stuff, since of course the only requirement that $(*)$ imposes is that $(\cdot,\cdot)$ is injective.