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Lecture 23 - Chapter 2: Commutative Monoidal Posets

Let's look at some examples of commutative monoidal posets! You probably know a lot of them without even realizing it. That's what category theory often does: make implicit knowledge explicit.

Last time we saw that many of our favorite number systems, like \(\mathbb{N}, \mathbb{Z}, \mathbb{Q}\) and \(\mathbb{R} \) are commutative monoidal posets with the usual \(\le\), with \(+\) as the monoid operation and \(0\) as the unit for this operation. You should be wondering this:

Puzzle 71. Can you make the complex numbers, \(\mathbb{C}\), into a commutative monoidal poset with the usual \(+\) and \(0\) and some concept of \(\le\)? If so, how many ways can you do this?

But more importantly, we can get a lot of commutative monoidal posets from logic. This may seem weird: we're supposed to be talking about resource theories, but now I'm talking about logic. But it's not weird: logic is about statements... and statements—or if you prefer, "information" or "data"—are a very important kind of resource!

We've already seen that different kinds of posets give different kinds of logic. For example, in Lecture 8 we took the collection of all subsets of a set \(X\) and defined \(S \le T\) if \(S \subseteq T\), getting a poset whose elements can be seen as statements about elements of \(X\). In Lecture 10 got another nice poset, and a different form of logic, by considering all the partitions of a set.

There are often two ways to create a commutative monoidal poset from a form of logic! In one the monoid operation is "join", \(\vee\), which often has the logical meaning of "or". In another the monoid operation is "meet", \( \wedge\), which has the logical meaning of "and".

Let's see how it works.

Posets with finite joins. Suppose \( (X, \le) \) is a poset for which every finite subset has a join. This is equivalent to saying that every pair of elements in \(X\) has a join and the empty set also has a join.

The join of two elements \(x,y \in X\) is called \(x \vee y\), and by definition it's the least element that's greater than or equal to both \(x\) and \(y\).

The join of the empty set is called the bottom element of \(X\), and denoted \(\bot\). By definition it's the least element that's greater than or equal to... everything in the empty set! In other words, it's the least element of \(X\).

From all this we get a commutative monoidal poset \( (X, \le, \vee, \bot) \). In logic, \(\vee\) often means "or" and \(\bot\) often means "false".

Puzzle 72. Prove this! To prove it's a monoidal poset, you must check that \(\vee\) is associative, that \(\bot \vee x = x = x \vee \bot\), and that

$$ x \le x' \textrm{ and } y \le y' \textrm{ imply } x \vee y \le x' \vee y' .$$ To prove it's commutative, you must check that

$$ x \vee y = y \vee x .$$ Puzzle 73. I claimed that every finite subset of a poset have a join if and only if every pair of elements has a join and the empty set has a join. Why is this true?

All this has a version for meets, too:

Posets with finite meets. Suppose \( (X, \le) \) is a poset for which every finite subset has a meet. This is equivalent to saying that every pair of elements in \(X\) has a meet and the empty set also has a meet.

The join of two elements \(x,y \in X\) is called \(x \wedge y\), and by definition it's the greatest element that's less than or equal to both \(x\) and \(y\).

The meet of the empty set is called the top element of \(X\), and denoted \(\top \). By definition it's the greatest element that's less than or equal to... everything in the empty set! In other words, it's the greatest element of \(X\).

From all this we get a commutative monoidal poset \( (X, \le, \wedge, \top) \). In logic, \(\wedge\) often means "and" and \(\top\) often means "true". (We are very lucky that in English "true" starts with the same letter as "top".)

Do we need to check all these claims in detail?

No! For any poset we can define a new poset, its opposite, where \(\le\) is defined to mean \(\ge\) in the old poset. Joins in the old poset are meets in the new poset, and vice versa. The bottom of the old poset is the top of the new poset, and vice versa. Using this trick, we can take all our claims about a poset with finite meets and turn them into the claims we've already seen about a poset with finite joins.

Up is down! Down is up! This trick is called duality, and I discussed it in Lecture 17.

The commutative monoidal posets we get from logic have some properties that are unusual when compared to, say, the real numbers:

Puzzle 74. Show that if \( (X, \le) \) is a poset with finite joins, we have

$$ \bot \le x $$ and

$$ x \vee x = x $$ for all \(x \in X \).

Similarly, if \( (X, \le) \) is a poset with finite meets, we have

$$ x \le \top $$ and

$$ x \wedge x = x $$ for all \(x \in X \).

These special properties say that information can be "deleted" and "duplicated" in certain special ways, that don't work for other resources. For example, \(x \wedge x = x\) says that knowing \(x\) and \(x\) is exactly the same as knowing \(x\). This is connected to the fact that we can take a piece of data, like the bits of information in a music file, and duplicate it. On the other hand, we can't usually duplicate money: it's not true that \( $ 10 + $10 = $10\). But this leads to big problems in the Information Age, where more and more good are being reduced to data. How can you can sell two copies of a music file for \($10 \) each, if the person you sell it to is free to duplicate it?

There are also issues with "deletion", but these are a bit subtler so I'll postpone them for later.

To read other lectures go here.

Comments

  • 1.
    edited May 14

    Puzzle 71. Can you make the complex numbers, ℂ, into a commutative monoidal poset with the usual + and 0 and some concept of ≤? If so, how many ways can you do this?

    We can always just do \(x \preceq_1 y \iff \mathfrak{Re}(x) \leq \mathfrak{Re}(y)\) where \(\mathfrak{Re}(x)\) is the real component of \(x\).

    But this is just a special case of looking at the magnitudes of the orthogonal projection of two points in \(\mathcal{C}\) onto a common point. Specifically, \(\mathfrak{Re}(x)\) is the magnitude of \(x\) projected onto \(1 + i0\). We could pick any point in \(\mathbb{C}\) and do this. The magnitude of the orthogonal projection of \(p + iq\) onto \(x + i y\) is \(p x + y q\), or the usual dot-product \((\cdot)\) from linear algebra.

    So we have \(2^{\aleph_0}\) other examples, each one corresponding to

    $$ x \preceq_p y \iff p \cdot x \leq p \cdot y $$ for each point \(p \in \mathbb{C}\).

    Comment Source:> **Puzzle 71.** Can you make the complex numbers, ℂ, into a commutative monoidal poset with the usual + and 0 and some concept of ≤? If so, how many ways can you do this? We can always just do \\(x \preceq_1 y \iff \mathfrak{Re}(x) \leq \mathfrak{Re}(y)\\) where \\(\mathfrak{Re}(x)\\) is the real component of \\(x\\). But this is just a special case of looking at the magnitudes of the [*orthogonal projection*](https://en.wikibooks.org/wiki/Linear_Algebra/Orthogonal_Projection_Onto_a_Line) of two points in \\(\mathcal{C}\\) onto a common point. Specifically, \\(\mathfrak{Re}(x)\\) is the magnitude of \\(x\\) projected onto \\(1 + i0\\). We could pick any point in \\(\mathbb{C}\\) and do this. The magnitude of the orthogonal projection of \\(p + iq\\) onto \\(x + i y\\) is \\(p x + y q\\), or the usual [dot-product \\((\cdot)\\)](https://en.wikipedia.org/wiki/Dot_product) from linear algebra. So we have \\(2^{\aleph_0}\\) other examples, each one corresponding to $$ x \preceq_p y \iff p \cdot x \leq p \cdot y $$ for each point \\(p \in \mathbb{C}\\). [wikipedia_norm]: https://en.wikipedia.org/wiki/Norm_(mathematics)
  • 2.
    edited May 14

    Thanks, Matthew, for \( 2^{\aleph_0} \) solutions to Puzzle 71.

    In layman's language: if we have two points \(x\) and \(y\) on the plane, we can decree that \(x \le y\) if \(x\) is no further north than \(y\), or no further southwest.... or no further in any chosen direction on the compass.

    This will get along with vector addition, meaning

    $$ x \le x' \textrm{ and } y \le y' \textrm{ imply } x + y \le x' + y' $$ for all points \(x,x',y,y'\) in the plane.

    But believe it or not, these are not all the solutions to Puzzle 71!

    Comment Source:Thanks, Matthew, for \\( 2^{\aleph_0} \\) solutions to Puzzle 71. In layman's language: if we have two points \\(x\\) and \\(y\\) on the plane, we can decree that \\(x \le y\\) if \\(x\\) is no further _north_ than \\(y\\), or no further _southwest_.... or no further in _any_ chosen direction on the compass. This will get along with vector addition, meaning \[ x \le x' \textrm{ and } y \le y' \textrm{ imply } x + y \le x' + y' \] for all points \\(x,x',y,y'\\) in the plane. But believe it or not, these are not all the solutions to Puzzle 71!
  • 3.
    edited May 14

    I expanded the last paragraph of this lecture after writing the rest. I decided it was good to say a bit about the economics of free duplication while the equation \(x \wedge x = x\) was still on people's minds!

    Comment Source:I expanded the last paragraph of this lecture after writing the rest. I decided it was good to say a bit about the economics of free duplication while the equation \\(x \wedge x = x\\) was still on people's minds!
  • 4.

    Puzzle 71:

    Take the distance from the origin and create a preordering from that.

    Comment Source:Puzzle 71: Take the distance from the origin and create a preordering from that.
  • 5.
    edited May 14

    Keith - Puzzle 71 asks you to make the complex numbers into a commutative monoidal poset with its usual \(+\) and \( 0 \) and some new notion of \(\le\). It's clearly a commutative monoid. You've described how to make it into a preorder, not a poset. But we can still ask: does your recipe give a commutative monoidal preorder?

    Comment Source:Keith - Puzzle 71 asks you to make the complex numbers into a commutative monoidal poset with its usual \\(+\\) and \\( 0 \\) and some new notion of \\(\le\\). It's clearly a commutative monoid. You've described how to make it into a preorder, not a poset. But we can still ask: does your recipe give a commutative monoidal preorder?
  • 6.
    edited May 14

    Puzzle 71. I would like to give a try. I guess we have to focus on modulus or real and imaginary parts separately, considering, in a geometrical sense, distances from the origin on the complex plane, and numbers as tuples of reals. Let \(z_1=a_1+i b_1\) and \(z_2=a_2+i b_2\) be two complex numbers. Let us consider, for example, \(|z_1|\le |z_2|\). Let \(z_3=a_3+i b_3\) and \(z_4=a_2+i b_4\) be another pair of complex numbers. Let us suppose that they verify \(|z_3|\le |z_4|\). Thus, we get \(|z_1|+|z_3|\le |z_2|+|z_4|\). The 0 could work both for modulus as well as for real and imaginary parts separately: \((a_1+0)=a_1=(0+a_1)\) and \((b_1+0)=b_1=(0+b_1)\); but also \(|z_1|+0=|z_1|=0+|z_1|\). I would say that \(|z_1|\le|z_2|\) and \(|z_2|\le|z_1|\) implies \(|z_1|=|z_2|\), and \((a_1)\le(a_2)\) and \((a_2)\le(a_1)\) implies \(a_1=a_2\); similarly for the imaginary part. Also, if \(a_1\le a_2\) and \(a_2\le a_3\), thus \(a_1\le a_3\) -- and \(b_1\le b_2\), \(b_2\le b_3\) implies \(b_1\le b_3\). Moreover, \((a_1+a_2)+a_3=a_1+(a_2+a_3)\), as well as \((b_1+b_2)+b_3=b_1+(b_2+b_3)\). This is also true for the modulus: \(|z_1|\le |z_2|\) and \(|z_2|\le |z_3|\) implies \(|z_1|\le |z_3|\), and \((|z_1|+ |z_2|)+|z_3|=|z_1|+(|z_2|+|z_3|)\). Finally, \(a_1+a_2=a_2+a_1\), \(b_1+b_2=b_2+b_1\), and \(|z_1|+|z_2|=|z_2|+|z_1|\).

    We might just work with real and imaginary parts separately. We can write \(z_1\) and \(z_2\) as \((a_1,b_1)\) and \((a_2,b_2)\), respectively. If, comparing the real part of the first number with the real part of the second number, and the imaginary part of the first number with the imaginary part of the second number, we get \(a_1 \le a_2\) and \(b_1 \le b_2\), it implies that \((a_1,b_1)\le(a_2,b_2)\). This would be graphically equivalent to considering the distance of points in the complex plane from the origin. Let \(z_3=a_3+i b_3\) and \(z_4=a_4+i b_4\) be other two complex numbers, that we can write as \((a_3,b_3)\) and \((a_4,b_4)\), respectively. If \(a_3\le a_4\) and \(b_3\le b_4\), then \((a_3,b_3)\le(a_4,b_4)\). We can compare the first pair of numbers, \(z_1,z_2\), with the second pair of numbers, \(z_3,z_4\). In particular, if \((a_1+a_3)\le(a_2+a_4)\) and \((b_1+b_3)\le(b_2+b_4)\), then \(((a_1+a_3),(b_1+b_3))\le((a_2+a_4),(b_2+b_4))\).

    Comment Source:Puzzle 71. I would like to give a try. I guess we have to focus on modulus or real and imaginary parts separately, considering, in a geometrical sense, distances from the origin on the complex plane, and numbers as tuples of reals. Let \\(z_1=a_1+i b_1\\) and \\(z_2=a_2+i b_2\\) be two complex numbers. Let us consider, for example, \\(|z_1|\le |z_2|\\). Let \\(z_3=a_3+i b_3\\) and \\(z_4=a_2+i b_4\\) be another pair of complex numbers. Let us suppose that they verify \\(|z_3|\le |z_4|\\). Thus, we get \\(|z_1|+|z_3|\le |z_2|+|z_4|\\). The 0 could work both for modulus as well as for real and imaginary parts separately: \\((a_1+0)=a_1=(0+a_1)\\) and \\((b_1+0)=b_1=(0+b_1)\\); but also \\(|z_1|+0=|z_1|=0+|z_1|\\). I would say that \\(|z_1|\le|z_2|\\) and \\(|z_2|\le|z_1|\\) implies \\(|z_1|=|z_2|\\), and \\((a_1)\le(a_2)\\) and \\((a_2)\le(a_1)\\) implies \\(a_1=a_2\\); similarly for the imaginary part. Also, if \\(a_1\le a_2\\) and \\(a_2\le a_3\\), thus \\(a_1\le a_3\\) -- and \\(b_1\le b_2\\), \\(b_2\le b_3\\) implies \\(b_1\le b_3\\). Moreover, \\((a_1+a_2)+a_3=a_1+(a_2+a_3)\\), as well as \\((b_1+b_2)+b_3=b_1+(b_2+b_3)\\). This is also true for the modulus: \\(|z_1|\le |z_2|\\) and \\(|z_2|\le |z_3|\\) implies \\(|z_1|\le |z_3|\\), and \\((|z_1|+ |z_2|)+|z_3|=|z_1|+(|z_2|+|z_3|)\\). Finally, \\(a_1+a_2=a_2+a_1\\), \\(b_1+b_2=b_2+b_1\\), and \\(|z_1|+|z_2|=|z_2|+|z_1|\\). We might just work with real and imaginary parts separately. We can write \\(z_1\\) and \\(z_2\\) as \\((a_1,b_1)\\) and \\((a_2,b_2)\\), respectively. If, comparing the real part of the first number with the real part of the second number, and the imaginary part of the first number with the imaginary part of the second number, we get \\(a_1 \le a_2\\) and \\(b_1 \le b_2\\), it implies that \\((a_1,b_1)\le(a_2,b_2)\\). This would be graphically equivalent to considering the distance of points in the complex plane from the origin. Let \\(z_3=a_3+i b_3\\) and \\(z_4=a_4+i b_4\\) be other two complex numbers, that we can write as \\((a_3,b_3)\\) and \\((a_4,b_4)\\), respectively. If \\(a_3\le a_4\\) and \\(b_3\le b_4\\), then \\((a_3,b_3)\le(a_4,b_4)\\). We can compare the first pair of numbers, \\(z_1,z_2\\), with the second pair of numbers, \\(z_3,z_4\\). In particular, if \\((a_1+a_3)\le(a_2+a_4)\\) and \\((b_1+b_3)\le(b_2+b_4)\\), then \\(((a_1+a_3),(b_1+b_3))\le((a_2+a_4),(b_2+b_4))\\).
  • 7.

    Keith - Puzzle 71 asks you to make the complex numbers into a commutative monoidal poset with its usual \(+\) and \( 0 \) and some new notion of \(\le\). It's clearly a commutative monoid. You've described how to make it into a preorder, not a poset. But we can still ask: does your recipe give a commutative monoidal preorder?

    My answer technically gives a preorder rather than a poset too :-(

    I believe I see how to do a monoidal poset, but I will hold back a bit...

    Comment Source:> Keith - Puzzle 71 asks you to make the complex numbers into a commutative monoidal poset with its usual \\(+\\) and \\( 0 \\) and some new notion of \\(\le\\). It's clearly a commutative monoid. You've described how to make it into a preorder, not a poset. But we can still ask: does your recipe give a commutative monoidal preorder? My answer technically gives a preorder rather than a poset too :-( I believe I see how to do a monoidal poset, but I will hold back a bit...
  • 8.
    edited May 14

    As a monoid, \(\langle \mathbb{C}, +, 0 \rangle\) is isomorphic to \(\langle \mathbb{R}^2, +, (0, 0) \rangle\), which has a natural partial order given by the product on \(\langle \mathbb{R}, \le \rangle\) -- in other words, we have \((x, y) \le (x', y')\) iff \(x \le x' \land y \le y'\). The monoid structure respects this order, since if \(x \le x'\) and \(r\) is any real number, we definitely have \(x + r \le x' + r\) and \(r + x \le r + x'\); this lifts to the product straightforwardly. We can carry this order across the isomorphism back to the complex numbers, giving a monoidal poset structure on \(\mathbb{C}\).

    This is much like Maria's construction in #6, but -- if I understand correctly -- we avoid identifying numbers whose components have the same magnitude, e.g. \(1 + i\) and \(-1 - i\).

    We can incorporate Matthew's ideas as true posets here: since rotations about the origin are automorphisms of the complex plane, we can similarly rotate our partial order. For instance, a 90-degree rotation would give us a poset where \(-1 + i \le -2 + 2i\). For every line of the complex plane passing through the origin, we can obtain a rotation that maps this line to the \(x\)-axis; every such line gives another induced partial order.

    Comment Source:As a monoid, \\(\langle \mathbb{C}, +, 0 \rangle\\) is isomorphic to \\(\langle \mathbb{R}^2, +, (0, 0) \rangle\\), which has a natural partial order given by the product on \\(\langle \mathbb{R}, \le \rangle\\) -- in other words, we have \\((x, y) \le (x', y')\\) iff \\(x \le x' \land y \le y'\\). The monoid structure respects this order, since if \\(x \le x'\\) and \\(r\\) is any real number, we definitely have \\(x + r \le x' + r\\) and \\(r + x \le r + x'\\); this lifts to the product straightforwardly. We can carry this order across the isomorphism back to the complex numbers, giving a monoidal poset structure on \\(\mathbb{C}\\). This is much like Maria's construction in #6, but -- if I understand correctly -- we avoid identifying numbers whose components have the same magnitude, e.g. \\(1 + i\\) and \\(-1 - i\\). We can incorporate Matthew's ideas as true posets here: since rotations about the origin are automorphisms of the complex plane, we can similarly rotate our partial order. For instance, a 90-degree rotation would give us a poset where \\(-1 + i \le -2 + 2i\\). For every line of the complex plane passing through the origin, we can obtain a rotation that maps this line to the \\(x\\)-axis; every such line gives another induced partial order.
  • 9.
    edited May 15

    Maria wrote in #6:

    Let us consider, for example, \(|z_1|\le |z_2|\).

    I am not sure this works.

    If we define \(x \preceq y \iff |x| \leq |y|\), then we have \(1/2 \preceq 1\) and \(-1 \preceq -1\), but sadly not \(1/2 + (- 1) \preceq 1 + (-1)\). This is because \(|1/2 + (- 1)| = |1/2|\) so \(0 \prec 1/2 + (- 1)\).

    Comment Source:[Maria wrote in #6](https://forum.azimuthproject.org/discussion/comment/18073/#Comment_18073): > Let us consider, for example, \\(|z_1|\le |z_2|\\). I am not sure this works. If we define \\(x \preceq y \iff |x| \leq |y|\\), then we have \\(1/2 \preceq 1\\) and \\(-1 \preceq -1\\), but sadly **not** \\(1/2 + (- 1) \preceq 1 + (-1)\\). This is because \\(|1/2 + (- 1)| = |1/2|\\) so \\(0 \prec 1/2 + (- 1)\\).
  • 10.

    My answer was going to be the same as Maria's, so now I'm confused.

    However, I do know of a really easy and boring way to make the complex numbers, \(\mathbb{C}\), into a commutative monoidal poset with the usual \(+\) and \(0\) and some concept of \(\le\).

    Take the zero function,

    $$ F(z) = 0*z $$ then all the required axioms will hold trivially.

    Comment Source:My answer was going to be the same as Maria's, so now I'm confused. However, I do know of a really easy and boring way to make the complex numbers, \\(\mathbb{C}\\), into a commutative monoidal poset with the usual \\(+\\) and \\(0\\) and some concept of \\(\le\\). Take the zero function, $$ F(z) = 0*z $$ then all the required axioms will hold trivially.
  • 11.
    edited May 15

    Keith (#10), it isn't clear from your description how the \(\le\) relation is to be defined.

    Comment Source:Keith (#10), it isn't clear from your description how the \\(\le\\) relation is to be defined.
  • 12.

    Keith, Matthew's solution is the zero function when \( p = 0+0i \)

    Comment Source:Keith, Matthew's solution is the zero function when \\( p = 0+0i \\)
  • 13.
    edited May 15

    Jonathan - Keith's proposed concept of \(\le\) in #4 seemed clear to me: given two points in the complex we say \(z \le z'\) if \(z\) is closer to the origin than \(z'\), i.e.

    $$ |z| \le |z'| .$$ This defines a preorder, not a partial order. So my next question was: does this make \((\mathbb{C}, +, 0) \) into a monoidal preorder?

    (I know the answer; I'm just seeing if he checked the compatibility of this concept of \(\le\) with addition.)

    Comment Source:Jonathan - Keith's proposed concept of \\(\le\\) in #4 seemed clear to me: given two points in the complex we say \\(z \le z'\\) if \\(z\\) is closer to the origin than \\(z'\\), i.e. \[ |z| \le |z'| .\] This defines a preorder, not a partial order. So my next question was: does this make \\((\mathbb{C}, +, 0) \\) into a monoidal preorder? (I know the answer; I'm just seeing if he checked the compatibility of this concept of \\(\le\\) with addition.)
  • 14.
    edited May 15

    John, I'm referring to #10, where \(F(z) = 0 * z\) is defined. This would suggest \(z \le z'\) iff \(F(z) \le F(z')\), which is the codiscrete preorder; but this connective step wasn't obvious.

    Comment Source:John, I'm referring to #10, where \\(F(z) = 0 * z\\) is defined. This would suggest \\(z \le z'\\) iff \\(F(z) \le F(z')\\), which is the codiscrete preorder; but this connective step wasn't obvious.
  • 15.

    Jonathan - oh, okay. Never mind. I'm still trying to get Keith to provide details on his earlier proposal.

    Comment Source:Jonathan - oh, okay. Never mind. I'm still trying to get Keith to provide details on his earlier proposal.
  • 16.
    edited May 15

    Hmm, so it is.

    I still don't understand why \( \{ \mathbb{C},+,0, \le, ||z|| \} \) doesn't count as a commutative monoidal poset.

    To me, and I'm sorry if I'm skipping ahead, the modulus is a homomorphism from \( \{ \mathbb{C},+,0, \le, ||z|| \} \) to \( \{ \mathbb{R}_+,+,0, \le, r \} \), ie,

    $$ ||z|| :: \mathbb{C} \rightarrow \mathbb{R}_+ . $$

    Comment Source:Hmm, so it is. I still don't understand why \\( \\{ \mathbb{C},+,0, \le, ||z|| \\} \\) doesn't count as a commutative monoidal poset. To me, and I'm sorry if I'm skipping ahead, the modulus is a homomorphism from \\( \\{ \mathbb{C},+,0, \le, ||z|| \\} \\) to \\( \\{ \mathbb{R}_+,+,0, \le, r \\} \\), ie, $$ ||z|| :: \mathbb{C} \rightarrow \mathbb{R}_+ . $$
  • 17.
    edited May 15

    Keith, that's because according to the measures given in #4 and #10, \(1 + 1i\) and \(1 - 1i\) are cyclically related (\(x \le y\) and \(y \le x\)), but are not equal to each other. The defining characteristic of a partial order, as opposed to a preorder, is that if \(x\) and \(y\) are cyclically related in this way, then \(x = y\).

    As John notes, however, these are perfectly serviceable preorders.

    Comment Source:Keith, that's because according to the measures given in #4 and #10, \\(1 + 1i\\) and \\(1 - 1i\\) are cyclically related (\\(x \le y\\) and \\(y \le x\\)), but are not equal to each other. The defining characteristic of a partial order, as opposed to a preorder, is that if \\(x\\) and \\(y\\) are cyclically related in this way, then \\(x = y\\). As John notes, however, these are perfectly serviceable _preorders_.
  • 18.

    Or, if I understand correctly, and correct me if I'm wrong, the problem here is that the concentric circles around the origin produce an equivalence class of numbers that are the same distance from 0.

    Comment Source:Or, if I understand correctly, and correct me if I'm wrong, the problem here is that the concentric circles around the origin produce an equivalence class of numbers that are the same distance from 0.
  • 19.

    That's exactly it.

    Comment Source:That's exactly it.
  • 20.

    Now that I think about it, the modulus can never form a monoidal poset.

    Using some advanced abstract nonsense, one can see the modulus forms a coequalizer, which I believe is enough to show that any two complex numbers sharing the same modulus must be in an equivalence relationship.

    Comment Source:Now that I think about it, the modulus can never form a monoidal poset. Using some advanced abstract nonsense, one can see the modulus forms a coequalizer, which I believe is enough to show that any two complex numbers sharing the same modulus must be in an equivalence relationship.
  • 21.
    edited May 15

    Keith wrote:

    I still don't understand why \( \{ \mathbb{C},+,0, \le, ||z|| \} \) doesn't count as a commutative monoidal poset.

    The main reason is that a commutative monoidal poset consists of 4 things:

    • a set
    • a binary operation on that set
    • an element of that set
    • and a binary relation on that set

    obeying some rules. You've listed 5. There's no role for that fifth item in a commutative monoidal poset.

    Comment Source:Keith wrote: > I still don't understand why \\( \\{ \mathbb{C},+,0, \le, ||z|| \\} \\) doesn't count as a commutative monoidal poset. The main reason is that a commutative monoidal poset consists of 4 things: * a set * a binary operation on that set * an element of that set * and a binary relation on that set obeying some rules. You've listed 5. There's no role for that fifth item in a commutative monoidal poset.
  • 22.
    edited May 15

    Is it correct that if \( \otimes \) is commutative, we only have to prove that:

    $$ \forall (x, x', y), x \le x' \textrm{ implies } x \otimes y \le x' \otimes y $$ By using different variables and then using commutativity, we get:

    $$ \forall (y, y', x'), y \le y' \textrm{ implies } y \otimes x' \le y' \otimes x' \textrm{ implies } x' \otimes y \le x' \otimes y' $$ And then chaining the two and using transitivity:

    $$ \forall (x, x', y, y'), x \le x' \textrm{ and } y \le y' \textrm{ implies } x \otimes y \le x' \otimes y \le x' \otimes y' $$ If so, this helps a lot with visualizing Puzzle 71. For all \( x, x', y \in \mathbb{C} \) the values used in the equation (i.e. \(x, x', x+y, x'+y)\) form a parallelogram, and \( x \le x' \textrm{ implies } x + y \le x' + y \). This is to say that opposite edges of any parallelogram must have the same "orientation" if we imagine edges pointing from the smaller to the larger value.

    Anyways, using the bullet-proof method of imagining parallelograms sliding around on the surface of a 3D plot, I think a superset of Matthew's set of solutions is

    $$ x \preceq_{p,F} y \iff F(p \cdot x) \leq F(p \cdot y) $$ for each point \(p \in \mathbb{C}\) and monotone function \(F : \mathbb{R} \to \mathbb{R} \). I think this is \(2^{2^{\aleph_0}}\) solutions.

    Comment Source:Is it correct that if \\( \otimes \\) is commutative, we only have to prove that: \[ \forall (x, x', y), x \le x' \textrm{ implies } x \otimes y \le x' \otimes y \] By using different variables and then using commutativity, we get: \[ \forall (y, y', x'), y \le y' \textrm{ implies } y \otimes x' \le y' \otimes x' \textrm{ implies } x' \otimes y \le x' \otimes y' \] And then chaining the two and using transitivity: \[ \forall (x, x', y, y'), x \le x' \textrm{ and } y \le y' \textrm{ implies } x \otimes y \le x' \otimes y \le x' \otimes y' \] If so, this helps a lot with visualizing **Puzzle 71**. For all \\( x, x', y \in \mathbb{C} \\) the values used in the equation (i.e. \\(x, x', x+y, x'+y)\\) form a parallelogram, and \\( x \le x' \textrm{ implies } x + y \le x' + y \\). This is to say that opposite edges of any parallelogram must have the same "orientation" if we imagine edges pointing from the smaller to the larger value. Anyways, using the bullet-proof method of imagining parallelograms sliding around on the surface of a 3D plot, I think a superset of Matthew's set of solutions is \[ x \preceq_{p,F} y \iff F(p \cdot x) \leq F(p \cdot y) \] for each point \\(p \in \mathbb{C}\\) and monotone function \\(F : \mathbb{R} \to \mathbb{R} \\). I think this is \\(2^{2^{\aleph_0}}\\) solutions.
  • 23.
    edited May 15

    Jonathan wrote:

    As John notes, however, these are perfectly serviceable commutative monoidal preorders.

    I didn't say that. I said:

    given two points in the complex let us say that \(z \le z'\) if \(z\) is closer to the origin than \(z'\), i.e.

    $$ |z| \le |z'| .$$ This defines a preorder, not a partial order. So my next question was: does this make \((\mathbb{C}, +, 0) \) into a monoidal preorder?

    There's something that needs to be checked here. If it works, we've got a monoidal preorder. If it doesn't, we don't.

    Comment Source:Jonathan wrote: > As John notes, however, these are perfectly serviceable commutative monoidal preorders. I didn't say that. I said: > given two points in the complex let us say that \\(z \le z'\\) if \\(z\\) is closer to the origin than \\(z'\\), i.e. > \[ |z| \le |z'| .\] > This defines a preorder, not a partial order. So my next question was: does this make \\((\mathbb{C}, +, 0) \\) into a monoidal preorder? There's something that needs to be checked here. If it works, we've got a monoidal preorder. If it doesn't, we don't.
  • 24.

    Right, I meant to say "preorder" -- which is an awfully long typo to make. I'll go back and edit.

    Comment Source:Right, I meant to say "preorder" -- which is an awfully long typo to make. I'll go back and edit.
  • 25.

    In answer to John's question posed in Comment 23: To prove that \((\mathbb{C}, + , 0)\) is a monoidal preorder, we would need to check that if \(y \leq y' \) and \(z \leq z'\) then \(y + z \leq y' + z'\). Unfortunately, this doesn't hold! For example using the "distance to the origin" definition of \(\leq \) $$1 \leq 1 \text{ and }1 \leq i \text{ but } 2 \nleq 1 + i.$$

    Comment Source:In answer to John's question posed in Comment 23: To prove that \\((\mathbb{C}, + , 0)\\) is a <b>monoidal</b> preorder, we would need to check that if \\(y \leq y' \\) and \\(z \leq z'\\) then \\(y + z \leq y' + z'\\). Unfortunately, this doesn't hold! For example using the "distance to the origin" definition of \\(\leq \\) $$1 \leq 1 \text{ and }1 \leq i \text{ but } 2 \nleq 1 + i.$$
  • 26.

    Something I'm noticing is that many of these preorders involve some sort of function \( f: \mathbb{C} \to \mathbb{R}\). Then we use the familiar ordering from \(\mathbb R\) to give us an ordering on \(\mathbb{C}\) by saying that $$ z \leq z' \text{ iff } f(z) \leq f(z').$$

    To make monoidal-ness of the ordering on \(\mathbb C\) follow from the monoidal-ness of \(\mathbb R\), we would need that \( f(z + z') = f(z) + f(z')\). In other words, \(f\) must be a homomorphism!

    I suspect that we could play this trick with other monoidal pre-orders besides \(\mathbb R\).

    Comment Source:Something I'm noticing is that many of these preorders involve some sort of function \\( f: \mathbb{C} \to \mathbb{R}\\). Then we use the familiar ordering from \\(\mathbb R\\) to give us an ordering on \\(\mathbb{C}\\) by saying that $$ z \leq z' \text{ iff } f(z) \leq f(z').$$ To make monoidal-ness of the ordering on \\(\mathbb C\\) follow from the monoidal-ness of \\(\mathbb R\\), we would need that \\( f(z + z') = f(z) + f(z')\\). In other words, \\(f\\) must be a homomorphism! I suspect that we could play this trick with other monoidal pre-orders besides \\(\mathbb R\\).
  • 27.
    edited May 15

    Sophie, that's similar to the approach I attempted with #8. Does it look right to you? I carry the order on \(\mathbb{R}^2\) back to \(\mathbb{C}\) by a monoid isomorphism, and observe that monoid automorphisms \(f : \mathbb{C} \to \mathbb{C}\) let us produce more partial orders.

    Comment Source:Sophie, that's similar to the approach I attempted with [#8](https://forum.azimuthproject.org/discussion/comment/18077/#Comment_18077). Does it look right to you? I carry the order on \\(\mathbb{R}^2\\) back to \\(\mathbb{C}\\) by a monoid isomorphism, and observe that monoid automorphisms \\(f : \mathbb{C} \to \mathbb{C}\\) let us produce more partial orders.
  • 28.
    edited May 15

    Hi Matthew, yes, your correction in #9 makes sense! But actually I did not mean \(x \preceq y \iff |x| \leq |y|\); I just meant instead a comparison between moduli only, as well as between imaginary and real parts separately. Also, we should not identify numbers that have the same magnitude, as Jonathan pointed out. If we do, we get an equivalence class of numbers, as Keith suggested. Thank you all for such an insightful exchange of ideas :)

    Comment Source:Hi Matthew, yes, your correction in [#9](https://forum.azimuthproject.org/discussion/comment/18078/#Comment_18078) makes sense! But actually I did not mean \\(x \preceq y \iff |x| \leq |y|\\); I just meant instead a comparison between moduli only, as well as between imaginary and real parts separately. Also, we should not identify numbers that have the same magnitude, as Jonathan pointed out. If we do, we get an equivalence class of numbers, as Keith suggested. Thank you all for such an insightful exchange of ideas :)
  • 29.
    edited May 15

    @Jonathan, thank you for pointing me towards your post! I had a few questions as I was reading it and was wondering if you could clarify.

    First you wrote

    in other words, we have \((x, y) \le (x', y')\) iff \(x \le x' \land y \le y'\)

    I assume you used \(\land\) to mean "and" and not "join". Is that correct?

    Second I am still unclear about what ordering you are getting on \(\mathbb R ^2 \) from product. You wrote:

    this lifts to the product straightforwardly

    Any chance you could make that explicit?

    Comment Source:@Jonathan, thank you for pointing me towards your post! I had a few questions as I was reading it and was wondering if you could clarify. First you wrote > in other words, we have \\((x, y) \le (x', y')\\) iff \\(x \le x' \land y \le y'\\) I assume you used \\(\land\\) to mean "and" and not "join". Is that correct? Second I am still unclear about what ordering you are getting on \\(\mathbb R ^2 \\) from product. You wrote: > this lifts to the product straightforwardly Any chance you could make that explicit?
  • 30.
    edited May 15

    Sophie wrote:

    Something I'm noticing is that many of these preorders involve some sort of function \( f: \mathbb{C} \to \mathbb{R}\). Then we use the familiar ordering from \(\mathbb R\) to give us an ordering on \(\mathbb{C}\) by saying that $$ z \leq z' \text{ iff } f(z) \leq f(z').$$

    To make monoidal-ness of the ordering on \(\mathbb C\) follow from the monoidal-ness of \(\mathbb R\), we would need that \( f(z + z') = f(z) + f(z')\). In other words, \(f\) must be a homomorphism!

    I suspect that we could play this trick with other monoidal pre-orders besides \(\mathbb R\).

    Excellent observations! The true mathematician wants to generalize. You've inspired me to pose these puzzles:

    Puzzle 75. Suppose \( (Y, \le_Y) \) is a preorder, \(X\) is a set and \(f : X \to Y\) is any function. Define a relation \(\le_X\) on \(X\) by

    $$ x \le_X x' \textrm{ if and only if } f(x) \le_Y f(x') .$$ Show that \( (X, \le_X ) \) is a preorder.

    Puzzle 76. Now suppose \( (Y, \le_Y) \) is a poset. Under what conditions on \(f\) can we conclude that \( (X, \le_X ) \) defined as above is a poset?

    Puzzle 77. Now suppose that \( (Y, \le_Y, \otimes_Y, 1_Y) \) is a monoidal preorder, and \( (X,\otimes_X,1_X ) \) is a monoid. Define \(\le_X\) as above. Under what conditions on \(f\) can we conclude that \( (X,\le_X\otimes_X,1_X) \) is a monoidal preorder?

    Comment Source:Sophie wrote: > Something I'm noticing is that many of these preorders involve some sort of function \\( f: \mathbb{C} \to \mathbb{R}\\). Then we use the familiar ordering from \\(\mathbb R\\) to give us an ordering on \\(\mathbb{C}\\) by saying that $$ z \leq z' \text{ iff } f(z) \leq f(z').$$ > To make monoidal-ness of the ordering on \\(\mathbb C\\) follow from the monoidal-ness of \\(\mathbb R\\), we would need that \\( f(z + z') = f(z) + f(z')\\). In other words, \\(f\\) must be a homomorphism! > I suspect that we could play this trick with other monoidal pre-orders besides \\(\mathbb R\\). Excellent observations! The true mathematician wants to generalize. You've inspired me to pose these puzzles: **Puzzle 75.** Suppose \\( (Y, \le_Y) \\) is a preorder, \\(X\\) is a set and \\(f : X \to Y\\) is any function. Define a relation \\(\le_X\\) on \\(X\\) by \[ x \le_X x' \textrm{ if and only if } f(x) \le_Y f(x') .\] Show that \\( (X, \le_X ) \\) is a preorder. **Puzzle 76.** Now suppose \\( (Y, \le_Y) \\) is a poset. Under what conditions on \\(f\\) can we conclude that \\( (X, \le_X ) \\) defined as above is a poset? **Puzzle 77.** Now suppose that \\( (Y, \le_Y, \otimes_Y, 1_Y) \\) is a monoidal preorder, and \\( (X,\otimes_X,1_X ) \\) is a monoid. Define \\(\le_X\\) as above. Under what conditions on \\(f\\) can we conclude that \\( (X,\le_X\otimes_X,1_X) \\) is a monoidal preorder?
  • 31.
    edited May 15

    In comment #1, Matthew Doty found lots of ways we can define \(\le\) for complex numbers that make \((\mathbb{C}, \le, +, 0)\) into a monoidal poset. However, there are lots of other ways!

    A bunch of these are fairly easy to find. You just need to take Matthew's idea and generalize it a little. So, I bet you folks can find them!

    There are also a lot of others, which are much harder to find. These difficult ones require the axiom of choice. In fact, using the axiom of choice, you can even find nonstandard ways to define \(\le\) for real numbers ways that give a monoidal poset \( (\mathbb{R}, \le, +, 0)\)!

    Comment Source:In [comment #1](https://forum.azimuthproject.org/discussion/comment/18065/#Comment_18065), Matthew Doty found lots of ways we can define \\(\le\\) for complex numbers that make \\((\mathbb{C}, \le, +, 0)\\) into a monoidal poset. However, there are lots of _other_ ways! A bunch of these are fairly easy to find. You just need to take Matthew's idea and generalize it a little. So, I bet you folks can find them! There are also a lot of _others_, which are much harder to find. These difficult ones require the axiom of choice. In fact, using the axiom of choice, you can even find nonstandard ways to define \\(\le\\) for _real_ numbers ways that give a monoidal poset \\( (\mathbb{R}, \le, +, 0)\\)!
  • 32.

    Alex wrote:

    Anyways, using the bullet-proof method of imagining parallelograms sliding around on the surface of a 3D plot, I think a superset of Matthew's set of solutions is

    $$ x \preceq_{p,F} y \iff F(p \cdot x) \leq F(p \cdot y) $$ for each point \(p \in \mathbb{C}\) and monotone function \(F : \mathbb{R} \to \mathbb{R} \). I think this is \(2^{2^{\aleph_0}}\) solutions.

    How does the function \(F\) affect \(\preceq_{p,F}\)? For example, \(F(x) = 2x\) gives the same partial order as \(F(x) = e^x \).

    Comment Source:Alex wrote: > Anyways, using the bullet-proof method of imagining parallelograms sliding around on the surface of a 3D plot, I think a superset of Matthew's set of solutions is > \[ x \preceq_{p,F} y \iff F(p \cdot x) \leq F(p \cdot y) \] > for each point \\(p \in \mathbb{C}\\) and monotone function \\(F : \mathbb{R} \to \mathbb{R} \\). I think this is \\(2^{2^{\aleph_0}}\\) solutions. How does the function \\(F\\) affect \\(\preceq_{p,F}\\)? For example, \\(F(x) = 2x\\) gives the same partial order as \\(F(x) = e^x \\).
  • 33.
    edited May 15

    A bunch of these are fairly easy to find. You just need to take Matthew's idea and generalize it a little. So, I bet you folks can find them!

    I mentioned yesterday that I had in mind a proper poset for an answer, rather than a preorder. Jonathan gave a poset - the product order of \(\langle R, \leq \rangle\) with itself.

    Another monoidal poset that works is the lexicographic order of \(\langle R, \leq \rangle\) with itself.

    $$ x + i y \leq_{lex} p + i q \iff x < p \text{ or } (x = p \text{ and } y \leq_{\mathbb{R}} q) $$ We can again get \(2^{\aleph_0}\) answers from this by just observing a different trick - this answer works even if we translate the complex plane around:

    $$ a \leq_{lex, k} b \iff a + k \leq_{lex} b + k $$ I have a little hunch this might be part of what John has in mind...

    Comment Source:> A bunch of these are fairly easy to find. You just need to take Matthew's idea and generalize it a little. So, I bet you folks can find them! I mentioned yesterday that I had in mind a proper poset for an answer, rather than a preorder. Jonathan gave a poset - the [product order](https://en.wikipedia.org/wiki/Product_order) of \\(\langle R, \leq \rangle\\) with itself. Another monoidal poset that works is the [lexicographic order](https://en.wikipedia.org/wiki/Lexicographical_order) of \\(\langle R, \leq \rangle\\) with itself. $$ x + i y \leq_{lex} p + i q \iff x < p \text{ or } (x = p \text{ and } y \leq_{\mathbb{R}} q) $$ We can again get \\(2^{\aleph_0}\\) answers from this by just observing a different trick - this answer works even if we *translate* the complex plane around: $$ a \leq_{lex, k} b \iff a + k \leq_{lex} b + k $$ I have a little hunch this might be part of what John has in mind...
  • 34.
    edited May 15

    Matthew - whoops, a bunch of my easy-to-find monoidal poset structures on \( (\mathbb{C}, +, 0 ) \) were actually monoidal preorders. It sounds like maybe you know these.

    By the way, any monoidal preorder on a group (like the complex numbers with \(+\) as its monoid operation) obeys

    $$ a \le b \iff a + k \le b + k .$$ So I don't think your "translated lexicographic orders" are different from the ordinary lexicographic order. But I seem to be making mistakes...

    Comment Source:Matthew - whoops, a bunch of my easy-to-find monoidal poset structures on \\( (\mathbb{C}, +, 0 ) \\) were actually monoidal preorders. It sounds like maybe you know these. By the way, any monoidal preorder on a group (like the complex numbers with \\(+\\) as its monoid operation) obeys \[ a \le b \iff a + k \le b + k .\] So I don't think your "translated lexicographic orders" are different from the ordinary lexicographic order. But I seem to be making mistakes...
  • 35.
    edited May 15

    Sophie wrote:

    in other words, we have \((x, y) \le (x', y')\) iff \(x \le x' \land y \le y'\)

    I assume you used \(\land\) to mean "and" and not "join". Is that correct?

    I can definitely see how that's ambiguous! I did mean "and". (Though technically, "and" is a join on some lattice of propositions, so it's not an incorrect reading, as long as the right parentheses are inferred.)

    (EDIT: Also, lest I fall back into bad habits, I think \(\wedge\) is the symbol for “meet”, the greatest lower bound. I do find it more natural sometimes to think of “true” as the informationless bottom element though.)

    Second I am still unclear about what ordering you are getting on \(\mathbb R ^2 \) from product. You wrote:

    this lifts to the product straightforwardly

    Any chance you could make that explicit?

    Sure thing. Generally when working with products, things "lift" componentwise. So we say \((x, y) \le (x', y')\) iff both \(x \le x'\) and \(y \le y'\). The lifting I'm referring to in the quote is that the fact that \(+\) respects the order on \(R\) lifts in the “usual” sense to how the component-wise \(+\) respects the order on \(R^2\). Explicitly, if \((x, y) \le (x', y')\) and we have some \((r, s) \in \mathbb{R}^2\), we know that \((x, y) + (r, s) \le (x', y') + (r, s)\) because \(x + r \le x' + r\) and \(y + s \le y' + s\) individually.

    Comment Source:[Sophie wrote](https://forum.azimuthproject.org/discussion/comment/18107/#Comment_18107): > > in other words, we have \\((x, y) \le (x', y')\\) iff \\(x \le x' \land y \le y'\\) > > I assume you used \\(\land\\) to mean "and" and not "join". Is that correct? I can definitely see how that's ambiguous! I did mean "and". (Though technically, "and" is a join on some lattice of propositions, so it's not an _incorrect_ reading, as long as the right parentheses are inferred.) (EDIT: Also, lest I fall back into bad habits, I think \\(\wedge\\) is the symbol for “meet”, the greatest lower bound. I do find it more natural sometimes to think of “true” as the informationless bottom element though.) > Second I am still unclear about what ordering you are getting on \\(\mathbb R ^2 \\) from product. You wrote: > > this lifts to the product straightforwardly > > Any chance you could make that explicit? Sure thing. Generally when working with products, things "lift" componentwise. So we say \\((x, y) \le (x', y')\\) iff both \\(x \le x'\\) and \\(y \le y'\\). The lifting I'm referring to in the quote is that the fact that \\(+\\) respects the order on \\(R\\) lifts in the “usual” sense to how the component-wise \\(+\\) respects the order on \\(R^2\\). Explicitly, if \\((x, y) \le (x', y')\\) and we have some \\((r, s) \in \mathbb{R}^2\\), we know that \\((x, y) + (r, s) \le (x', y') + (r, s)\\) because \\(x + r \le x' + r\\) and \\(y + s \le y' + s\\) individually.
  • 36.
    edited May 15

    So I don't think your "translated lexicographic orders" are different from the ordinary lexicographic order. But I seem to be making mistakes...

    Oops, thanks for the correction.

    I was thinking about translation because I was noticing that if \(\leq\) is a monoidal poset then for every affine transformation \(A\) there corresponds a monoidal preorder \(\preceq_A\), where

    $$ x \preceq_A y \iff A(x) \leq A(y) $$ I was thinking of this because affine transforms preserve addition and little else. Since affine transformations can all be expressed as \(A(x) = \mathbf{a} x + \mathbf{b}\), I was erroneously thinking that translation would lead to distinct posets/preorders.

    An affine transformation without a translation term is just a linear map.

    So now I am wondering: if I have two isomorphic monoidal posets \(\langle \mathbb{C}, +, 0, \leq_1\rangle\) and \(\langle \mathbb{C}, +, 0, \leq_2\rangle\), then is their isomorphism function \(\phi : \mathbb{C} \to \mathbb{C}\) a linear map?

    Comment Source:> So I don't think your "translated lexicographic orders" are different from the ordinary lexicographic order. But I seem to be making mistakes... Oops, thanks for the correction. I was thinking about translation because I was noticing that if \\(\leq\\) is a monoidal poset then for every [affine transformation](https://en.wikipedia.org/wiki/Affine_transformation) \\(A\\) there corresponds a monoidal preorder \\(\preceq_A\\), where $$ x \preceq_A y \iff A(x) \leq A(y) $$ I was thinking of this because affine transforms preserve addition and little else. Since affine transformations can all be expressed as \\(A(x) = \mathbf{a} x + \mathbf{b}\\), I was erroneously thinking that translation would lead to distinct posets/preorders. An affine transformation without a translation term is just a [linear map](https://en.wikipedia.org/wiki/Linear_map). So now I am wondering: if I have two isomorphic monoidal posets \\(\langle \mathbb{C}, +, 0, \leq_1\\rangle\\) and \\(\langle \mathbb{C}, +, 0, \leq_2\\rangle\\), then is their isomorphism function \\(\phi : \mathbb{C} \to \mathbb{C}\\) a linear map?
  • 37.

    Puzzle 73

    I claimed that every finite subset of a poset have a join if and only if every pair of elements has a join and the empty set has a join. Why is this true?

    The empty set has a join because that is stipulated in the requirement. For any non-empty subset \( A \in (P,\le) \) we must show that for \( a,b,c,...,z \in A \), \( \bigvee A = \bigvee ( a,b,c...,z ) = a \vee b \vee c \vee d \vee ... \vee y \vee z = (a \vee b) \vee (c \vee d) \vee ... \vee (y \vee z) \), i.e that we can split the join into seperate pairs, for which the joins exist. To do this we need to show that we can freely set parentheses, which is equivalent to showing that joins are associative. Therefore, we want to prove that \( ( a \vee b) \vee c = a \vee (b \vee c) \)

    A join is the least upper bound. Since it is an upper bound \( a \le a \vee b \le (a \vee b) \vee c\) and \( c \le b \vee c \le a \vee (b \vee c)\).

    Since it is the least upper bound \( c \le a \vee (b \vee c) \) implies \( (a \vee b) \vee c \le a \vee (b \vee c) \). Likewise, \( a \le (a \vee b) \vee c \) implies \( a \vee (b \vee c) \le (a \vee b) \vee c \).

    By proving the inequality from both sides we have shown that \( (a \vee b) \vee c = a \vee (b \vee c) \) as desired \( \square \)

    The proof that meets are associative follows by duality.

    Comment Source:**Puzzle 73** > I claimed that every finite subset of a poset have a join if and only if every pair of elements has a join and the empty set has a join. Why is this true? The empty set has a join because that is stipulated in the requirement. For any non-empty subset \\( A \in (P,\le) \\) we must show that for \\( a,b,c,...,z \in A \\), \\( \bigvee A = \bigvee ( a,b,c...,z ) = a \vee b \vee c \vee d \vee ... \vee y \vee z = (a \vee b) \vee (c \vee d) \vee ... \vee (y \vee z) \\), i.e that we can split the join into seperate pairs, for which the joins exist. To do this we need to show that we can freely set parentheses, which is equivalent to showing that joins are associative. Therefore, we want to prove that \\( ( a \vee b) \vee c = a \vee (b \vee c) \\) A join is the least upper bound. Since it is an upper bound \\( a \le a \vee b \le (a \vee b) \vee c\\) and \\( c \le b \vee c \le a \vee (b \vee c)\\). Since it is the least upper bound \\( c \le a \vee (b \vee c) \\) implies \\( (a \vee b) \vee c \le a \vee (b \vee c) \\). Likewise, \\( a \le (a \vee b) \vee c \\) implies \\( a \vee (b \vee c) \le (a \vee b) \vee c \\). By proving the inequality from both sides we have shown that \\( (a \vee b) \vee c = a \vee (b \vee c) \\) as desired \\( \square \\) The proof that meets are associative follows by duality.
  • 38.

    Puzzle 75.

    Suppose \( (Y, \le_Y) \) is a preorder, \(X\) is a set and \(f : X \to Y\) is any function. Define a relation \(\le_X\) on \(X\) by

    $$ x \le_X x' \textrm{ if and only if } f(x) \le_Y f(x') .$$ Show that \( (X, \le_X ) \) is a preorder.

    Our relation turns \( f \) into a monotone map. For all \( x \in X \) we have \( f(x) \le_Y f(x)\) and thus \( x \le_X x\) satisfying reflexivity. Similarly, for all \( x,y,z \in X \), \( f(x) \le_Y f(y)\) and \( f(y) \le_Y f(z)\) implies \( f(x) \le_Y f(z)\) and thus \( x \le_X y\) and \( y \le_X z\) implies \( x \le_X z\) satisfying transitivity. This gives us a preorder on X.

    Puzzle 76.

    Now suppose \( (Y, \le_Y) \) is a poset. Under what conditions on \(f\) can we conclude that \( (X, \le_X ) \) defined as above is a poset?

    Since we don't want to induce any equivalent elements in \(X\), \( f \) must be injective.

    Puzzle 77.

    Now suppose that \( (Y, \le_Y, \otimes_Y, 1_Y) \) is a monoidal preorder, and \( (X,\otimes_X,1_X ) \) is a monoid. Define \(\le_X\) as above. Under what conditions on \(f\) can we conclude that \( (X,\le_X\otimes_X,1_X) \) is a monoidal preorder?

    We need to assure that our induced preorder structure is compatible with our monoidal structure. To this end we require our monotone map \( f \) to be a monoidal monotone for which \( 1_Y \le_Y f(1_X) \) and \( f(x) \otimes_Y f(y) \le_Y f(x \otimes_X y) \)

    Regarding Puzzle 71, does this mean we simply need to find injective monoidal monotones to other commutative monoidal posets (e.g \( (\mathbb{R}, \le, +, 0 )\)) or do we need stricter requirements to preserve the commutative sturcture (e.g \( 1_Y = f(1_X) \) and \( f(x) \otimes_Y f(y) = f(x \otimes_X y) \))?

    I'm off to bed, so maybe someone else can continue my thought process...

    Comment Source:**Puzzle 75.** >Suppose \\( (Y, \le_Y) \\) is a preorder, \\(X\\) is a set and \\(f : X \to Y\\) is any function. Define a relation \\(\le_X\\) on \\(X\\) by >\[ x \le_X x' \textrm{ if and only if } f(x) \le_Y f(x') .\] >Show that \\( (X, \le_X ) \\) is a preorder. Our relation turns \\( f \\) into a *monotone map*. For all \\( x \in X \\) we have \\( f(x) \le_Y f(x)\\) and thus \\( x \le_X x\\) satisfying reflexivity. Similarly, for all \\( x,y,z \in X \\), \\( f(x) \le_Y f(y)\\) and \\( f(y) \le_Y f(z)\\) implies \\( f(x) \le_Y f(z)\\) and thus \\( x \le_X y\\) and \\( y \le_X z\\) implies \\( x \le_X z\\) satisfying transitivity. This gives us a preorder on X. **Puzzle 76.** >Now suppose \\( (Y, \le_Y) \\) is a poset. Under what conditions on \\(f\\) can we conclude that \\( (X, \le_X ) \\) defined as above is a poset? Since we don't want to induce any equivalent elements in \\(X\\), \\( f \\) must be injective. **Puzzle 77.** >Now suppose that \\( (Y, \le_Y, \otimes_Y, 1_Y) \\) is a monoidal preorder, and \\( (X,\otimes_X,1_X ) \\) is a monoid. Define \\(\le_X\\) as above. Under what conditions on \\(f\\) can we conclude that \\( (X,\le_X\otimes_X,1_X) \\) is a monoidal preorder? We need to assure that our induced preorder structure is compatible with our monoidal structure. To this end we require our *monotone map* \\( f \\) to be a *monoidal monotone* for which \\( 1_Y \le_Y f(1_X) \\) and \\( f(x) \otimes_Y f(y) \le_Y f(x \otimes_X y) \\) Regarding Puzzle 71, does this mean we simply need to find injective *monoidal monotones* to other commutative monoidal posets (e.g \\( (\mathbb{R}, \le, +, 0 )\\)) or do we need stricter requirements to preserve the commutative sturcture (e.g \\( 1_Y = f(1_X) \\) and \\( f(x) \otimes_Y f(y) = f(x \otimes_X y) \\))? I'm off to bed, so maybe someone else can continue my thought process...
  • 39.

    Hi Marius, thanks for your answers to puzzles 75/76/77! Since John rewrote these puzzles in this post I copied your answers to that thread as well. Hope that's okay with you!

    Comment Source:Hi Marius, thanks for your answers to puzzles 75/76/77! Since John rewrote these puzzles in <a href = "https://forum.azimuthproject.org/discussion/2089/no-lecture-today-but-puzzles"> this post </a> I <a href = "https://forum.azimuthproject.org/discussion/comment/18138/#Comment_18138">copied your answers</a> to that thread as well. Hope that's okay with you!
  • 40.

    Thanks for the answers, Marius! I'll discuss them over here, where Sophie copied them. I have some qualms about your answer to Puzzle 77.

    Comment Source:Thanks for the answers, Marius! I'll discuss them [over here, where Sophie copied them](https://forum.azimuthproject.org/discussion/comment/18138/#Comment_18138). I have some qualms about your answer to Puzzle 77.
  • 41.
    edited May 23

    Puzzle 71 I think the problem comes down to whether there is a mapping from $$ \mathbb {C} \rightarrow \mathbb {R} \times \mathbb {R} \rightarrow \mathbb {R} $$ Clearly \( \mathbb {R} \) forms a nice poset. One nice poset is the total order.

    I know how to do this with... $$ \mathbb {Q} \rightarrow \mathbb {Z} \times \mathbb {Z} \rightarrow \mathbb {Z} $$ Here is an idea, interleave the digits of imaginary and real parts of complex number to form a real number. There are as many interleavings as there are bases with is \(\mathbb{N}\).

    e.g. $$ (25.647\dots, 0.023\dots) \rightarrow 2050.604273\dots $$ Another set of ways would be to perform \(\le\) in two stages. First look at the \(\mathfrak{Re}\) as @MatthewDoty suggested and if that is not strictly less then, and only then, compare \(\mathfrak{Img}\). This approach can be extended to @KeithEPeterson approach, first evaluating the distance from the origin and then comparing the angle, effectively making a spiral.

    These ideas involve producing a total order over \( \mathbb{C} \) but there are still lots of posets that are not total orders. The approaches that produce a preorder really only have a problem with the \( = \) bit of \( \le \). If we do not require every element to be comparable, not a total order, then we are free to adapt @MatthewDoty approach by allowing \( = \) only if the elements are identical. e.g. \( (2.3 + 1.1i) \le (3.7 - 10.1i) \) but there is no relation between \( (2.3 + 1.1i) \text{ and } (2.3 - 10.1i) \).

    Do these approaches work?

    Comment Source:**Puzzle 71** I think the problem comes down to whether there is a mapping from $$ \mathbb {C} \rightarrow \mathbb {R} \times \mathbb {R} \rightarrow \mathbb {R} $$ Clearly \\( \mathbb {R} \\) forms a nice poset. One nice poset is the total order. I know how to do this with... $$ \mathbb {Q} \rightarrow \mathbb {Z} \times \mathbb {Z} \rightarrow \mathbb {Z} $$ Here is an idea, interleave the digits of imaginary and real parts of complex number to form a real number. There are as many interleavings as there are bases with is \\(\mathbb{N}\\). e.g. $$ (25.647\dots, 0.023\dots) \rightarrow 2050.604273\dots $$ Another set of ways would be to perform \\(\le\\) in two stages. First look at the \\(\mathfrak{Re}\\) as @MatthewDoty suggested and if that is not strictly less then, and only then, compare \\(\mathfrak{Img}\\). This approach can be extended to @KeithEPeterson approach, first evaluating the distance from the origin and then comparing the angle, effectively making a spiral. These ideas involve producing a total order over \\( \mathbb{C} \\) but there are still lots of posets that are not total orders. The approaches that produce a preorder really only have a problem with the \\( = \\) bit of \\( \le \\). If we do not require every element to be comparable, not a total order, then we are free to adapt @MatthewDoty approach by allowing \\( = \\) only if the elements are identical. e.g. \\( (2.3 + 1.1i) \le (3.7 - 10.1i) \\) but there is no relation between \\( (2.3 + 1.1i) \text{ and } (2.3 - 10.1i) \\). Do these approaches work?
  • 42.
    edited July 1

    Typos: more and more good are being reduced to data -- goods; How can you can sell -- repeated can.

    Comment Source:Typos: more and more good are being reduced to data -- goods; How can you can sell -- repeated can.
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