It looks like you're new here. If you want to get involved, click one of these buttons!

- All Categories 2.3K
- Chat 499
- Study Groups 18
- Petri Nets 9
- Epidemiology 3
- Leaf Modeling 1
- Review Sections 9
- MIT 2020: Programming with Categories 51
- MIT 2020: Lectures 20
- MIT 2020: Exercises 25
- MIT 2019: Applied Category Theory 339
- MIT 2019: Lectures 79
- MIT 2019: Exercises 149
- MIT 2019: Chat 50
- UCR ACT Seminar 4
- General 67
- Azimuth Code Project 110
- Statistical methods 3
- Drafts 2
- Math Syntax Demos 15
- Wiki - Latest Changes 3
- Strategy 113
- Azimuth Project 1.1K
- - Spam 1
- News and Information 147
- Azimuth Blog 149
- - Conventions and Policies 21
- - Questions 43
- Azimuth Wiki 708

Options

What is the result of joining the following two systems?

(This question only makes sense if you read the book!)

## Comments

We would get the partition {{11, 12, 22, 23, 13}, {21}}.

`We would get the partition {{11, 12, 22, 23, 13}, {21}}.`

Sorry for the quality.

`![12](https://image.ibb.co/iXi4h7/12.png) Sorry for the quality.`

here is my entry, shows construction of the solution (visually)

`here is my entry, shows construction of the solution (visually) <a href="http://ibb.co/i18Uzn"><img src="http://preview.ibb.co/nGmGen/book_chapter1_exersize02_sm.jpg" alt="book_chapter1_exersize02_sm" border="0"></a>`

Great! These pictures depict

partitionsof the set {1,2,3,4,5,6}: that is, ways of writing it as a disjoint union of nonempty subsets calledblocks.The collection of all partitions of a set is partially ordered by 'refinement': we say one partition \(P\) is

finerthan another partition \(P'\) if every block of \(P'\) is a union of blocks of \(P\). (One could say "finer than or equal to", but people say just "finer".) We also say, in this situation, that \(P'\) iscoarserthan \(P\).We write \(P \le P'\) if \(P\) is finer than \(P'\).

Here's a picture of all 15 partitions of a 4-element set, with finer ones nearer the bottom, and edges pointing up from each partition to the next coarser ones. This is called a Hasse diagram:

The picture was drawn by Tilman Piesk on Wikicommons. Click on it for a lot more information.

For any two partitions \(P,P'\) of a set \(S\), their

meet\(P \vee P'\) is the finest partition that is coarser than both. That's what this problem is about.`Great! These pictures depict **[partitions](https://en.wikipedia.org/wiki/Partition_of_a_set)** of the set {1,2,3,4,5,6}: that is, ways of writing it as a disjoint union of nonempty subsets called **blocks**. The collection of all partitions of a set is partially ordered by 'refinement': we say one partition \\(P\\) is **finer** than another partition \\(P'\\) if every block of \\(P'\\) is a union of blocks of \\(P\\). (One could say "finer than or equal to", but people say just "finer".) We also say, in this situation, that \\(P'\\) is **coarser** than \\(P\\). We write \\(P \le P'\\) if \\(P\\) is finer than \\(P'\\). Here's a picture of all 15 partitions of a 4-element set, with finer ones nearer the bottom, and edges pointing up from each partition to the next coarser ones. This is called a [Hasse diagram](https://en.wikipedia.org/wiki/Hasse_diagram): <a href = "https://blogs.ams.org/visualinsight/2015/06/15/lattice-of-partitions/"> <img width = "450" src = "https://blogs.ams.org/visualinsight/files/2015/06/lattice_of_partitions.png"> </a> The picture was drawn by [Tilman Piesk](https://commons.wikimedia.org/wiki/File:Set_partitions_4;_Hasse;_circles.svg) on Wikicommons. Click on it for a lot more information. For any two partitions \\(P,P'\\) of a set \\(S\\), their **meet** \\(P \vee P'\\) is the finest partition that is coarser than both. That's what this problem is about.`

That’s a great picture John, thanks for posting it! I started doing this exercise on my own and understood how to construct the diagram. However, I found it difficult to anticipate exactly how to best lay out the elements to make the structure clear. Are there any techniques available for doing this that are better than experimentation?

`That’s a great picture John, thanks for posting it! I started doing this exercise on my own and understood how to construct the diagram. However, I found it difficult to anticipate exactly how to best lay out the elements to make the structure clear. Are there any techniques available for doing this that are better than experimentation?`

Eye crossing works to get a quick view of the union.

`Eye crossing works to get a quick view of the union.`

John #4: After seeing your separate comment that you sometimes confuse meet and join (as in John #4 above) because their standard English meanings are similar, I began confusing them too:-)

I'm trying out the mnemonic that the M in Meet contains two \(\land\)s, and \(\land\) points upward to the subset elements; while the J in Join is vaguely \(\vee\)-shaped, and \(\vee\) points downward.

`John #4: After seeing your separate comment that you sometimes confuse meet and join (as in John #4 above) because their standard English meanings are similar, I began confusing them too:-) I'm trying out the mnemonic that the M in Meet contains two \\(\land\\)s, and \\(\land\\) points upward to the subset elements; while the J in Join is vaguely \\(\vee\\)-shaped, and \\(\vee\\) points downward.`

I think of meet as what is common to the two -- the point at which they meet -- so that's the intersection. Joining implies combining them to make something bigger, which is the union.

`I think of meet as what is common to the two -- the point at which they meet -- so that's the intersection. Joining implies combining them to make something bigger, which is the union.`

At first I was overthinking the solution and came up with this:

But after reading the definition of the \/ operator again, realized that since there is a path from 12 to 13 namely 12, 22, 23, 13 that there would, by definition, be a connection between the two.

`At first I was overthinking the solution and came up with this: <img src="http://image.ibb.co/bXKYhH/ch1_ex2_soln.png"/> But after reading the definition of the \/ operator again, realized that since there is a path from 12 to 13 namely 12, 22, 23, 13 that there would, by definition, be a connection between the two.`

My revised mnemonics are:

`My revised mnemonics are: > "meet up (^) and glob intersection" - MUAGI > "join down (v) or lub union" - JoDOLU`

Jerry wrote:

See what a great teacher I am? I can not only transmit my knowledge, I can also transmit my confusion!

David wrote:

Yes, this feels like the original motivation of those terms. It should help me keep them straight.

`[Jerry wrote](https://forum.azimuthproject.org/discussion/comment/16625/#Comment_16625): > After seeing your separate comment that you sometimes confuse meet and join (as in John #4 above) because their standard English meanings are similar, I began confusing them too:-) See what a great teacher I am? I can not only transmit my knowledge, I can also transmit my confusion! [David wrote](https://forum.azimuthproject.org/discussion/comment/16638/#Comment_16638): > I think of meet as what is common to the two -- the point at which they meet -- so that's the intersection. Joining implies combining them to make something bigger, which is the union. Yes, this feels like the original motivation of those terms. It should help me keep them straight.`

Matthew Johnson wrote:

I don't know! But when I was a grad student I had a roommate who worked on VLSI design - the design of very large integrated circuits. Part of this amounts to taking a graph and trying to draw it in the plane with the fewest crossing. He developed algorithms for doing this sort of thing. Perhaps some of these could be used for drawing pictures like this... but those algorithms weren't developed for visual beauty, so I don't know if they'd create nice pictures. So, I think people just try stuff and see what works.

`[Matthew Johnson wrote](https://forum.azimuthproject.org/discussion/comment/16597/#Comment_16597): > That’s a great picture John, thanks for posting it! I started doing this exercise on my own and understood how to construct the diagram. However, I found it difficult to anticipate exactly how to best lay out the elements to make the structure clear. Are there any techniques available for doing this that are better than experimentation? I don't know! But when I was a grad student I had a roommate who worked on [VLSI](https://en.wikipedia.org/wiki/Very-large-scale_integration) design - the design of very large integrated circuits. Part of this amounts to taking a graph and trying to draw it in the plane with the fewest crossing. He developed algorithms for doing this sort of thing. Perhaps some of these could be used for drawing pictures like this... but those algorithms weren't developed for visual beauty, so I don't know if they'd create nice pictures. So, I think people just try stuff and see what works.`

I'm a bit late to the party! The resulting system would be split into two partitions, {11, 12, 13, 22, 23} and {21}.

`I'm a bit late to the party! The resulting system would be split into two partitions, {11, 12, 13, 22, 23} and {21}.`

By definition points \(x\) and \(y\) are connected if there are points \(z_{1}...z_{n}\) where \(x\) is connected to \(z_{1}\), \(z_{i}\) is connected to \(z_{i+1}\), and \(z_{i+1}\) is connected to \(y\). Applying this to the problem: \(11\) and \(12\) are connected in first set \(12\) and \(22\) are connected in second set \(\implies\) \(11\) and \(22\) are connected

\(22\) and \(23\) are connected in first set \(23\) and \(13\) are connected in second set \(\implies\) \(13\) and \(22\) are connected

\(21\) is isolated in both sets

Hence the join operation will result in two partitions, which are {11,12,13,22,23} and {21}

`By definition points \\(x\\) and \\(y\\) are connected if there are points \\(z_{1}...z_{n}\\) where \\(x\\) is connected to \\(z_{1}\\), \\(z_{i}\\) is connected to \\(z_{i+1}\\), and \\(z_{i+1}\\) is connected to \\(y\\). Applying this to the problem: \\(11\\) and \\(12\\) are connected in first set \\(12\\) and \\(22\\) are connected in second set \\(\implies\\) \\(11\\) and \\(22\\) are connected \\(22\\) and \\(23\\) are connected in first set \\(23\\) and \\(13\\) are connected in second set \\(\implies\\) \\(13\\) and \\(22\\) are connected \\(21\\) is isolated in both sets Hence the join operation will result in two partitions, which are {11,12,13,22,23} and {21}`

I'm having trouble understanding the definition of the join operation. Near the bottom of page 3 of the book, there is an example showing the join of system A with three "unconnected" elements and system B with two connected elements and an unconnected element, *. The result of joining A with B is given as a two-element system with two the connected elements and one individual element, the *.

A join is defined as: "That is, we shall say the join of systems A and B, denote it A ∨ B, has a connection between points x and y if there are some points z1, . . . , zn such that, in at least one of A or B, it is true that x is connected to z1, zi is connected to zi+1, and zn is connected to y."

So why is * in the resulting joined system?

My working hypothesis is that in order to satisfy the definition of the join, we must assume that * is joined to itself, i.e., in the definition, the point x and the point y are the same point, namely *. In other words, the connection is reflexive in addition to being symmetric and transitive.

Since nobody seems to be having any problems with this, I suspect my hypothesis is unnecessary and I am overlooking some obvious, simple answer to my problem. Maybe I'm misinterpreting the definition of a join.

`I'm having trouble understanding the definition of the join operation. Near the bottom of page 3 of the book, there is an example showing the join of system A with three "unconnected" elements and system B with two connected elements and an unconnected element, *. The result of joining A with B is given as a two-element system with two the connected elements and one individual element, the *. A join is defined as: "That is, we shall say the join of systems A and B, denote it A ∨ B, has a connection between points x and y if there are some points z1, . . . , zn such that, in at least one of A or B, it is true that x is connected to z1, zi is connected to zi+1, and zn is connected to y." So why is * in the resulting joined system? My working hypothesis is that in order to satisfy the definition of the join, we must assume that * is joined to itself, i.e., in the definition, the point x and the point y are the same point, namely *. In other words, the connection is reflexive in addition to being symmetric and transitive. Since nobody seems to be having any problems with this, I suspect my hypothesis is unnecessary and I am overlooking some obvious, simple answer to my problem. Maybe I'm misinterpreting the definition of a join.`

\(\ast\) is

notconnected in the joined system to either \(\circ\) or \(\bullet\).Each system has 3 elements and some number of connections. \(A\) has 3 elements and 0 connections (each element is unconnected from each other element). \(B\) has 3 elements and 1 connection (\(\bullet\) and \(\circ\) are connected, \(\ast\) is unconnected from them). \(C = A \vee B\) also has 3 elements, and 1 connection (the same as \(B\)).

Another thing to consider is what is changed by the join? That is, given two systems, do the

elementschange, or theconnectionswhen you join them? (NB: The connections do not have to change, \(A\vee A = A\)).Originally I used "partition", that's because of how it's drawn. I've updated my description to be about elements and connections.

`\\(\ast\\) is *not* connected in the joined system to either \\(\circ\\) or \\(\bullet\\). Each system has 3 elements and some number of connections. \\(A\\) has 3 elements and 0 connections (each element is unconnected from each other element). \\(B\\) has 3 elements and 1 connection (\\(\bullet\\) and \\(\circ\\) are connected, \\(\ast\\) is unconnected from them). \\(C = A \vee B\\) also has 3 elements, and 1 connection (the same as \\(B\\)). Another thing to consider is what is changed by the join? That is, given two systems, do the *elements* change, or the *connections* when you join them? (NB: The connections do not have to change, \\(A\vee A = A\\)). ----- Originally I used "partition", that's because of how it's drawn. I've updated my description to be about elements and connections.`

I think Jared answered Charlie's question just fine. I find terminology like "systems" and "connections" a bit annoying here because later in the book one sees a more precise treatment of these ideas using more standard, and more precisely defined, terminology.

As Jared already knows, when we get serious we use the term "partition" instead of "system". Charlie: we'll explore partitions in detail here:

However, it would be unwise to read these lectures until you've read the previous ones!

`I think Jared answered Charlie's question just fine. I find terminology like "systems" and "connections" a bit annoying here because later in the book one sees a more precise treatment of these ideas using more standard, and more precisely defined, terminology. As Jared already knows, when we get serious we use the term "partition" instead of "system". Charlie: we'll explore partitions in detail here: * [Lecture 10 - Chapter 1: The Logic of Partitions](https://forum.azimuthproject.org/discussion/1963/lecture-10-the-logic-of-partitions/p1) * [Lecture 11 - Chapter 1: The Poset of Partitions](https://forum.azimuthproject.org/discussion/1991/lecture-11-chapter-1-the-poset-of-partitions/p1) * [Lecture 12 - Chapter 1: Generative Effects](https://forum.azimuthproject.org/discussion/1999/lecture-12-chapter-1-generative-effects/p1) * [Lecture 13 - Chapter 1: Pulling Back Partitions](https://forum.azimuthproject.org/discussion/2008/lecture-13-chapter-1-pulling-back-partitions/p1) However, it would be unwise to read these lectures until you've read the previous ones!`

Jared - Thank you for the insights. My hesitation was caused by the fact that the definition of join doesn't seem to require that \(C = A \vee B\) must have three elements -- it seemed that I had the option of omitting * from C so that C would have only two (connected) elements. Now I see that if A, B, and C all have the same number of elements (i.e., the elements do not change), then the join is much easier to understand. I seem to have a "talent" for overly complicating solutions to relatively straightforward problems. I'll catch on ...

John - Thanks for the preview!

`Jared - Thank you for the insights. My hesitation was caused by the fact that the definition of join doesn't seem to require that \\(C = A \vee B\\) must have three elements -- it seemed that I had the option of omitting * from C so that C would have only two (connected) elements. Now I see that if A, B, and C all have the same number of elements (i.e., the elements do not change), then the join is much easier to understand. I seem to have a "talent" for overly complicating solutions to relatively straightforward problems. I'll catch on ... John - Thanks for the preview!`

Charles - aim for simplicity; it's more likely to be right. As Einstein said:

`Charles - aim for simplicity; it's more likely to be right. As Einstein said: > _Everything should be made as simple as possible, but not simpler._`

The join {{11, 12}, {13}, {21}, {22, 23}} v {{11}, {12, 22}, {13, 23}, {21}} is {{11, 12, 13, 22, 23}, {21}}

`The join {{11, 12}, {13}, {21}, {22, 23}} v {{11}, {12, 22}, {13, 23}, {21}} is {{11, 12, 13, 22, 23}, {21}}`