Talk:Join (mathematics)

Latest comment: 14 years ago by Tobias Bergemann in topic Merge proposal

changed edit

I changed this from a redirect to Lattice (order) to an article. This does induce some material reduplication; but in general, in these articles a binary operation is distinguished from a set with this binary operation. For consistency; in order to be able to include join in the category of binary operations; and since joins are found not only in lattices but also in semilattices, I thought it better to include a brief stand-alone article.

I did the same for Meet (mathematics). The two articles are ridicuously similar; and I think it is high time to describe the basic duality of the two lattice operations more explicitly. JoergenB 20:03, 21 September 2006 (UTC)Reply

Should the name be changed? edit

Considering the join concept in Relational algebra, perhaps it is somewhat pretentious to retain the title Join (mathematics) for this article? JoergenB 20:19, 21 September 2006 (UTC)Reply

Also the join operation in topology? 74.255.70.210 (talk) 21:06, 9 July 2008 (UTC)Reply

Ick, not cut and paste! edit

You should not do that -- sooner or later someone will change one article without changing the other (or, even worse, make non-dual changes to both articles). Also, you haven't been particularly explicit about their duality on the actual pages.

P.S. Maybe semilattice is a better place to talk about meet and join, hmm? User:SamB 13 November 2006

Join & Meet examples edit

The meet and join pages would both be much clearer with some examples! Harry Metcalfe 13:09, 19 January 2007 (UTC)Reply

Problematic definitions edit

As currently written, the partial order approach is either wrong or unconvincing. In particular, the definition of the join, point 2, states "for any w in A, such that x ≤ w and y ≤ w, we have z ≤ w (i.e., z is less than any other upper bound of x and y)". First off, the parenthesis should read "z is less than or equal to any other upper bound of x and y" (emphasis added). Second, this being a partial ordering, point 2 is undecidable because you are bound to run into at least one w for which z ≤ w is unspecified (if it were otherwise, we would have a complete ordering, no?).

Second, the "proof" of the uniqueness of the join seems incorrect. If we've surmounted the decidability problem of point 2 above, the relation z ≤ z' ≤ z, whence z = z' only establishes that z and z' are equal in order, not that they are identical to each other.

Can anyone address these concerns? Urhixidur (talk) 15:58, 21 August 2008 (UTC)Reply

I'll answer myself partially: point 2 is possible within a partial ordering because w is not "any member of A", but belongs to the subset of members of A for which the relations x ≤ w and y ≤ w are defined. Nevertheless, the issue that z ≤ w may be undefined for some w remains. Urhixidur (talk) 16:20, 21 August 2008 (UTC)Reply

Here's a Hasse diagram of what is indeed a lattice, but where the join of at least one pair is not unique:

    H
  / | \
 /  |  \
E   F===G
|   |\  |
|   | \ |
|   |  \|
B   C   D
 \  |  /
  \ | /
    A

The arrowheads (not shown) flow from A to H overall. Note in particular the flow F to G and G to F. The join of (F, G) is not unique: it matches both F and G. F and G are of equal rank, but they are not identical since there is no C-G ordering defined (actually C-G is defined transitively from C-F-G). Urhixidur (talk) 17:53, 21 August 2008 (UTC)Reply

Edges in Hasse diagrams do not need arrowheads (and are in fact usally drawn without them) as the direction of the relation is represented by the vertical position of the concerned nodes. Hasse diagrams cannot contain horizontal lines because, per definition, any partial order must be antisymmetric: if a ≤ b and b ≤ a then a = b. I am afraid I do not really understand what you try to demonstrate with your "pseudo Hasse diagram". Please elaborate. — Tobias Bergemann (talk) 11:10, 22 August 2008 (UTC)Reply

In that diagram, we have F <= G and G <= F (which indeed means F = G, ordering-wise). What is their join? It is F or G, equivalently. The join is not "unique" in the sense that there are two elements that match it. The join's value (rank, order) is unique, on the other hand. See my point? Urhixidur (talk) 20:01, 14 October 2008 (UTC)Reply

I'm not at all sure what you're arguing here. (1) Under which definition do you think the above diagram is a lattice? (2) In the partial order definition of a join, if there are two incomparable upper bounds for a pair of elements {a,b}, then those elements {a,b} have no least upper bound and thus no join. — Carl (CBM · talk) 23:19, 14 October 2008 (UTC)Reply

Merge proposal edit

I think it makes no sense to maintain two articles that are exact duals of each others. Let's combine meet (mathematics) and join (mathematics) into meet and join. --Hans Adler (talk) 18:49, 15 April 2009 (UTC)Reply

Seconded. — Matt Crypto 19:53, 8 June 2009 (UTC)Reply
I have started to merge both articles by moving Meet (mathematics) to Join and meet and by turning Join (mathematics) into a redirect to Join and meet. —Tobias Bergemann (talk) 20:57, 8 November 2009 (UTC)Reply