Talk:Ascending chain condition

Commutative rings

The first sentence of the article appears to suggest that all commutative rings satisfy the ascending chain condition. Is that true? Noetherian ring#Introduction suggests the opposite. Hashproduct (talk) 17:42, 4 January 2010 (UTC)

That's not true, the sentence was formulated in a confusing way. Hopefully it is more clear now. — Emil J. 17:55, 4 January 2010 (UTC)

Correctness of definition

A partially ordered set (poset) P is said to satisfy the ascending chain condition (ACC) if every strictly ascending sequence of elements eventually terminates. Equivalently, given any sequence

$a_{1}\,\leq \,a_{2}\,\leq \,a_{3}\,\leq \,\cdots ,$

there exists a positive integer n such that

$a_{n}=a_{n+1}=a_{n+2}=\cdots .$

How is that equivalent? Consider the sequence $a_{i}$ = $a_{1}$ for all $i$. It clearly satisfies the conditions, therefore any nonempty poset would satisfy ACC by that (incorrect) definition. --Daira Hopwood ⚥ (talk) 15:00, 10 September 2013 (UTC)

I was mistaken; it says any sequence, so the condition isn't satisfied by there being one such sequence. --Daira Hopwood ⚥ (talk) 15:15, 10 September 2013 (UTC)

First Comment Bullet

In the first bullet of the Comments section, a new set is defined to be the "disjoint union" of the infinite collection of sets {0}, {0,1}, {0,1,2}, ... . Since this is an infinite union, I think we can only assume it is well defined if we invoke the Axiom of Choice. But how will we define it? The only logical way it seems is to start taking a disjoint union (call this union A1) of any two elements of this sequence, and then taking the union of A_1 with one other element not already "unioned", giving a set A_2, and so on, without end. It seems logical (I hope) that no matter how we do this, we will always end up a limiting set (A_n as n -> infinity) equal to the set of nonnegative integers.

Next the comment says this set (non-negative integers) satisfies the ascending chain condition (ACC). This is false, since the nonnegative integers, the even integers, and an infinite number of other unbounded chains inside it, do not terminate.

Also, a lot of this bullet does not make sense to me. It is not clear what is being discussed.

— Preceding unsigned comment added by 146.142.1.10 (talk) 19:01, 23 March 2015

A subtly different and stronger condition than "containing no infinite ascending/descending chains" is "contains no arbitrarily long ascending/descending chains (optionally, 'based at a given element')". For instance, the disjoint union of the posets {0}, {0,1}, {0,1,2}, etc., satisfies both the ACC and the DCC, but has arbitrarily long chains. If one further identifies the 0 in all of these sets, then every chain is finite, but there are arbitrarily long chains based at 0.

Agree with IP. If the posets are taken as P_n and if A_n = Disjoint Union (P₀ ... P_n) then the A_i are isomorphic to N, which satisfies DCC but not ACC. If something else is meant, then it is not clear what it is. Bullet moved to talk page until it is clarified. All the best: Rich Farmbrough, 19:05, 24 September 2015 (UTC).

Assessment comment

The comment(s) below were originally left at Talk:Ascending chain condition/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

Perhaps some background on what this is, something more than "something in mathematics" which is essentially what is said even a broad category would be better--Cronholm144 23:05, 14 May 2007 (UTC) I've addressed it and uprated, didn't want to skip a class on myself, but I can't see how it might be further expanded. Arcfrk 05:42, 24 May 2007 (UTC)

Last edited at 05:42, 24 May 2007 (UTC). Substituted at 01:46, 5 May 2016 (UTC)