Talk:Filtration (mathematics)

Sigma-algebra case
The concept is more general than the sigma-algebra case, though.

Charles Matthews 08:28, 15 Jun 2004 (UTC)

Not that I know of, so feel free to expand the article to the general case. Bryan Barnard 12:29, 3 Jul 2004 (UTC)

Need clarification
There appears to be an anomaly in the article. The definition mentions

"... subject only to the condition that if i ≤ j in I then Si is a subset of Sj"

whereas in the groups subsection of Examples in Algebra mentions

"A filtration of a group G, is then a sequence Gn of normal subgroups of G which is decreasing in the sense that for any n, we have Gn+1 ⊆ Gn."

According to the first statement, it should have been Gn ⊆ Gn+1.

Tsachin 13:16, 16 May 2007 (UTC)Tsachin

Let I be N with the reverse ordering. You get the same concept. Joeldl 13:47, 16 May 2007 (UTC)

Is the definition of filtration given in the "Groups" section the "only" definition in this context? It seems like a decreasing sequence of subgroups (not necessarily normal) or an increasing sequence of (normal) subgroups would also be consistent with the general definition given in the introduction. Are these also considered filtrations? ATC2 (talk) 18:11, 15 March 2011 (UTC)

I-stable module filtrations
It would be nice to have some mention of I-stable module filtrations (for some ideal I), and how any two I-stable filtrations have bounded difference and hence induce the same topology (which is, of course, the I-topology). 128.148.41.116 (talk) 23:39, 5 May 2008 (UTC)
 * That is the Artin-Rees lemma. Joeldl (talk) 06:44, 6 May 2008 (UTC)

Modal logic connection?
I have come across filtrations in proofs of the decidability of modal logics (in Chellas's book, e.g.). Are they an example of the same thing? — Preceding unsigned comment added by 81.131.24.208 (talk) 17:14, 30 January 2013 (UTC)

Right-continuous filtration
Is it still right if $$t \in \{0,1,\dots,\infty\}$$? The whole idea is in infinitesimal increase. --Igor Yalovecky (talk) 15:12, 5 February 2018 (UTC)

Strict Subset
The introduction currently has the definition:


 * if $$i\leq j$$ in $$I$$, then $$S_i\subset S_j$$

but doesn't the correct definition use $$\subseteq$$ instead? That's even used further down in the mention of a descending filtration.

Gfredericks756 (talk) 14:58, 8 August 2021 (UTC)

"Filtered ring" listed at Redirects for discussion
The redirect [//en.wikipedia.org/w/index.php?title=Filtered_ring&redirect=no Filtered ring] has been listed at redirects for discussion to determine whether its use and function meets the redirect guidelines. Readers of this page are welcome to comment on this redirect at  until a consensus is reached. Jay 💬 09:54, 13 April 2024 (UTC)