Talk:Pretopological space

Dubious

I stuck a dubious tag on one of the assertions. The book in front of me states that the collection of open sets generated by the praclosure operator is a full-fledged, real topology, and not something less than that. Unless I flubbed something, this seems easy enough to prove. So I don't understand why praclosure leads to something less than a full topology. Am I missing something? (And, in general, I find this article confusing, and that's no help). linas 02:26, 25 November 2006 (UTC)

See comment at Talk:Praclosure operator. Those axioms don't give you binary unions of open sets. So I think you must be misquoting: generated by must mean taking all possible unions, first. You seem to think 'generated by' is innocuous. So I'm taking your tag down. Charles Matthews 15:01, 26 November 2006 (UTC)

I replied at that talk page, which now contains a proof of the main lemma needed to show that binary unions of open sets are open. So I still find this article dubious. linas 04:06, 28 November 2006 (UTC)

I cut the following out of the article:

In order to define a topology on X, the closure operator must also be idempotent; that is, it must satisfy for all subsets A of X:

cl (cl (A)) = cl (A).

For explanation why this is a necessary and sufficient condition, see Kuratowski closure axioms.

Complaints: 1) article on Kuratowski closure axioms fails to explain nec & suff. (as do all three of my books on topology). 2) the talk page of praclosure now contains a proof that the praclosure can generate a topology. In that topology, praclosure is idempotent on closed sets (that is how a closed set is defined).

Proposed solution: the topology generated by the praclosure is a topology, which can be quite strange, but not the usual topology; and it seems to me that the "usual" topology does require the fourth, idempotency, axiom. Does this now all make sense? linas 14:18, 28 November 2006 (UTC)

You can define a topology for a pretopological space, but it doesn't the right notion of convergence. The closure operator on a topological space is idempotent on all sets; praclosure operators are idempotent on just closed sets, not all sets. I'll add back in the information about idempotency. -- Walt Pohl 18:56, 9 August 2007 (UTC)

It's a while back, Linas, but you didn't convince me. Charles Matthews 21:29, 15 August 2007 (UTC)

this is a special case

AFAIK a pretopology is defined by a pseudo-closure operator a that only needs to satisfy the Kuratowski closure axioms 1 and 4; then the pretopology is called

• of V-type iff A c B => a( A ) c a( B )
• of V_D type iff a( A u B ) = a( A ) u a( B )
• of V_S type iff a( A ) = union(x in A) a({x})

Thus, in particular the K-axiom 3 is not necessarily satisfied.— MFH:Talk 15:56, 18 June 2008 (UTC) PS: references:

• Belmandt, Z.: Manuel de prétopologie et ses applications. Hermes, 1993.
• http://dx.doi.org/10.1016/S1571-0653(04)00011-3
• http://dx.doi.org/10.1016/S0020-0255(02)00189-5
• http://www.stat.unipg.it/iasc/Proceedings/2006/COMPSTAT_Satellites/KNEMO/Lavori/Papers%20CD/Le%20Lamure.pdf
• http://ieeexplore.ieee.org/xpl/freeabs_all.jsp?tp=&arnumber=4223063&isnumber=4223036

A question about continuous functions

http://math.stackexchange.com/q/1855076/4876

It is known that for a function f from a topological space to interval $[0;1]$ to be continuous, it is enough that preimages $f^{-1}]a;1]$ and $f^{-1}[a;1[$ be open for every $a$ in our interval.

Now let a function $f$ is from a pretopological space to interval $[0;1]$. What conditions are sufficient for $f$ to be continuous? — Preceding unsigned comment added by VictorPorton (talk • contribs) 16:24, 10 July 2016 (UTC)

Motivation

The only reason we are given for having a notion of "pretopology" is this:

… a pretopological space is a generalization of the concept of topological space.

How is this useful? The generalisation is achieved by dropping one of four axioms; but clearly we could make many other generalisations by dropping another one or more of those axioms, and call the resulting system a "pretopology". Yet many of the important results (theorems) of topology will fail to hold (or be much more difficult to prove) in the absence of those axioms.

So why choose just this one axiom to drop? My intuition, FWIW, is that specific desirable theorems of topology (or some more general analogues of them) will still hold in this particular kind of pretopology.

The article needs to clarify:

1. the history of the term "pretopology"; and
2. the thinking — reasoning and motivation — that lead to the current definition; and
3. which important theorems of topology are also theorems of pretopology; and
4. which important theorems of topology are not also theorems of pretopology.

yoyo (talk) 13:30, 25 November 2018 (UTC)