Talk:Open and closed maps: Difference between revisions

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Sufficient for continuity?

Regarding "...a function f : X → Y is continuous...if the preimage of every closed set of Y is closed in X." at the end of the second paragraph:

Forgive me if I'm mistaken, but in general, preimages of closed sets being closed does not ensure continuity. E.g. for f: [0,1] -> (0,1] where f(x) = x (for x != 0) and f(0) = 1/2, f is not continuous. Perhaps something is assumed that I missed? 203.150.100.189 08:38, 20 March 2007 (UTC)[reply]

Please read again: "if the preimage of every open set of Y is open in X". For your example, the preimage of ]1/4,3/4[ is then {0}U]1/4,3/4[ which is not open.--133.11.80.84 07:05, 3 September 2007 (UTC)[reply]

203: Pass to complements. Suppose f is continuous and take a closed set V. Then the complement of V, which I'll denote C(V), is open, so it pulls back to an open set. But the preimage of V is the complement of the preimage of C(V), so it's closed. The other way: Suppose f pulls back closed sets to closed sets. A similar argument gives you that f pulls back open sets to open sets, so f is continuous. HTH. Druiffic (talk) 04:26, 15 November 2008 (UTC)Druiffic[reply]

Yeah, but Druiffic, 203's example proves you wrong! consider the closed set V=[1/8,1/4] u [3/4,1] Then C(V) = (0,1/8) u (1/4,3/4) is clearly open. Next, f^-1(C(V))= (0,1/8) u (1/4,3/4) u {0} = [0,1/8) u (1/4,3/4) is clearly NOT open! Thus, although f pulls back closed sets just fine, this f is not continuous and clearly fails to pull back open complements into open complements.67.198.37.16 (talk) 17:01, 30 June 2016 (UTC)[reply]

Silly me -- I am wrong. In the target space, the interval (0,1/8) is NOT open -- it is half-closed. That is, in the target space, the interval (0,1/8] is fully closed, and it does not pull back to a closed set. This is a subtle point: the definition of a closed set in the target space is NOT the same as the definition of a closed set in the natural topology on the reals -- in particular, the entire target space is considered to be closed. Its an interesting, illuminating failed counter-example, illustrating that one mst be careful to use the definition of "open" and "closed" correctly. 67.198.37.16 (talk) 18:17, 30 June 2016 (UTC)[reply]

Surjective claim

Unless I'm going crazy, "An open map is also closed if and only if it is surjective" is definitely not true. Just include X into two (disjoint) copies of itself. —Preceding unsigned comment added by 24.19.0.156 (talk) 00:54, 8 February 2011 (UTC)[reply]

This statement appears to have been removed from the article with this edit, about a week after you complained. It was added in this edit, about 3 weeks before you complained. So it was a short-lived mistake. 67.198.37.16 (talk) 18:27, 30 June 2016 (UTC)[reply]

Example

I was wondering if isometries in metric spaces are open. — Preceding unsigned comment added by Noix07 (talk • contribs) 15:35, 16 December 2013 (UTC)[reply]

Well, wouldn't an isometry be a homeomorphism? If so, then note that homeomorphisms are open.67.198.37.16 (talk) 18:40, 30 June 2016 (UTC)[reply]

The floor function is open?

A sentence in the "Examples" section says that the floor function is open. This seems doubtful; the image of open interval (1/3, 2/3) under this map is the singleton set {0}, which is not open. (Right?) If I'm right, would someone please correct that line?

Norbornene (talk) 20:29, 3 January 2017 (UTC)[reply]

Whether a given subset is open or not depends on the topology used, and that specific paragraph in the Examples section talks about the floor function as a map from the reals, equipped with the usual topology induced by the absolute difference, to the integers, equipped with the discrete topology. And in the discrete topology the singleton set {0} (as a subset of Z) is indeed open (as are all other subsets of Z). – Tea2min (talk) 08:29, 4 January 2017 (UTC)[reply]

Unclear definition of relatively open

As the definition is currently written, a relatively open map seems like it's the same thing as an open map. (This makes the crazy capital letters "WARNING" a little hard to understand.)

My guess is that it should say $\operatorname {Im} f$ is the image of $f$ *considered as a topological space under the subspace topology*, which would make sense and would make the two definitions different. However, I can't currently check the reference to be sure. Is this correct?

Nathaniel Virgo (talk) 07:09, 27 October 2020 (UTC)[reply]

@@ Line 36: / Line 36: @@
 As the definition is currently written, a relatively open map seems like it's the same thing as an open map. (This makes the crazy capital letters "WARNING" a little hard to understand.)
-My guess is that it should say $\operatorname{Im}f$ is the image of $f$ *considered as a topological space under the subspace topology*, which would make sense and would make the two definitions different. However, I can't currently check the reference to be sure. Is this correct?
+My guess is that it should say <math>\operatorname{Im}f</math> is the image of <math>f</math> *considered as a topological space under the subspace topology*, which would make sense and would make the two definitions different. However, I can't currently check the reference to be sure. Is this correct?
 [[User:Nathanielvirgo|Nathaniel Virgo]] ([[User talk:Nathanielvirgo|talk]]) 07:09, 27 October 2020 (UTC)