Talk:Open and closed maps

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)DruifficReply

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

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

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

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

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 ${\displaystyle \operatorname {Im} f}$  is the image of ${\displaystyle f}$  *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

The article says:

"3. A map ${\displaystyle f:X\to Y}$  is called an open map or a strongly open map if it satisfies any of the following equivalent conditions: ... For every ${\displaystyle x\in X}$  and every neighborhood ${\displaystyle N}$  of ${\displaystyle x}$  (however small), ${\displaystyle f(N)}$  is a neighborhood of ${\displaystyle f(x)}$ .

• Either instance of the word "neighborhood" in this statement can be replaced with "open neighborhood" and the resulting statement would still characterize strongly open maps."

This is how I understand the above: By replacing one or two of the occurrences of "neighborhood" with "open neighborhood" in the above statement we can get 4 different statements:

A) For every ${\displaystyle x\in X}$  and every neighborhood ${\displaystyle N}$  of ${\displaystyle x}$ , ${\displaystyle f[N]}$  is a neighborhood of ${\displaystyle f(x)}$ .

B) For every ${\displaystyle x\in X}$  and every open neighborhood ${\displaystyle N}$  of ${\displaystyle x}$ , ${\displaystyle f[N]}$  is a neighborhood of ${\displaystyle f(x)}$ .

C) For every ${\displaystyle x\in X}$  and every neighborhood ${\displaystyle N}$  of ${\displaystyle x}$ , ${\displaystyle f[N]}$  is an open neighborhood of ${\displaystyle f(x)}$ .

D) For every ${\displaystyle x\in X}$  and every open neighborhood ${\displaystyle N}$  of ${\displaystyle x}$ , ${\displaystyle f[N]}$  is an open neighborhood of ${\displaystyle f(x)}$ .

Indeed A), B) and D) are equivalent to the main definition of (strongly) open map, which is that any open subset of ${\displaystyle X}$  is mapped to an open subset of ${\displaystyle Y}$ . However C) is definitely not equivalent to ${\displaystyle f}$  being an open map. As a counterexample take the real line ${\displaystyle \mathbb {R} }$  endowed with the Euclidean topology and the identity function ${\displaystyle id(x)=x}$ . Obviously ${\displaystyle id}$  is an open map, but C) does not hold for ${\displaystyle id}$ . For example ${\displaystyle [-1,1]}$  is a neighborhood of ${\displaystyle 0}$  but ${\displaystyle id}$  maps that interval to itself and it is not an open neighborhood of ${\displaystyle 0}$ .

So I would suggest removing the sentence "* Either instance of the word "neighborhood" in this statement can be replaced with "open neighborhood" and the resulting statement would still characterize strongly open maps." and just listing conditions A), B) and D) explicitly and saying that they are equivalent to the main definition of (strongly) open map. Vesselin.atanasov (talk) 03:14, 10 March 2023 (UTC)Reply