Open and closed maps: What good for?

Question

I'm wondering what open mappings are actually good for (except for inverse becomes continuous)???

My irritation came since, people stress that an open mapping not necessarily preserves closed sets (well, sure, I mean closed maps are some totally different subject since they don't describe neigbborhoods).

I cannot imagine any other purpose despite quotient maps or homeomorphisms but thats again about some continuity issues; maybe you know some problem where this becomes really important...

Open Mapping Theorem in functional analysis? The fact that a continuous and open bijection is a homeomorphism? — Asaf Karagila, Feb 02 '14 at 05:31
Open maps are very useful for finding directions in a strange place. — copper.hat, Feb 02 '14 at 05:33
Well, yes, but why would you consider then closed maps in general! as well? — C-star-W-star, Feb 02 '14 at 05:34
@copper.hat: Yeah, Google Maps! on my smartphone[s] saved me several times when I was lost in foreign cities! (And thank goodness for free WiFi in pubs around Vienna...) — Asaf Karagila, Feb 02 '14 at 05:35
@Freeze_S: Sorry, I was having a little fun at your question's expense! — copper.hat, Feb 02 '14 at 05:37
My irritation was basically about why then consider closed maps as well? I mean what we're (google maps) are interested into is neigbborhoods (not necessarily open but always containing open ones) but never interested in closed sets (not conrtaining nbhds) — C-star-W-star, Feb 02 '14 at 05:38
@copper: I was about to reply "xD" however not sure wether u would have taken it the wrong intension ^^ — C-star-W-star, Feb 02 '14 at 05:39
Google maps: Closed set=far away ...best way to escape more far away ^^ — C-star-W-star, Feb 02 '14 at 05:41

score 8 · Answer 1 · answered Feb 02 '14 at 06:38

8

How about this: as soon as you know a mapping is open, you know that maxima can't be achieved on the interior of sets (if our codomain is $\mathbb{R}$ or $\mathbb{C}$ for example): consider $f: X \rightarrow Y$ an open map. Then $f(\Omega)$ is open in $Y$, so $f(\Omega)$ cannot contain an element with maximal norm.

answered Feb 02 '14 at 06:38

That is a really nice idea ...turning the argument around a closed but not necessarily open nor continuous real valued map either achieves its maximum and minimum or is unbounded ...and the nice thing is this does not need compactness as in the extreme value theorem ...oups, that is trivially true oO – C-star-W-star Feb 02 '14 at 07:19

C-star-W-star · Accepted Answer · 2014-06-16T13:01:04.003

Open and closed maps become useful when combined with continuity!

Open/Closed Maps:

For a continuous and open/closed map we have:

If it is injective then it is an embedding.
If it is surjective then it is a quotient map.
If it is bijective then it is a homeomorphism.

Note: Neither embeddings nor quotient maps are necessarily open/closed
but homeomorphisms are necessarily open/closed.

Closed Map Lemma:

Continuous maps from compact spaces to Hausdorff spaces are closed.

score 0 · Answer 3 · edited Apr 13 '17 at 12:20

0

A result in complex analysis says that all (non-constant) holomorphic functions are open maps. This allows for all sorts of nifty results, such as this one: an injective holomorphic function can never have zero derivative.

edited Apr 13 '17 at 12:20

Community

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answered Jun 16 '14 at 13:03

mval

353

Open and closed maps: What good for?

3 Answers3

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