Mathematics desk
< June 26	<< May \| June \| Jul >>	Current desk >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

June 27[edit]

Continuous on Closed Cover[edit]

What properties does a topological space X need to satisfy so that if f: X -> X is continuous on each set in a closed cover of X, f is continuous on X? This is not a homework question, I just can't remember and, for some reason, can't seem to figure it out at the moment. Thank you for any help :-) 209.252.235.206 (talk) 07:08, 27 June 2012 (UTC)[reply]

I doubt there are any properties that non-trivially imply this. Euclidean space lacks this property, for example, because the collection of singletons form a closed cover, and every function is continuous on a singleton.--2601:9:1300:5B:21C:B3FF:FEC6:8DA8 (talk) 08:55, 27 June 2012 (UTC)[reply]

Wow, I'm getting sloppy these days! Completely overlooked that; how about a closed cover containing 2 sets E and F? I'm thinking more along the lines of cutting an interval in half and showing f is continuous on each part; you can even assume that each is homeomorphic X. 209.252.235.206 (talk) 09:07, 27 June 2012 (UTC)[reply]

I don't think there's a property for spaces that works for every closed cover, but there is a criterion for the closed cover: it needs to be locally finite (every point has a neighborhood that intersect only finitely many closed sets in the cover). Money is tight (talk) 11:18, 27 June 2012 (UTC)[reply]

Thank you:-) — Preceding unsigned comment added by 209.252.235.206 (talk) 04:43, 29 June 2012 (UTC)[reply]

The answer is very simple for T1 spaces X: in this case, the condition is equivalent to: X has the discrete topology. Indeed, if it has the discrete topology, every map on X is continuous and the property trivially holds; conversely, since in a T1 space X points are closed sets, they constitue a closed covering; thus the property implies that every map f: X -> X is continuous. This also implies that any subset of X is closed (there are maps which are constant on it and on its complement). The answer is less simple for general topological spaces, not necessarily T1; one should argue similarly using the covering made by the closures of single points, and obtain a characterization in terms of the structure of the lattice of the closed sets of X. --pm a 10:48, 29 June 2012 (UTC)[reply]