Talk:Open and closed maps: Difference between revisions
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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? |
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? |
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[[User:203.150.100.189|203.150.100.189]] 08:38, 20 March 2007 (UTC) |
[[User:203.150.100.189|203.150.100.189]] 08:38, 20 March 2007 (UTC) |
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More to the point, "...a function f : X → Y is continuous if the preimage of every open set of Y is open in X" is candidly false! |
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consider f:R->R, x|->x^2. then the preimage of (-1,1) (an open set in Y) is [0,1) (not an open set in X). |
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Will delete this phrase. |
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[[User:HyDeckar|HyDeckar]] 01:13, 4 July 2007 (UTC) |
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Revision as of 01:14, 4 July 2007
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)