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f:R->R satisfies that the image of every open interval is an open interval. Does this imply that f is continuous on R? ALTERNATIVE STATEMENT: f:(a,b)->R(where (a,b) is obviously in R) satisfies: "(c,d) is a subset of (a,b) implies that f((c,d)) is an open interval." Does this imply that f is continuous on R? N.B.:The open intervals mentioned above are all finite in the sense that they are of the form (x,y),where x,y are real numbers.

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 You should ask this on math.stackexchange.com, or one of the other sites mentioned in the FAQ; here, your question is off-topic, as the FAQ itself explains. – Mariano Suárez-Alvarez Oct 12 2011 at 4:05
This website is for more advanced topics. – Keshav Srinivasan Oct 12 2011 at 4:32
At the beginning of Allen Hatcher's Lecture notes on point set topology (freely available on his Cornell webpage) it is explained very well why this definition corresponds exactly to our intuitive notion (on p.2) (and thus the other definition) of continuity. – sisn Oct 12 2011 at 6:59

closed as too localized by Will Jagy, Gjergji Zaimi, Ricky Demer, Alex Bartel, Andres Caicedo Oct 12 2011 at 4:50

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