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$ f : \mathbb R ^a \to \mathbb R ^b $

If the image by $f$ of any compact is compact and the image of any connected set is connected, then $f$ is continuous.

(If only one of the two conditions is true, the result is false.)

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closed as off-topic by Ittay Weiss, B. Mehta, qwr, Rebellos, Stefan4024 Nov 28 '17 at 22:35

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  • 1
    $\begingroup$ Please show some effort rather than just dump a question on us and indicate how,nice,it would be to see several approaches. $\endgroup$ – Ittay Weiss Nov 28 '17 at 18:04
  • $\begingroup$ Ok, sorry. I will try to find something by myself, but it seems quite difficult. $\endgroup$ – user371663 Nov 28 '17 at 18:19
  • $\begingroup$ So, please, dont solve the entire problem immediately. Rather, I would like a hint. $\endgroup$ – user371663 Nov 28 '17 at 18:22
  • $\begingroup$ what is the counterexample if you remove the assumption that the image of compact is compact? $\endgroup$ – Yanko Nov 28 '17 at 18:30
  • $\begingroup$ in R->R : f(x) = sin(1/x) with f(0) = 0. I'm thinking about a general counter-example. Maybe it is possible to use this function as coordinated applications. $\endgroup$ – user371663 Nov 28 '17 at 18:33