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

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

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.)

## closed as off-topic by Ittay Weiss, B. Mehta, qwr, Rebellos, Stefan4024Nov 28 '17 at 22:35

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Ittay Weiss, qwr, Rebellos, Stefan4024
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• Please show some effort rather than just dump a question on us and indicate how,nice,it would be to see several approaches. – Ittay Weiss Nov 28 '17 at 18:04
• Ok, sorry. I will try to find something by myself, but it seems quite difficult. – user371663 Nov 28 '17 at 18:19
• So, please, dont solve the entire problem immediately. Rather, I would like a hint. – user371663 Nov 28 '17 at 18:22
• what is the counterexample if you remove the assumption that the image of compact is compact? – Yanko Nov 28 '17 at 18:30
• 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. – user371663 Nov 28 '17 at 18:33