The problem I'm going to post, may appear a bit routine at first sight but it is not so! Suppose that a,b are two real numbers and $f:(a,b)\rightarrow \mathbb{R}$ satisfies: $f((c,d))$ is a bounded open interval for EVERY subinterval $(c,d)$ of $(a,b)$. Can we conclude that $f$ is continuous on $(a,b)$ then? A little thought will reveal that the problem is equivalent to asking whether we can conclude that $f$ is monotone on $(a,b)$, under the same hypotheses. But I am finding it as much impossible to prove that $f$ is continuous, as to find a discontinuous counterexample, satisfying the hypotheses!!
A strange characterization of continuity
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continuity
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2Essentially, your map is open. Look here: http://math.stackexchange.com/questions/75589/open-maps-which-are-not-continuous – 2012-05-12
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0[This](http://mathforum.org/library/drmath/view/62395.html) seems to furnish a counterexample. Note the $f$ constructed here does satisfy the boundedness requirement. – 2012-05-12