I really cannot formalize it: what I'd say is that the complement of cantor set ($E^c$) is open, so the preimage with $\chi$ of each neighbourhood of $1$ (which is open for def.) is the set $E$, which is closed. since the preimage of an open set is not open $\chi$ is discontinuous. can it holds? thanks!
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What is $\chi$? – 5xum Jun 09 '16 at 06:53
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2I'm guessing it is the characteristic function of the Cantor set. – Funktorality Jun 09 '16 at 06:54
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yes, I've changed the title. thanks for hint. – kauser bi Jun 09 '16 at 06:55
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You've proved that if $U$ is open in $\mathbb R$ then $\chi^{-1}(U)$ is closed which proves that $\chi$ is not continuous. However I believe this does not prove it is discontinuous at every point. To argue this you would say that if $x\in C$ and $V$ is a neighborhood of $x$, then $V$ contains a point not in $C$ (since $C$ has empty interior) and so $\chi(V)=\{0,1\}$. That contradicts the definition here of continuity at $x$, which would require that for every neighborhood $U$ of $\chi(x)$, there is a neighborhood $V$ of $x$ so that $\chi(V)\subset U$.
Funktorality
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thanks. I couldn't find a proof with the definition of continuity using preimages! – kauser bi Jun 11 '16 at 20:25