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Prove or give a counterexample. let $f \colon R \to R$ such that $f$ is continuous on $D \subseteq R$ and $f$ is continuous on $S \subseteq R$. Then f is continuous on $D \cup S$. I am kind of puzzled with notation ($f$ is continuous on $D \subseteq R$ and $f$ is continuous on $S \subseteq R$. Then $f$ is continuous on $D \cup S$) Can you give me an example?

What if I consider this example: f(x) = x if x is rational and f(x) = 0 if x is irrational. Then I will have that f is continuous only at 0.( a counterexample)?

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    What part of the notation puzzles you? (At the moment, you simply restate verbatim the question.)2012-03-08
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    Suppose $f$ is continuous on $(-1,0]$ and that $f$ is continuous on $(0,1)$. Does $f$ have to be continuous on $(-1,1)$?2012-03-08
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    Or maybe do an example like this: $f$ is continuous on $\mathbb R \setminus \{0\}$ and $f$ is continuous on $\{0\}$.2012-03-08
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    There is a semantic problem here that has not been addressed so far: A function $f:\ {\mathbb R}\to{\mathbb R}$ has a *set of continuity* $C_f:=\{x\in{\mathbb R}\ |\ f\ {\rm is\ continuous\ at}\ x\}$. When $D\subset C_f$ and $S\subset C_f$ then obviously $D\cup S\ \subset C_f$. On the other hand such an $f$ can be restricted to some given subset $D\subset{\mathbb R}$ which is now a topological space in its own right. When $f_D\ :\ D\to{\mathbb R}, \quad x\mapsto f(x)$ is continuous and similarly $f_S$ is continuous then $f_{D\cup S}$ need not be continuous on $D\cup S$.2012-03-08

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