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consider the simple function $f(x) = \left\{ \begin{array}{ll} 1 & \mbox{if $x \in [0,1]$}\\ 10 & \mbox{if $x \in \mathbb{Q} \cap [1,2]$}\\ -10 & \mbox{if $x \in \mathbb{Q^c} \cap [1,2]$}\\ 2 & \mbox{if $x \in [2,3]$}\ \end{array} \right.$

is it correct to say that f is continuous on $[0,1] \cup [2,3]$?

4 Answers 4

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As others have pointed out, it seems reasonable to assume that you meant to define $f$ like this: $f = \chi_{[0,1]} + 10\chi_{\mathbb{Q}\cap (1,2)} -10\chi_{(1,2)\setminus\mathbb{Q}} +2\chi_{[2,3]}, $ where $\chi_A$ is the indicator function on $A$.

And it is true that $f|_{[0,1]\cup [2,3]}$ is continuous since it's constant on each of its domain's connected components.

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    @Ross: Thanks for letting me know about that typo. That's part of the reason why my usual preamble contains definitions of the commands `\union` and `\intersect`. It's much harder to accidentally mess those up2011-02-27
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Hint: Your function is not well defined at 1 and 2, as those $x$ values each meet two of your criteria. Having fixed that, say by making the interval in the second line $(1,2)$, what happens near 1 and 2?

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This function is not well-defined at some points. For example $f(1) = 1$ since $1 \in [0,1]$, but also $f(1) = 10$ because $1 \in \mathbb{Q} \cap [1,2]$ as well (similarly for $x=2$).

Once you correct this, note that if you define $f(1)=1$, then arbitrarily close to $1$ you can find numbers that are rational (or irrational) and greater than one. Here the function evaluates to $10$ (or $-10$), and so it will not be continuous at $1$ (nor at $2$ for the same reason).

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(Assume the definition of the function is corrected so that $f(1)=1$, $f(2)=2$.)

$f$ is a function on $[1,3]$. The points where $f$ is continuous are $[0,1) \cup (2,3]$. In that sense, one would not say that $f$ is continuous on $[0,1] \cup [2,3]$.

However, what is true is that the restriction $f|_{[0,1] \cup [2,3]}$ of $f$ to $[0,1] \cup [2,3]$ is a continuous function on all of its domain. But strictly speaking this is a different function from $f$.