Can Dirichlet's function be modified in such a way that it is continuous at some real number? For instance, as $xD(x)$ is continuous at $x=0$, is it possible that $(x-1)D(x)$ is continuous at $x=1$? Here $D(x)$ is the Dirichlet function.
Can it be modified to be continuous at finitely many points in $\mathbb{Q}$?
I am aware that there can't be a function continuous only on rational numbers, but with the same approach as $2$, if possible, why isn't it possible to construct such a function?
Thanks in advance!
Answer
Given numbers $x_1,\ldots,x_n$, you can define an "accordion function" $g(x)$ by piecing together absolute values so that $g(x_j)=0$ and $g((x_j+x_{j+1})/2)=1$. Then $g(x)D(x)$ is continuous at $x_1,\ldots,x_n$ and discontinuous everywhere else.
The same approach wouldn't work for all the rationals because they are dense, and so there is no room for the function to "raise up" and "come down".
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