We will show that the function is not continuous at any point "x \\in \\mathbb{R}." Consider an arbitrary point "x \\in \\mathbb{R}." It is well known that the set of rational numbers "\\mathbb{Q}" is everywhere dense in "\\mathbb{R}." Therefore there exists a sequence "\\{r_n\\}_{n=1}^\\infty \\subset \\mathbb{Q}" such that "r_n \\to x" for "n \\to \\infty". Hence "f(r_n)=2 \\to 2" for "n \\to \\infty". Also it is known that the set of irrational numbers "I=\\mathbb{R} \\setminus \\mathbb{Q}" is everywhere dense in "\\mathbb{R}". Therefore there exists a sequence "\\{x_n\\}_{n=1}^\\infty \\subset I" such that "x_n \\to x" for "n \\to \\infty". Hence "f(x_n)=4 \\to 4" for "n \\to \\infty."
So, thus we found two sequences "r_n \\to x \\gets x_n," but "|f(r_n)-f(x_n)| \\not \\to 0" for "n \\to \\infty". Therefore "f" is not continuous at "x \\in \\mathbb{R}." Now if "B \\subset \\mathbb{R}" is any subset, then "f" is not continuous on "B" , because "f" is discontinuous at every point.
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