Discussion
The function \(f(x)\) defined as :
\(f(x)\) = 0, if x is irrational
= 1, if x is rational,
is :
Answer: D
The function is discontinuous at all points of R.
This is Dirichlet's function. Rationals and irrationals are both dense in R, so every interval around any point \(a\) contains points where f is 0 and points where f is 1.
If \(a\) is rational, \(f(a) = 1\), but a sequence of irrationals tending to \(a\) has \(f = 0\) all along, so \(f(x_{n}) \not\to f(a)\). If \(a\) is irrational, \(f(a) = 0\), and a sequence of rationals tending to \(a\) has \(f = 1\). Either way f is discontinuous at \(a\).
So it is continuous nowhere: not on R, not at rational points, and not at irrational points. (Continuity at the irrationals only is the property of Thomae's function, which takes the value \(1/q\) at \(p/q\).)
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