Let [math: f(x)] be the function that is equal to 0 for all rational [math: x] and 1 for every irrational [math: x]. Is [math: f(x)] continuous at [math: x=π]x=\pi?

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This question asks about the continuity of a classic example: the Dirichlet function, defined as f(x) = 0 if x is rational and f(x) = 1 if x is irrational.

Option a: 'No.' If we examine continuity at x = π, we note a crucial fact: every open interval around π contains both rational and irrational numbers. Consequently, as x approaches π through rational values, f(x) = 0, and as x approaches π through irrational values, f(x) = 1. The limit would have to be the same along all sequences approaching π, but we obtain two different limit values (0 and 1) depending on the path of approach. Thus the limit as x → π does not exist. Since π is itself irrational, f(π) = 1. For continuity, the limit as x → π must equal f(π); however, the non-existent limit already fails this prerequisite, so the function is not continuous at π.

Option b: 'Yes.' This statement claims continuity at π. The above reasoning shows that the limit does not exist due to the alternating values 0 and 1 in every neighborhood of π, which already violates the basic requirement for continuity. Moreover, even if one tries to apply an epsilon-delta criterion, choosing ε = 1/2 would force any δ-neighborhood around π to contain points x with f(x) = 0 and points with f(x) = 1, making |f(x) − f(π)| equal to 1 for some x within δ, which cannot be kept below ε. Therefore, this option is incorrect since the function fails the continuity test at π.

In summary, the key idea is that the Dirichlet function oscillates between 0 and 1 in any interval, so no point, including π, can be a point of continuity. The correct assessment is that the function is not continuous at π, which contradicts the statement in option b.

Let [math: f(x)] be the function that is equal to 0 for all rational [math: x] and 1 for every irrational [math: x]. Is [math: f(x)] continuous at [math: x=π]x=\pi?

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