In Spivak's Calculus (4th ed.), on pg. 99, he claims/proves that the following function converges to 0 at every a in the interval (0, 1): f(x) = 0, for x irrational ......= 1/q, for x rational, where x = p/q in lowest terms. His proof goes as follows: "For any number a, with 0 < a < 1, the function f approaches 0 at a. To prove this, consider any number ε > 0. Let n be a natural number so large that 1/n ≤ ε. Note that the only numbers x for which |f(x) - 0| < ε could be false are: 1/2; 1/3, 2/3; 1/4, 3/4; 1/5, 2/5, 3/5, 4/5; ...; 1/n, ..., (n-1)/n. (*) (If a is rational, then a might be one of these numbers). However many of these numbers there may be, there are, at any rate, only finitely many. [emphasis mine] Therefore, of all these numbers, one is closest to a; that is, |p/q - a| is smallest for one p/q among these numbers. (If a happens to be one of these numbers, then consider only the values |p/q - a| for p/q ≠ a.) This closest distance may be chosen as the δ. For if 0 < |x - a| < δ, then x is not one of 1/2, ..., (n-1)/n and therefore |f(x) - 0| < ε is true. This completes the proof." * * * Here's my issue with the proof: isn't his list (*) missing a bunch of numbers? For example, let p be some number relatively prime to n+1. Shouldn't p/(n+1) be on the list? My understanding of his logic is that the list ends at n because any number smaller (e.g. n+1) doesn't need to be considered, since 1/(n+1) ≤ ε. But we don't know that p/(n+1) ≤ ε if p is large enough. So isn't the proof incorrect? Aren't there infinitely many numbers to worry about? [Just to clarify, I'm not suggesting that the main assertion of it approaching 0 is wrong - just that the proof is flawed.] EDIT: I haven't tried reproving it correctly yet, since I wanted to make sure the actual proof was wrong before I spent too much time doing that (especially if it's possible it doesn't approach 0 like it claims).