Example 6.2.4(a):
- If f(x) = 5x - 6, prove that f is continuous in its domain.
- If f(x) = 1 if x is rational and f(x) = 0 if x is irrational, prove that x is not continuous at any point of its domain.
Pick any > 0. Take any sequence { xn } converging to c. Then there exists an integer N such that
| xn - c | < / 5for n > N. Then
| f(xn) - (5 c - 6) | = | 5 xn - 6 - 5 c + 6 | = 5 | xn - c | <
for n > N. But then the sequence {f(xn)} converges to 5 c - 6, or in other words: if a sequence {xn} converges to c, then f(xn) converges to f(c). That proves continuity of the first function.
As for the second one: if c is any real number we can find a sequence of rational numbers an converging to c, as well as another sequence of irrational numbers bn also converging to c. But then the sequence {f(an)} is identically 1, and the sequence {f(bn)} is identically 0. But then f does not have a limit at c, and hence can not be continuous at c either (we have seen this argument - more formally - in a previous example already).