## 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 *{ x _{n} }* converging to

*c*. Then there exists an integer

*N*such that

for| x_{n}- c | < / 5

*n > N*. Then

| f(x_{n}) - (5 c - 6) | = | 5 x_{n}- 6 - 5 c + 6 | = 5 | x_{n}- c | <

for *n > N*. But then the sequence
*{f(x _{n})}* converges to

*5 c - 6*, or in other words: if a sequence

*{x*converges to

_{n}}*c*, then

*f(x*converges to

_{n})*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 *a _{n}* converging
to

*c*, as well as another sequence of irrational numbers

*b*also converging to

_{n}*c*. But then the sequence

*{f(a*is identically

_{n})}*1*, and the sequence

*{f(b*is identically

_{n})}*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).