Real Analysis with functions?

Question

Prove that the fuction f: R ----> R given below is not continuous at any x0 in the real numbers.

so f(x) is a peicewise with
-1 if x in Q
1 if x is not in Q

Usually to prove this, you use the definition of continous functions with sequences...but i'm not sure where to start...to prove that this is not continuous

Answer 1

Let x_0 ∈ R. For every δ > 0, (x_0 - δ , x_0 + δ) contains infinitely many rationals and infinitely many irrationals (because the rationals and the irrationals are dense in R). Therefore, if x_0 is rational, we can take an irrational x_1 in (x_0 - δ , x_0 + δ) to get



|f(x_1) - f(x_0)| = |1 - (-1)| = 2



and if x_0 is irrational, we can take a rational x_2 in (x_0 - δ , x_0 + δ) to get



|f(x_2) - f(x_0)| = |-1 - 1| = 2



Hence, for every δ > 0, we can find an u in (x_0 - δ , x_0 + δ) such that



|f(u) - f(x_0)| = 2 > 0



which shows f is discontinuous at every real x_0.



This can also be done with sequences:



If x_0 is rational, take a sequence u_n of irrationals that converges to x_0. Then, f(u_n) → 1 ≠ -1 = f(x_0), which shows f is discontinuous at x_0.

If x_0 is irrational, take a sequence vu_n of rationals that converges to x_0. Then, f(v_n) → -1 ≠ 1 = f(x_0), which shows f is discontinuous at x_0.



I like the 1st proof best.

Answer 2

x0 is in Q, then we can find a sequence x_n entirely in (not Q) converging to x0, and vice versa if x0 is in (not Q). this is true because both Q (the rationals) and (not Q) (the irrationals) are dense in the set of real numbers. in either case, lim f(x_n) does not --> to f(x0) even though x_n --> x0.

Answer 3

Take any sequence of rational numbers which converges to an irrational number. (For example, the successive decimal approximations of pi approach pi:



x_1 = 3

x_2 = 3.1

x_3 = 3.14

.

.

.



Then f( x_n) ---> -1 as n ---> inf (this is a constant sequence!)



but lim f (x_n) is not equal to f ( lim x_n) = f(pi) = +1.