Real Analysis - Prove f is discontinuous

[steelphantom]

Homework Statement

Let f(x) = 1 for rational numbers x and f(x) = 0 for irrational numbers. Show that f is discontinuous at every x in R.

Homework Equations

Definition of continuity.

The Attempt at a Solution

I want to find a sequence (x_n) that converges to x_0 but that x_n is rational for even n, and irrational for odd n. This will show that (f(x_n)) cannot converge, since it will alternate between 0 and 1, and thus f is discontinuous. My problem is that I can't think of a sequence that does this!

Thanks for any help.

 

[NateTG]

Step 1: Find a rational sequence that converges to $x$.
Step 2: Find an alternating rational/irrational sequence that converges to $0$.
(Hint: I suggest something of the form $0,y_1,0,y_3...$.)
Step 3: Add the two together.

 

[steelphantom]

How about (x_n) = x, and (y_n) = 0, 1/pi, 0, 1/pi^2, ... ? This should work, right? Thanks.

 

[lurflurf]

forget seqences
let U be an open set containing x
show f(U)={0,1}

If you must procede on the original path...
consider a standard rational sequence like
x_n=(n^-1)(floor(x*n)+(1/2)(1+(-1)^n))
altered like
x_n=(n^-1)(floor(x*n)+(1/sqrt(2))(1+(-1)^n))

 

[lurflurf]

How about (x_n) = x, and (y_n) = 0, 1/pi, 0, 1/pi^2, ... ? This should work, right? Thanks.

If you know pi is irrational
maybe it is 4...