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Real Analysis - Prove f is discontinuous

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1. 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.

2. Homework Equations
Definition of continuity.


3. 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. :smile:
 

Answers and Replies

NateTG
Science Advisor
Homework Helper
2,448
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Step 1: Find a rational sequence that converges to [itex]x[/itex].
Step 2: Find an alternating rational/irrational sequence that converges to [itex]0[/itex].
(Hint: I suggest something of the form [itex]0,y_1,0,y_3...[/itex].)
Step 3: Add the two together.
 
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How about (x_n) = x, and (y_n) = 0, 1/pi, 0, 1/pi^2, ... ? This should work, right? Thanks.
 
lurflurf
Homework Helper
2,419
122
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
Homework Helper
2,419
122
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...
 

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