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Continuity of x^2 and -x^2

by bertram
Tags: continuity
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bertram
#1
Apr16-08, 10:59 AM
P: 2
1. The problem statement, all variables and given/known data
f(x) = {x^2 [tex]x \in Q[/tex]
-x^2 [tex]x \in R/Q [/tex]
At what points is f continuous?

2. Relevant equations

continuity: for every [tex]\epsilon > 0 [/tex] there exists [tex]\delta > 0 d(f(x),f(p)) < \epsilon[/tex] for all points [tex]x\inE[/tex] for which d(x,p) < [tex]\delta[/tex]

3. The attempt at a solution
Alright my initial thought was that it would not be continuous at any point in Q, because for any two rationals there is an irrational between them (this is correct?), but then it would be continuous at all irrationals from a theorem (4.6 in Rudin) for [tex]p\in Q[/tex], lim(x-> p) f(x) = -p^2 [tex]\neq[/tex] p^2 = f(p)
However, then this function is continuous at irrationals. For [tex]p\in R/Q[/tex], lim(x-> p) f(x) = -p^2 = f(p)

is this reasoning sound ok?
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quasar987
#2
Apr16-08, 11:57 AM
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I didn't bother understanding what you tried (it's not very clear)... but you seem to be playing with the right idea, using the density of Q and R\Q to show failure of continuity. You should be able to use this reasoning to show that f is discontinuous everywhere but at one point.
HallsofIvy
#3
Apr16-08, 01:22 PM
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You can say that f(x) is continuous at x0 if and only if lim f(xn)= f(x0) for every sequence {xn} converging to x0. But you can't restrict that to rational numbers only or irrational numbers only.

Notice that the f(xn) will be close to f(x0) for both rational and irrational xn if and only if x02= -x02. For what x0 is that true?


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