March 21st, 2018, 05:49 AM  #1 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2  Continuity of a Function
The points of continuity of the function $f:R > R$ defined by $f(x) = ${ $x^21,$ if $x$ is irrational $0$, if $x$ is rational are a) $x= 1, x=0, x=1$ b) $x=1, x=1$ c) $x=1, x=0$ d) $x=0, x=1$ How to solve this type of question when rationals & irrationals are given as the condition. Thank you 
March 21st, 2018, 08:36 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,367 Thanks: 1272 
1) use the usual epsilondelta proof 2) note that due to denseness of the reals in the rationals (and vice versa), rational and irrational numbers can be found arbitrarily close together. 3) Thus the only points where $f$ will be continuous is where the value of $f$ is equal for both rationals and irrationals 4) There are only two points where this will be true, and they are rather trivial to find. See if you can figure it out. 
March 22nd, 2018, 03:42 AM  #3  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2  Quote:
The points of continuity are $x=1$ & $x=1$  

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