$f:(0,1)\rightarrow \mathbb{R}$ and $f(x)= x^{2} $ if $x$ is rational and $2-x^{2}$ if x is irrational. Where is f continuous?

Question

Is this function not Continuous anywhere in $(0,1)$, because the value of the function will keep oscillating between $ x^{2} $(for rational) and $2-x^{2}$ (for irrational) values,

So there will be breaks in the function??

I think it is discontinuous on $(0, 1)$. First we check $x^2 \ne 2 - x^2$ for all $x \in (0, 1)$. Then by density of $\mathbb{Q}$ and $\mathbb{R} \backslash \mathbb{Q}$... — Alex Vong, Apr 25 '17 at 06:23

marwalix · Accepted Answer · 2017-04-25T06:31:11.257

1

Hint: the only place where $f$ can be continuous is where $x^2=2-x^2$ i.e $x=\pm 1\notin (0,1)$

edited Apr 25 '17 at 06:31

answered Apr 25 '17 at 06:24

marwalix

16,773

SO the distinction based on rational and irrational does not maek any diference here? – onelessproblem Apr 25 '17 at 06:28
Also, Thanks so much! – onelessproblem Apr 25 '17 at 06:28

score 1 · Answer 2 · answered Apr 25 '17 at 06:45

We can use squeeze rule to find the points of continuity.

Consider two functions $g(x) = \min\{x^2, 2-x^2\}$ and $h(x) = \max\{x^2, 2-x^2\}$

Clearly, $g(x) \leq f(x) \leq h(x)$ for all $x$ and the following holds:

(1) $g(x), h(x)$ are continuous at $x = 1$

(2) $g(1) = h(1) = f(1) = 1$

Therefore, $f(x)$ is continuous at $x= 1$. Similarly, we can also show that $f(x)$ is continuous at $x = -1$.

Now, the domain in the question is $(0,1)$. Therefore, $f(x)$ is discontinuous in $(0,1)$

$f:(0,1)\rightarrow \mathbb{R}$ and $f(x)= x^{2} $ if $x$ is rational and $2-x^{2}$ if x is irrational. Where is f continuous?

2 Answers2