Problem:
let $f:[0,1]\longrightarrow\mathbb R$ be defined by $f(x) = \left\{ \begin{array}{lr} 2x-1 & \text{ if } x \notin \mathbb{Q}\\ x^2 & \text{ if } x \in \mathbb{Q} \end{array} \right.$
Determine the points where $f$ is continuous
Solution:
I think this is continuous either everywhere or only at $x=1$.
Any ball of radius $\delta$ around any point $x$ contains both irrational and rational numbers.
thus there are only 3 cases:
- $|f(x)-f(y)|$ S.T. $x\in Q, y\notin Q$
- $|f(x)-f(y)|$ S.T. $x\in Q, y\in Q$
- $|f(x)-f(y)|$ S.T. $x\notin Q, y\notin Q$
In each of these cases we have :
$|2x-1-x^2|=|(-x+1)(x-1)|\\ |x^2-x^2|=|0|\\ |2x-1-(2x-1)|=|0|$
Since any ball around any $x$ contains both rationals and irrationals, we have to look at case $1$. In this case, $\forall x,y \in B_\delta(x)$, if $x=1$ then this is continuous, because $\forall \epsilon >0$, $|f(x)-f(y)|\rightarrow |0|=0 < \epsilon$
But is it continuous everywhere? Because this function, $|-x^2+2x-1|$ will grow monotonically as a function of $x$. Therefore, for whatever $\epsilon$, cant one find a $\delta$ where $|-x^2+2x-1|<\epsilon$ if $|x-y|<\delta$
Choose $x^2<\epsilon$, so am I right that it is continuous everywhere?