Answer to Question #239363 in Real Analysis for dakj

Question

Show that the function defined as 𝑓(𝑥) = { 𝑥, 𝑥 ∈ ℚ

𝑥 ² , 𝑥 ∈ ℝ − ℚ is continuous at 1 and discontinuous at 2.

Answer 1

$f(x)=\begin{cases} x ~~~~~ \text{if } x\in \mathbb{Q}\\ x^2 ~~~ \text{if } x \in \R-\mathbb{Q} \end{cases}\\ \text{To show that } f\text{ is continuous at 1. We must show that } \\ \lim_{x\to 1} f(x)=f(1)\\ \text{Suppose } f(x)=x\\ \ \lim_{x\to 1} f(x)=1\\ \text{Suppose } f(x)=x^2\\ \ \lim_{x\to 1} f(x)=1 \text{ Since } f(x)=1, \text{ then } f \text{ is continuous at } 1.\\\\$

b.

$\text{Consider the neighborhood of 2}\\(2-\frac{1}{n}, 2+\frac{1}{n}) n\in \N\\ \text{Then the neighborhood contain an irrational number } x_n n\in \N. \text{ Since irrational number is dense in } \R.\\ i.e. x_n\in (2-\frac{1}{n}, 2+\frac{1}{n})\\ \implies 2-\frac{1}{n}<x_n<2+\frac{1}{n}\\ \implies -\frac{1}{n}<x_n - 2<\frac{1}{n}\\ \implies |x_n-2|<\frac{1}{n}\\ \implies x_n \to 2 \text{ as } n\to \infty\\ \text{Now, } f(x_n)=x_n^2 ~~\forall n\in \N \text{ as } x_n \text{ is irrational}\\ \therefore \lim_{n\to \infty}f(x_n)= \lim_{n\to \infty} x_n^2=(\lim_{n\to \infty} x_n)^2=2^2=4\neq f(2)=2\\ \text{Thus, } f \text{ is discontinuous at } 2$