Proving of continuous functions

[Joe20]

Appreciate the help needed for the attached question. Thanks!

 

[Evgeny.Makarov]

Suppose that $f$ is continuous at some rational point $x_0\in\mathbb{Q}$. Then for $\varepsilon=1/2$ there exists a $\delta>0$ such that for an open interval $U=(x_0-\delta,x_0+\delta)$ we have $f(U)\subseteq(1/2,3/2)=(f(x_0)-1/2,f(x_0)+1/2)$. But this is impossible since every neighborhood of $x_0$, including $U$, contains an irrational point that is mapped to $0$ by $f$. The case of irrational $x_0$ is considered similarly.

 

[Joe20]

Suppose that $f$ is continuous at some rational point $x_0\in\mathbb{Q}$. Then for $\varepsilon=1/2$ there exists a $\delta>0$ such that for an open interval $U=(x_0-\delta,x_0+\delta)$ we have $f(U)\subseteq(1/2,3/2)=(f(x_0)-1/2,f(x_0)+1/2)$. But this is impossible since every neighborhood of $x_0$, including $U$, contains an irrational point that is mapped to $0$ by $f$. The case of irrational $x_0$ is considered similarly.

Hi, may I ask how do you know that f(U)⊆(1/2,3/2) and how do u know ε=1/2? How do I answer to the f(x) = 1 and 0? I am confused.

 

[HOI]

He doesn't "know" that [tex]\epsilon= \frac{1}{2}[/tex], he gave it that value. To prove that a function is continuous at x= a you must show that [tex]\lim_{x\to a}f(x)= f(a)[/tex]. And to show that you must show that "given any [tex]\epsilon> 0[/tex] there exist [tex]\delta> 0[/tex] such that …". This must be true for any [tex]\epsilon[/tex].

But Evgeny Makarov was showing that this is not true. It was sufficient to show there there is some value of [tex]\epsilon[/tex] for which this is not true. He showed it was not true for [tex]\epsilon= \frac{1}{2}[/tex] and that is sufficient to show that the function is not continuous.