MHB Is Showing One ε Enough to Prove Discontinuity?

[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 \epsilon= \frac{1}{2}, he gave it that value. To prove that a function is continuous at x= a you must show that \lim_{x\to a}f(x)= f(a). And to show that you must show that "given any \epsilon> 0 there exist \delta> 0 such that …". This must be true for any \epsilon.

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