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.