Exercise 4.6.11

Question

Show that if $f$ is not continuous at $x$, then $f$ is not $\alpha$-continuous for some $\alpha>0$. Now explain why this guarantees that
  $$
  D_{f}=\bigcup_{n=1}^{\infty} D_{f}^{\alpha_{n}}
  $$
  where $\alpha_{n}=1 / n$.

Ulisse Mini · Accepted Answer

Negating the definition of $f$ being continuous at $x$ , we see $f$ is not continuous at $x$ iff there exists an $𝜖_{0} > 0$ such that no $δ > 0$ satisfies $| f (x) - f (y) | < 𝜖$ for all $0 < | x - y | < δ$ . Once $α_{n} < 𝜖_{0}$ (i.e. $n > 1 ∕ 𝜖_{0}$ ) we will have $x \in D_{f}^{α_{n}}$ .

(This completes the proof that $D_{f}$ is an $F_{σ}$ set!)

Exercise 4.6.11

Answers

Comments

Exercise 4.6.11

Answers

Comments

Add answer