Superior limit of functions

Question

Give me help with this exercise please.

Let $f:[0,1]\to\mathbb R$ be a bounded function. Given $\varepsilon>0$, let $F_{\varepsilon}\subseteq{[0,1]}$ the set of $x\in [0,1]$ such that there is a sequence $(x_n)$ with $\lim_{n\to+\infty} x_n=x,\quad \lim sup_{n}|f(x_{n+1})-f(x_n)| \geq \varepsilon.$

Say if each of the following sentences is true or false. Justify your answer.

(a) For every function $f$, the set $F_{\varepsilon}$ is closed.

(b) If $x_0\in F_{\varepsilon}$, the $f$ is discontinuous at $x_0$.

(c) If $f$ is discontinuous at $x_0\in[0,1]$, then there is $\varepsilon >0$ such that $x_0\in F_{\varepsilon}$.

For (a), i thought taking $a\in\overline{F_{\varepsilon}}$, then there is $a_k\in F_{\varepsilon}$ such that $a_k\rightarrow{ a}$.

As $a_k\in F_{\varepsilon}$, then there is $(x_{k_n})$ such that

$\lim_{n\to+\infty} x_{k_n}=a_k,\quad \lim sup_{n}|f(x_{k_{n+1}})-f(x_{k_n})| \geq \varepsilon.$

Then, let $m=m(k,n)=k+n$, and define $b_m=a_{k_n}$, so $b_m\rightarrow{x}$ and $\lim sup_{n}|f(b_{m+1})-f(b_m)| =\lim sup_{n}|f(x_{k_{n+1}})-f(x_{k_n})| \geq \varepsilon$

Therefore,$ x\in F_{\varepsilon}$. But i am not sure, because i have not used the fact that $f$ is bounded.

(b) Is TRUE, by the sequence criteria for continuous functions.

I think (c) is true, but i'm not so sure.

Answer 1

For starter:

(2) Given the sequence $(x_n)$ such that $\lim\sup_n |f(x_{n+1 }) - f(x_n)| \geq \epsilon$, we can consider taking a subsequence $(x_{\varphi(n)})$ such that $\lim_n |f(x_{\varphi(n+1)}) - f(x_{\varphi(n)})| \geq \epsilon$. If $f$ is continuous at $x_0$, then given any $x_n \longrightarrow x_0$, one has $f(x_n) \longrightarrow f(x_0)$, such that $\lim_n |f(x_{n+1 }) - f(x_n)| = 0$. Therefore if $x_0 \in F_{\epsilon}$, $f$ is discontinuous at $x_0$.

(3) Suppose $f$ is discontinuous as $x_0$. There exists a sequence $x_n \longrightarrow x_0$ such that $f(x_n) \nrightarrow f(x_0)$. Then consider the sequence $\{x_1, x_0, x_2, x_0, x_3, x_0, \dots\}$